Non-commutative Algebra

Patrick Da Silva

Freie Universität Berlin

June 2, 2017

Table of Contents

Page

1 Introduction to unital rings 51.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Centralizers and bicentralizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.3 Tensor products and Hom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.4 Isotypical modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2 Finiteness conditions on rings and modules 262.1 Artinian, noetherian and finite length modules . . . . . . . . . . . . . . . . . . . . . . . . 262.2 Artinian and noetherian rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.3 Simple modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.4 Completely reducible modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.5 Isotypical components of completely reducible modules . . . . . . . . . . . . . . . . . . . 402.6 Centralizer and bicentralizer of a completely reducible module . . . . . . . . . . . . . . . 422.7 Centralizer of a simple module . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3 Simple and semisimple rings 463.1 Left & right vector spaces over a skew field . . . . . . . . . . . . . . . . . . . . . . . . . . 463.2 Centralizer of a completely reducible module . . . . . . . . . . . . . . . . . . . . . . . . . 493.3 Semisimple rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4 Radicals 594.1 Radical of a module . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594.2 Radical of a ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 624.3 Radical of artinian rings/modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 684.4 Modules over an artinian ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

5 Tensor products of K-algebras 745.1 Modules over tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 745.2 Tensor product of K-fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805.3 Tensor product of completely reducible modules . . . . . . . . . . . . . . . . . . . . . . . 825.4 Separable algebras and modules over a field . . . . . . . . . . . . . . . . . . . . . . . . . . 87

3

Chapter 1

Introduction to unital rings

We assume the reader is familiar with a bit of algebra (at least elementary group theory). All our rings areassumed unital, namely multiplication admits a unique neutral element which will always be denoted by 1.This set of notes will serve as a reference to other sets of notes when proofs are necessary, and thus is notmeant to be introductory. The set of non-negative integers N is equal to 0, 1, · · · , n, · · · (this has theadvantage of turning it into a monoid).

1.1 Generalities

Definition 1.1. Let A be an abelian group denoted additively, i.e. via the symbol + and with neutralelement 0. We say that A is a (unital) ring when it is equipped with a multiplication (usually denoted byjuxtaposition) · : A×A→ A satisfying the following axioms :

• Distributivity : For all a1, a2, a3 ∈ A,

a1(a2 + a3) = a1a2 + a1a3, (a1 + a2)a3 = a1a3 + a2a3.

• Associativity : For all a1, a2, a3 ∈ A,

a1(a2a3) = (a1a2)a3.

• Neutral element : There exists a unique element, call the unit element (or simply 1) and denoted by1 ∈ A, such that for all a ∈ A,

1a = a = a1.

A subring of A is a subset of A which contains 1 and is closed under addition and multiplication, and thusbecomes a ring of its own.

Remark 1.2. If M is an abelian group, the collection of all abelian group endomorphisms of M , denotedby EndZ(M), is a ring ; the identity endomorphism of M serves as unit element, addition is performedpointwise and multiplication is given by composition.

Definition 1.3. Let A,B be rings.

(i) A morphism of rings is a map ϕ : A→ B such that for all a1, a2 ∈ A,

ϕ(a1 + a2) = ϕ(a1) + ϕ(a2), ϕ(a1a2) = ϕ(a1)ϕ(a2), ϕ(1) = 1.

It is clear under this definition that the image imϕdef= ϕ(A) is a subring of B.

5

Chapter 1

(ii) A left A-module is an abelian group M together with a morphism of rings ϕ : A → EndZ(M).This morphism defines a left action of A on M , namely a map A × M → M , whose output is

usually denoted by juxtaposition, and is given by amdef= ϕa(m), where ϕa

def= ϕ(a). The assumptions

on a morphism of rings shows that ϕ1 = idM , ϕa is an endomorphism of abelian groups andϕa1a2 = ϕa1 ϕa2 .

(iii) Given a ring A, the opposite ring Aopp is a ring whose underlying abelian group is equal to A but ifwe denote multiplication in Aopp by · and that of A by juxtaposition, for a1, a2 ∈ Aopp, we set

a1 · a2def= a2a1.

Reading the axioms of a ring in the multiplicatively reversed order shows that Aopp is a ring and hasthe same unit element as A.

(iv) A right A-module is an abelian group N together with a morphism of rings ψ : A→ EndZ(M)opp.To make notation more natural, an element ϕ ∈ EndZ(M)opp acts on m ∈ M by setting theendomorphism on the right, not on the left ; in other words, for all m ∈M ,

ϕ ∈ EndZ(M) −→ ϕ(m),

ϕ ∈ EndZ(M)opp −→ (m)ϕ.

This morphism defines a right action of A on M , namely a map M × A→ M , whose output is also

denoted by juxtaposition and is given by madef= ϕa(m). This is indeed a right action of A on M :

m(a1a2)def= (m)ϕa1a2 = ((m)ϕa1)ϕa2 .

The order has indeed been reversed, since in EndZ(M), the composition ϕa1 ϕa2 applied to m ∈Mis equal to ϕa1(ϕa2(m)), where as in EndZ(M)opp, the composition ϕa1 ϕa2 applied to m equalsϕa2(ϕa1(m)) ; to emphasize the right action and hopefully minimize confusion, we chose the abovenotation.

(v) An (A,B)-bimodule is an abelian groupM together with two morphisms of rings, namely ϕL : A→EndZ(M) and ϕR : B → EndZ(M)opp, corresponding to left and right multiplication. The maps ϕL

and ϕR are required to turnM into a left A-module and a right B-module respectively, together withthe additional property that for all a ∈ A, b ∈ B and m ∈M ,

(ϕLa (m))ϕRb = ϕLa ((m)ϕRb ).

Note that for left and right A-modules M , we usually denote the action of a ∈ A on m ∈ M

by juxtaposition, namely amdef= ϕa(m). The axioms of an (A,B)-bimodule becomes essentially

“associativity”, namely a(mb) = (am)b. When A = B, we call an (A,A)-bimodule simply an A-bimodule.

(vi) If M is an abelian group which is a left A-module and a left B-module, we can canonically turn Minto a right Bopp-module. We say that the two module structures on M are compatible if they makeM a (A,Bopp)-module. This is equivalent to asking that

∀m ∈M,a ∈ A, b ∈ B, a(bm) = b(am).

Similarly, ifM is a right A-module and a right B-module, we say that these are compatible if it makesM a (Aopp, B)-bimodule, which is equivalent to asking that

∀m ∈M,a ∈ A, b ∈ B, (ma)b = (mb)a.

6

More generally, if Aii∈I and Bjj∈J are two families of rings and M is an (Ai, Bj)-bimodule foreach i ∈ I, j ∈ J , we say that these structures are compatible if the module structures given byAi, Ai′ are compatible for all i, i′ ∈ I and the module structures iven by Bj , Bj′ are compatible forall j, j′ ∈ J .

(vii) A morphism of left A-modules/right A-modules/(A,B)-bimodules is a morphism of abelian groupsbetween two left A-modules/right A-modules/(A,B)-bimodules which commutes with the action ofA on M . For instance, if ϕ : M → N is a morphism of abelian groups between two left A-modules,it is a morphism of left A-modules precisely when

∀a ∈ A, ∀m ∈M, ϕ(am) = aϕ(m).

Remark 1.4. The collection of (small) left A-modules together with their morphisms form a category, whichwe denote by A-Mod. Similarly, right A-modules and (A,B)-bimodules form categories ; we denotethem by Mod-A and A-Mod-B, respectively. Therefore, left and right A-modules inherit notions ofendomorphism, automorphism, monomorphism, epimorphism, isomorphism, etc. The composition of twomaps (which are morphisms of some type, most of the time) is denoted by (for example, ϕ ψ), exceptwhen the maps composed belong to some ring, in which case we might adopt juxtaposition as the symbolfor composition.

Definition 1.5. Let A be a ring and M be a left A-module. The endomorphism ring of M is thering EndA(M) consisting of all endomorphisms of M , i.e. morphisms of left A-modules ϕ : M → M .Addition is done element-wise and multiplication is given by composition. If M is a right A-module, wealso denote its endomorphism ring by EndA(M). When M is an (A,B)-bimodule, we denote the ring ofall (A,B)-module endomorphisms of M by End(A,B)(M).

Remark 1.6. A morphism of rings ϕ : A → B is equivalent to a morphism of rings ϕopp : Aopp → Bopp ;the underlying set maps ϕ and ϕopp are equal, but the requirement that ϕopp respects multiplication is themultiplicative axiom of ϕ read in reverse. It follows that a right A-module is nothing more but a left Aopp-module, so anything proven about all left A-modules for any ring A is also proven for right A-modules,when the statement is read appropriately.

To prevent ourselves from dealing with left and right A-modules all the time, unless we specifically needboth notions, we will stick to left A-modules. Given a left A-module M , the right Aopp-module structureon M corresponding to its left A-module structure will be denoted by Mopp (i.e. the morphism of ringsϕ : A → EndZ(M) corresponds to ϕ : Aopp → EndZ(M)opp, so it is given by the same data but in adifferent notation).

Definition 1.7. Let A be a ring and M a left A-module. A subset N ⊆M is called a left A-submodule ofM (or simply an A-submodule) if it is an abelian subgroup of M which is stable under the action of A, i.e.for all a ∈ A and n ∈ N , an ∈ N . We also say that N is stable under A to say that it is an A-submoduleofM . In this case, we write N ≤M to indicate that N is a submodule ofM . Given a family of submodulesNii∈I , we can define their sum as

∑i∈I

Nidef=

∗∑i∈I

ni

∣∣∣∣∣ni ∈ Ni

where the superscript ∗ indicates that only finitely many of the ni’s are non-zero. A family of submodulesNii∈I of M is said to be in a direct sum if they are in a direct sum as abelian groups, namely everym ∈

∑i∈I Ni can be written uniquely as a sum m =

∑∗i∈I ni, ni ∈ Ni. When

∑i∈I Ni = M , we write

M =⊕

i∈I Ni and we say that M is the direct sum of the Ni’s.

Remark 1.8. To check ifM is a direct sum of two of its submodules N1, N2, it suffices to check if N1 +N2 =M and N1 ∩N2 = 0, where 0 denotes the trivial submodule consisting only of the zero element of M . One

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Chapter 1

needs to be careful when trying to check that M is the direct sum of k submodules, k ∈ N ; in this case,one must verify that

k∑i=1

Ni = M, ∀1 ≤ i ≤ k, Ni ∩

k∑j=1

j 6=i

Nj

= 0.

Proposition 1.9. (Isomorphism theorems) Let A be a ring and ϕ : M → N be a morphism of left A-modules.

(i) If N ≤ M , the quotient module M/N is defined as the quotient abelian group. Its A-modulestructure is given by

a(m+N)def= am+N.

It is indeed a left A-module.

(ii) The kernel and the image of ϕ is defined as its kernel and image as abelian groups. We denote themby kerϕ ≤M and imϕ ≤ N . It induces a natural morphism ϕ : M/ kerϕ→ N which is onto imϕ,thus giving the first isomorphism theorem :

M/ kerϕ ' imϕ.

(iii) If N1, N2 ≤ M are two left A-submodules, then N1 + N2, N1 ∩ N2 ≤ M and we have the secondisomorphism theorem

(N1 +N2)/N1 ' N1/(N1 ∩N2).

(iv) If N1 ≤ N2 ≤M , then N1 is a submodule of M and

(M/N1)/(N2/N1) 'M/N2.

Proof. This is the same proof as in the commutative case, hence is omitted.

Definition 1.10. Let A be a ring. One can see A as an A-bimodule over itself by using multiplication, i.e.ϕLa1(a2) = a1a2 = (a1)ϕRa2 . A left ideal (resp. right ideal, two-sided ideal) of A is a left A-submodule(resp. right A-submodule, A-bisubmodule) of A when seen as an A-bimodule over itself.

Given a left A-moduleM and a subset S ⊆M , we can define the left A-submodule ofM generated byS, written A〈S〉, as the smallest A-submodule of M containing S ; it can be constructed as the intersectionof all A-submodules of M which contain S, since this collection is non-empty because M is in it. Similarly,one defines the right A-submodule of M generated by S as 〈S〉A when M is a right A-module. If B isa ring and M is an (A,B)-bimodule, then the (A,B)-bisubmodule of M generated by S is defined thesame way and is written A〈S〉B .

When A is seen as an A-bimodule over itself, we use a different notation. For S ⊆ A, the left ideal,right ideal and two-sided ideal generated by S are denoted respectively by A(S) , (S)A and A(S)A.

If S = m1, · · · ,mn is a finite subset of M , we write A〈m1, · · · ,mn〉 instead (similar notations whenM is a right A-module or an (A,B)-bimodule, or when A = M is an A-bimodule over itself and we speak

of ideals). If S = m is a singleton, we may write Amdef= A〈m〉, mA

def= 〈m〉A and AmB

def= A〈m〉B , so

that

A〈m1, · · · ,mn〉 =n∑i=1

Ami, 〈m1, · · · ,mn〉A =n∑i=1

miA, A〈m1, · · · ,mn〉B =n∑i=1

AmiB.

8

Remark 1.11. When ϕ : A → B is a morphism of rings, imϕ is a subring of B and kerϕ is a two-sidedideal of A, which allows to turn the quotient A/ kerϕ into an A-bimodule. This A-bimodule structure turnsA/ kerϕ into a ring of its own and A/ kerϕ ' imϕ.

Proposition 1.12. Let A be a ring andM a left A-module. An element π ∈ EndA(M) is called a projectionif π2 = π.

(i) If π is a projection, then we have M = kerπ ⊕ imπ.

(ii) If π is a projection, then so is 1− π (recall that 1 ∈ EndA(M) is the identity map of M ).

(iii) Wheneverm ∈M is written asm = m1 +m2 wherem1 ∈ imπ andm2 ∈ kerπ, we havem1 = π(m)and m2 = (1− π)(m).

(iv) We have kerπ = im (1− π) and imπ = ker(1− π).

(v) Let π ∈ EndA(M) be a projection. We have the identities

EndA(M)π = ϕ ∈ EndA(M) | kerϕ ⊇ kerπ, πEndA(M) = ϕ ∈ EndA(M) | imϕ ⊆ imπ.

(vi) IfM = N1⊕N2 is the direct sum of two left A-submodules, let π ∈ EndA(M) be the endomorphism

defined by π(n1 +n2)def= n1. Then π is the unique projection in EndA(M) which satisfies kerπ = N2

and imπ = N1 ; we call it the projection onto N1 and denote it by πN1 or simply π1 if the contextallows it. More generally, if M =

⊕i∈I Ni, there is a unique projection π ∈ EndA(M) satisfying

imπ = Ni and kerπ =⊕

j∈I\iNj .

Proof. For m ∈ M , write m = (m − π(m)) + π(m), so that M = kerπ + imπ. If m ∈ kerπ ∩ imπ,then m = π(m′) = π2(m′) = π(m) = 0, which proves (i). Note that π = (1− (1− π)), which proves (ii),(iii) and (iv) because m = (m− π(m)) + π(m).

To prove (v), we proceed with the first equality. If ϕ = ψπ, then π(m) = 0 implies ψ(π(m)) = 0,hence the inclusion (⊇) holds. For the reverse, if ϕ : M → M satisfies kerϕ ⊇ kerπ, then ϕ isentirely determined by the values it takes on imπ. But for m ∈ imπ, we have ϕ(m) = ϕ(π(m)), henceϕ = ϕπ ∈ EndA(M)π.

As for the second equality, it is clear that imπϕ ⊆ imπ, which gives (⊆). For the reverse inclusion, sup-pose ϕ ∈ EndA(M) satisfies imϕ ⊆ imπ. Then ϕ(m) ∈ imπ for all m ∈ M , hence π(ϕ(m)) = ϕ(m),which means ϕ = πϕ ∈ πEndA(M), as desired.

Finally, the endomorphism defined in (vi) is clearly a projection. Its unicity comes from part (iii).

Definition 1.13. Let A be a ring, M a left A-module and N ≤ M an A-submodule. We say that N is adirect summand if there exists a projection on M onto N . In other words, N is a direct summand if thereexists an A-submodule N ′ ≤M such that M = N ⊕N ′.

Definition 1.14. Let A be a ring and M,N be two left A-modules. The set HomA(M,N) is defined as theset of all morphisms of left A-modules ϕ : M → N . It is obviously an abelian group, and it is turned intoa left A-module by

(aϕ)(m)def= aϕ(m) = ϕ(am).

This construction is functorial in the following sense. Given morphisms ϕ : M2 → M1 and ψ : N1 → N2,we obtain a morphism HomA(ϕ,ψ) : HomA(M1, N1) → HomA(M2, N2) by sending α to ψ α ϕ. The

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Chapter 1

following diagram commutes :

HomA(M1, N1) HomA(M2, N1)

HomA(M1, N2) HomA(M2, N2)

HomA(ϕ,idN1)

HomA(idM1,ψ)

HomA(ϕ,ψ)HomA(idM2

,ψ)

HomA(ϕ,idN2)

We sometimes write HomA(M,ψ) instead of HomA(idM , ψ) : HomA(M,N1) → HomA(M,N2) whenψ : N1 → N2 is a morphism ; similarly for HomA(ϕ,N) : HomA(M1, N)→ HomA(M2, N).

Remark 1.15. As in the case of abelian groups, we have a notion of short exact sequences

0 M1 M2 M3 0.ϕ ψ

This sequence is said to be exact at M2 when kerψ = imϕ, and more generally, it is called a short exactsequence when ϕ is injective, ψ is surjective and kerψ = imϕ, in which case M2/ϕ(M1) ' M3 (this isequivalent to asking the sequence to be exact atM1,M2 andM3). Applying HomA(N,−) on this sequence,we obtain the left-exact sequence

0 HomA(N,M1) HomA(N,M2) HomA(N,M3)HomA(N,ϕ) HomA(N,ψ)

Left-exactness means that the sequence is exact at HomA(N,M1) and HomA(N,M2) but not necessarilyat HomA(N,M3). Applying HomA(−, N) on this sequence, we obtain the left-exact sequence

0 HomA(M3, N) HomA(M2, N) HomA(M1, N)HomA(ψ,N) HomA(ϕ,N)

(note the order inversion ; this is because HomA(−, N) is a contravariant functor, i.e. it reverses the orderin which morphisms are composed). See Proposition 1.16 for proofs.

When ψ is a projection, we say that the above exact sequence splits. When applying HomA(N,−) orHomA(−, N) to split exact sequences, the resulting exact sequence is also split. One easily sees that whenN ≤ M , N is a direct summand of M if and only if the identity map of M extends to a endomorphismπ ∈ EndA(M) with imπ = N , in which case M = N ⊕ kerπ.

Proposition 1.16. Let

0 M1 M2 M3 0ϕ ψ

be a sequence of left A-modules (not necessarily exact). The following are equivalent :

(i) The sequence above is right-exact

(ii) For all left A-modules N , the following sequence is exact :

0 HomA(M3, N) HomA(M2, N) HomA(M1, N)HomA(ϕ,N) HomA(ψ,N)

Also, the following are equivalent :

(i) The above sequence is left-exact

(ii) For all left A-modules N , the following sequence is exact :

0 HomA(N,M1) HomA(N,M2) HomA(N,M3)HomA(N,ϕ) HomA(N,ψ)

10

Proof. The fact that (i) implies (ii) is a consequence of the left-exactness of Hom and is proved in thesame way as in the case of abelian groups, so is left as an exercise. Conversely, in the second case, wecan take N = A and notice that we have the natural isomorphism HomA(A,Mi) ' Mi, so we recoverthe exactness required in (i).

In the first case, we have to work a bit more. LetNdef= M3/imψ. The projection map π : M3 →M3/imψ

is such that π ψ = 0 by definition, but since π ψ = HomA(ψ,N)(π) = 0 and HomA(ψ,N) isinjective, we get π = 0, i.e. imψ = M3, which means ψ is surjective. Let N = M3 and considerψ ∈ HomA(M2, N). Since HomA(ψ ϕ,N) = HomA(ψ,N) HomA(ϕ,N) = 0 for all left A-modulesN , we have ψ ϕ = 0 (take N = M3 and apply HomA(ψ ϕ,N) to idM3 ), which implies imϕ ⊆ kerψ.

Next, take Ndef= M2/imϕ. Since π : M2 →M2/imϕ lies in ker HomA(ϕ,N) = im HomA(ψ,N), there

exists ν : M3 → M2/imϕ such that π = HomA(ψ,N)(ν) = ν ψ. Therefore, kerψ ⊆ ker(ν ψ) =kerπ = imϕ, completing the proof.

Definition 1.17. Let A be a ring.

(i) An element e ∈ A is called an idempotent if e2 = e.

(ii) A family of idempotents eii∈I of A is called orthogonal if for any i, j ∈ I distinct, we have eiej = 0.

Corollary 1.18. Let A be a ring and M a left A-module. Consider a family of submodules Mii∈I suchthat

∑i∈IMi = M . Then the following are equivalent :

(i) We have M =⊕

i∈IMi

(ii) Each Mi is a direct summand and the family of idempotents πii∈I ⊆ EndA(M) is orthogonal.

Proof. Obvious.

Remark 1.19. The assumption that∑

i∈IMi = M is not vital ; we can always add an element ∗ to the setI such that M∗ =

⋂i∈I kerπi and it follows that M =

∑i∈I∪∗Mi. If M was already the direct sum of

the Mi’s, this does not change anything to the statement ; if it isn’t, we have to replace (i) by the equality

M ′def=∑

i∈IMi =⊕

i∈IMi and the πi by πi|M ′ , which shows that the πi|M ′ form a family of orthogonalidempotents.

1.2 Centralizers and bicentralizers

Definition 1.20. Let A be a ring and S ⊆ A be a subset. The centralizer of S in A is denoted by CA(S)and is defined as

CA(S)def= a ∈ A | ∀s ∈ S, as = sa.

Note that by definition, CA(S) is a subring of A, so we can consider its bicentralizer C2A(S)

def= CA(CA(S)).

The center of A is equal to its own centralizer and is denoted by

Z(A)def= CA(A) = z ∈ A | ∀a ∈ A, az = za.

We say that B is commutative if B = Z(B) (in which case we resolve to the theory of commutative rings).

Remark 1.21. It might be tempting to define recursively the kth centralizer by CkA(S)def= CA(Ck−1

A (S)), butthis notion is pointless, as we can see right away. It is clear that S ⊆ C2

A(S) by definition (the elementsof the centralizer commute with all elements of S, so the elements of S commute with the elements of

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Chapter 1

the centralizer). Letting CA(S) play the role of S in the latter inclusion gives CA(S) ⊆ C3A(S). When

S ⊆ T ⊆ A, we have CA(S) ⊇ CA(T ) (elements which commute with every element of T also commutewith every element of S), therefore S ⊆ C2

A(S) implies CA(S) ⊇ C3A(S), which gives CA(S) = C3

A(S).Therefore, the centralizer of the bicentralizer is the centralizer itself.

If B ⊆ A is a subring, then Z(B) = CA(B) ∩B. In particular, if B is commutative, then B ⊆ CA(B),thus C2

A(B) ⊆ CA(B). It follows that

Z(CA(B)) = CCA(B)(CA(B)) = CA(CA(B)) ∩ CA(B) = C2A(B) ∩ CA(B).

One also sees that C2A(A) = CA(Z(A)) = A follows from the definition.

Definition 1.22. Let A be a commutative ring. An associative A-algebra is a ring B equipped with amorphism ϕ : A→ B, called the structure map of the algebra, such that imA ⊆ Z(B).

More generally, an A-algebra is defined as an A-bimodule B equipped with an A-bilinear map (−,−) :B × B → B ; when this bilinear map satisfies (a, 1) = (1, a) = a and is associative (meaning that(b1, (b2, b3)) = ((b1, b2), b3) for all b1, b2, b3 ∈ B), we recover the notion of an associative A-algebra. In thisdocument, all our A-algebras are associative unless otherwise mentioned, so we will call them A-algebras.

If B1, B2 are two A-algebras, since A is commutative, we can consider the tensor product B1 ⊗A B2

which becomes a ring via the definition

(b1 ⊗ b2)(b′1 ⊗ b′2)def= (b1b

′1)⊗ (b2b

′2)

which is extended by A-bilinearity to all of B1 ⊗A B2. It canonically becomes an A-algebra via the mor-phism A→ B1 ⊗A B2 defined by a 7→ a⊗ 1 = 1⊗ a.

An A-subalgebra of a A-algebra B is a subring C ⊆ B such that the structure map ϕ : A → B of Asatisfies ϕ(A) ⊆ C , in which case C becomes an A-algebra using the same structure map with codomainrestricted to C since ϕ(A) ⊆ Z(B) ∩ C ⊆ Z(C).

Definition 1.23. A ring K in which for every a ∈ A, there exists a unique element a−1 ∈ A satisfyingaa−1 = a−1a = 1, is called a skew field or a division ring. If K is commutative, we call K a field. WhenK is a field and A is a K-algebra, A is in particular a K-vector space and thus we can apply the techniquesof linear algebra. One remark : since K has no nonzero ideals, a K-algebra always contains an isomorphiccopy of K as a K-subalgebra, so we usually write K ⊆ A with no regards to the structure map.

Proposition 1.24. Let K be a field and A,B be two K-algebras. If A′ ⊆ A and B′ ⊆ B are K-subalgebras,then CA(A′) is a K-subalgebra of A, CB(B′) is a K-subalgebra of B and

CA⊗KB(A′ ⊗K B′) = CA(A′)⊗K CB(B′).

As a corollary, Z(A⊗K B) = Z(A)⊗K Z(B).

Proof. Since A′ ⊆ A, we have K ⊆ Z(A) = CA(A) ⊆ CA(A′). Since K ⊆ A′ ⊆ C2A(A′), we have

K ⊆ CA(A′) ∩ C2A(A′) = Z(CA(A′)) by Remark 1.21, hence CA(A′) (and similarly CB(B′)) are

K-subalgebras.

12

As for the equality, note that the inclusion (⊇) is clear. Consider the following direct sum decompositions

A = CA(A′)⊕A′′, B = CB(B′)⊕B′′

=⇒ A⊗K B =

(CA(A′)⊗K CB(B′)

)⊕(

(A′′ ⊗K CB(B′))⊕ (CA(A′)⊗K B′′)⊕ (A′′ ⊗K B′′)

),

which shows that CA(A′)⊗K CB(B′) is a direct summand of the K-vector space A⊗K B and

CA(A′)⊗K CB(B′) =

((A⊗K CB(B′)

)∩(CA(A′)⊗K B

)).

By the symmetry of our argument, it will suffice to show that CA⊗KB(A′ ⊗K B′) ⊆ CA(A′) ⊗K B. Letz =

∑∗i∈I ai⊗ bi ∈ CA⊗KB(A′⊗K B′) where ai ∈ A, bi ∈ B. Without loss of generality, we can suppose

that the bi are linearly independent over K . For any a ∈ A′, we have

z(a⊗ 1) = (a⊗ 1)z =⇒∑i∈I

(aia− aai)⊗ bi = 0.

Since the bi are linearly independent, this means that aia− aai = 0 for all i, which means ai ∈ CA(A′),as desired.

Definition 1.25. Let A be a ring and M an A-module. The structure map ϕ : A → EndZ(M) is a mor-

phism of rings, so let AMdef= ϕ(A), which is a subring of EndZ(M). For each a ∈ A, we use the notation

aMdef= ϕa ∈ AM . The map ϕ : A → AM defined by a 7→ AM is onto and AM is called the ring of

homotheties of M (its elements are called homotheties of M ).

The centralizer and bicentralizer ofM are defined as the centralizer/bicentralizer of AM in EndZ(M) :

CA(M)def= CEndZ(M)(AM ), C2

A(M)def= C2

EndZ(M)(AM ).

By definition, the centralizer of M is the endomorphism ring of the A-module M since

CA(M) = ϕ ∈ EndZ(M) | ∀a ∈ A,ϕaM = aMϕ = EndA(M).

The countermodule1 of M , denoted by M/ is the abelian group M seen as a module over its centralizer,namely, the endomorphism ring EndA(M). Multiplication is defined the obvious way, namely if ϕ ∈EndA(M) andm ∈M/, we define ϕm

def= ϕ(m). Note that the A-module and EndA(M)-module structures

of M are compatible since ϕ(am) = aϕ(m) for all a ∈ A, ϕ ∈ EndA(M) and m ∈ M . We definethe double countermodule of M , denoted by M//, as the countermodule of its countermodule, namely

M// def= (M/)/.

Remark 1.26. Even when A is commutative, CA(M) = EndA(M) is, in general, a noncommutative ring.Therefore, results that we build in this document here can also be useful for the theory of commutative rings,since considering only commutative rings prevents the study of noncommutative rings such as EndA(M).

Note that compatibility of module structures is not a transitive notion, i.e. if M is a left module overthe three rings A,B,C , the module structures of A and B are compatible and the module structures of Band C are compatible, this does not mean that the structures of A and C are compatible. A particularlyrelevant example is that of the A-module, CA(M)-module and C2

A(M)-module structures on the A-moduleM . The first two and last two are compatible, but A and C2

A(M) give compatible structures on M if andonly if C2

A(M) ⊆ CA(M), which is not always the case (see Example 1.34).1The French term for this is contre-module, but we have not found an English translation which is commonly used.

13

Chapter 1

Definition 1.27. Let A be a ring. We have already seen that A can be interpreted as an A-bimodule overitself, but we can relax these assumptions and only consider the left A-module A, which we denote by A`.Equivalently, we can consider the right A-module A, which we denote by Ar.

Proposition 1.28. Let A be a ring.

(i) The centralizer of the left A-module A` is the opposite ring of homotheties of the right A-module Ar .In symbols,

CA(A`) = (AAr)opp = (Aopp)(Aopp)` ' Aopp.

This means that A`/ is canonically a left Aopp-module. Namely, as left Aopp-modules, A`/ ' (Aopp)`.

(ii) The centralizer of the right A-module Ar is the opposite ring of homotheties of the left A-module A`.In symbols,

CA(Ar) = (AA`)opp = (Aopp)(Aopp)r ' A

opp.

The same comments as in part (i) apply, i.e. Ar/ is a right Aopp-module satisfying Ar/ ' (Aopp)r .

Proof. We begin by proving (i). If ϕ ∈ EndZ(A`) = EndZ((Aopp)r) commutes with aA`for each a ∈ A,

thenϕ(a) = ϕ(a1) = (ϕ aA`

)(1) = (aA` ϕ)(1) = aϕ(1),

so that ϕ corresponds to multiplication on the right by the element ϕ(1). Conversely, any such endomor-phism (of the form aAopp

r) belongs to CA(A`) since multiplication on the left commutes with multiplication

on the right ; this is associativity of multiplication. Therefore, CA(A`) and (AAr)opp are identified asabelian groups. As for multiplication, if ψ,ϕ ∈ CA(A`), then

(ψ ϕ)(a) = ψ(ϕ(a)) = ψ(aϕ(1)) = aϕ(1)ψ(1) = a(ψ(1)Ar ∗ ϕ(1)Ar).

where ∗ stands for multiplication in (AAr)opp. Finally, we show that the identity map of A induces theisomorphism A`

/ ' (Aopp)`. For ϕ,ψ ∈ CA(A`), we have

(ψ ϕ)(a) = a(ψ(1) ∗ ϕ(1)) = ψ(1) ∗ ϕ(1) ∗ a.

The left-most term is the left CA(A`)-action on A`/, and identifying the action of CA(A`) with thatof Aopp via the map ϕ 7→ ϕ(1)(Aopp)` , this completes the proof. (Note that CA(A`) and (AA`

)opp

are not just isomorphic, they are equal ; this is because they are defined as the same subsets of EndZ(A).)

The statement of (ii) is proved along the lines of the proof of part (i), everything being written in reverseorder.

Remark 1.29. The centralizer of a left A-module M is equal to the ring of homotheties of M/ (in symbols,CA(M) = CA(M)M/ ). Since it is clear that CA(M)M/ ⊆ CA(M), this is the statement that for ϕ ∈CA(M) = CEndZ(M)(AM ), as endomorphisms of M , we have ϕM/ = ϕ. In other words, the map ϕ 7→ ϕMfrom CA(M) to EndZ(M) is an injective morphism of rings. This is clear because the endomorphism ofM/ which multiplies by ϕM is exactly ϕ. It follows that CA(M)M/ = CA(M) and

C2A(M) = C2

EndZ(M)(AM ) = CEndZ(M)(CEndZ(M)(AM )) = CEndZ(M)(CA(M)M/) = CCA(M)(M/),

thusC2A(M) = CCA(M)(M

/) = CCA(M)(M/)M// = C2

A(M)M// .

This means that C2A(M) is the ring of homotheties of the double countermodule M// and the centralizer

ofM/. The latter can be re-written as C2A(M) = EndEndA(M)(M

/), but since for any a ∈ A, the endomor-phism aM is an EndA(M)-linear map onM/ by definition of EndA(M), we obtain AM ⊆ C2

A(M) = AM// .Also note that since CEndZ(M)(AM ) = C3

EndZ(M)(AM ), the centralizer of M// is just the centralizer of M ,so that we do not speak of M///.

14

Corollary 1.30. Let A be a ring. We have

C2A(Al) = AA`

, C2A(Ar) = AAr .

Proof. It suffices to prove that the first equality holds. By using Proposition 1.28 and Remark 1.29, we seethat

C2A(A`) = CAopp(A`

/) = CAopp((Aopp)r) = AA`.

The proof of the second equality is done analogously.

Proposition 1.31. Let A be a ring and M a left A-module. Suppose N ≤M is an A-submodule which is adirect summand.

(i) N is stable under C2A(M), i.e. is also a submodule of M//. Letting N ′ be a complement for N , since

N ′ is also a direct summand, it is also stable under C2A(M), so N is also a direct summand of M//.

(ii) If N ′ is another direct summand of M , any A-linear map ϕ : N → N ′ is also C2A(M)-linear.

(iii) For every ψ ∈ C2A(M), the restriction of ψ to N lies in C2

A(N), i.e. ψM |N = (ψ|N )N . This impliesthat restriction to N is a morphism of rings (−)|N : C2

A(M)→ C2A(N).

Proof. Let πN ∈ EndA(M) be the projection of M onto N .

(i) For ψ ∈ C2A(M), since elements of C2

A(M) commute with elements of CA(M) = EndA(M) bydefinition, we see that

ψ(N) = ψ(π(M)) = π(ψ(M)) ⊆ π(M) = N.

(ii) Let ϕ : N → N ′ be an A-linear map. Then ϕ πN ∈ EndA(M) = CA(M), so for ψ ∈ C2A(M)

and n ∈ N , we have

ϕ(ψn) = ((ϕ πN ) ψ)(n) = (ψ (ϕ πN ))(n) = ψ(ϕ(n)).

(iii) Let ψ ∈ C2A(M) and ϕ ∈ CA(N) = EndA(N). By part (ii), if we set N ′ = N , we see that

ϕ ∈ EndC2A(M)(N), which means that ψ|N ϕ = ϕ ψ|N because ψ acts on N via ψ|N . This

implies ψ|N ∈ C2A(N), so we are done.

Definition 1.32. Let Aii∈I be a family of rings. Their product is the ring Adef=∏i∈I Ai with addition

and multiplication defined pointwise, so that its unit element is equal to (1)i∈I . Given a family Mii∈Iwhere each Mi is a left Ai-module, the abelian group

∏i∈IMi is a left

∏i∈I Ai-module via

(ai)i∈I(mi)i∈Idef= (aimi)i∈I .

Proposition 1.33. Let A1, A2 be rings and Mi be a left Ai-module, i = 1, 2. Then

CA1×A2(M1 ×M2) = CA1(M1)× CA2(M2), C2A1×A2

(M1 ×M2) = C2A1

(M1)× C2A2

(M2).

Proof. The second equality follows from the first since the bicentralizer is the centralizer of the counter-

15

Chapter 1

module, hence

C2A1×A2

(M1 ×M2) = CCA1×A2(M1×M2)(M1 ×M2)

= CCA1(M1)×CA2

(M2)(M1 ×M2)

= CCA1(M1)(M1)× CCA2

(M2)(M2)

= C2A1

(M1)× C2A2

(M2).

For the first, note thatM1 ×M2 = (M1 × 0)︸ ︷︷ ︸

def=M1

⊕ (0 ×M2)︸ ︷︷ ︸def=M2

.

The element ϕ ∈ EndZ(M1 ×M2) belongs in the set CA1×A2(M1 ×M2) if and only if for all (a1, a2) ∈A1×A2, we have ϕ(a1, a2)M = (a1, a2)Mϕ. Letting πi be the projection ontoMi, ϕ ∈ CA1×A2(M1×M2)implies that πiϕ ∈ CAi(Mi), so ϕ = π1ϕ+π2ϕ ∈ CA1(M1)×CA2(M2). The reverse inclusion is obvious,which proves equality.

Example 1.34. We give an example where the restriction morphism (−)|N : C2A(M) → C2

A(N) is neitherinjective or surjective. Consider a field K and the K-algebra EndK(K3). After fixing the standard basise1, e2, e3, this is canonically isomorphic to the ring of 3 × 3 matrices with coefficients in K , which wedenote by Mat3×3(K). Consider the K-subalgebra A of matrices of the forma 0 0

b c 00 0 a

, a, b, c ∈ K.

Considered as a left A-module, Mdef= K3 is the direct sum of the A-submodules N

def= K〈e1, e2〉 and

N ′def= K〈e3〉. Note that elements of CA(M) have to be K-linear maps since they have to commute with

elements of the form λM for λ ∈ K . The condition that ϕ ∈ EndZ(M) commutes with all matrices of theabove form gives linear conditions on the coefficients of the matrix form, so solving for it tells us that thematrix form of ϕ over the basis e1, e2, e3 is of the formα 0 0

0 α 0β 0 γ

One sees that this ring is of the same form as A, where the roles of e2 and e3 have been permuted. It followsthat A = C2

A(M). The map (−)|N : C2A(M)→ C2

A(N) is injective since it can be written asa 0 0b c 00 0 a

7−→ [a 0b c

]

However, one sees that CA(N) is the set of 2×2 scalar matrices, so C2A(N) is the full ring of 2×2 matrices,

which means that (−)|N is not surjective. Similarly, CA(N ′) = C2A(N ′) ' K , so (−)|N ′ is surjective but

not injective.

Note that A × A is a ring (its unit element is (1, 1)). Consider the left A × A-module M ×M with

action given by (a1, a2)(m1,m2)def= (a1m1, a2m2). The subset N × N ′ is a direct summand and by

Proposition 1.33, we see that C2A×A(M ×M) = A × A and CA×A(N × N ′) = CA(N) × CA(N ′). The

restriction map (−)|N×N ′ : C2A(M) × C2

A(M) → C2A(N) × C2

A(N ′) is neither injective or surjective sinceit is the product of the restriction maps (−)|N × (−)|N ′ on each component, the first not being surjectiveand the second not being injective.

16

Lemma 1.35. Let A be a ring, M a left A-module and N,P two A-submodules such that M = N ⊕ P .Suppose that there exists two families of A-submodules Pii∈I and Nii∈I of P and N respectively, suchthat P =

∑i∈I Pi and Pi ' N/Ni.

(i) We have CA(M)〈N〉 = M .

(ii) The restriction morphism (−)|N : C2A(M)→ C2

A(N) is injective.

(iii) If AN = C2A(N), then AM = C2

A(M).

Proof. (i) This is clear since CA(M)〈N〉 contains Pi by assumption ; just take the surjective mapN → N/Ni ' Pi and lift it to an endomorphism of M to see that CA(M)〈N〉 ⊇ Pi, so that

CA(M)〈N〉 ⊇ P .

(ii) Assume ϕ ∈ C2A(M) satisfies ϕ|N = 0. The composition

N → N/Ni ' Pi → P

is A-linear, thus C2A(M)-linear by Proposition 1.31 (ii). If ϕ|N = 0, then ϕ(Pi) = ϕ(N/Ni) = 0,

which means ϕ(P ) =∑

i∈I ϕ(Pi) = 0. This means ϕ(M) = ϕ(N) + ϕ(P ) = 0, i.e. ϕ = 0, so(−)|N is injective.

(iii) Let ϕ ∈ CA2 (M). Since AN = C2A(N), there exists a ∈ A such that ϕ|N = aN , i.e. (ϕ−aM )|N = 0.

Since restriction to N is injective by part (ii), we see that ϕ = aM ∈ AM . Since AM ⊆ CA2 (M), weare done.

Proposition 1.36. Let A be a ring and M a left A-module with an A-submodule such that M ' A` ⊕ N(in other words, we assume that M admits a direct summand isomorphic to A`). Then AM = C2

A(M).

Proof. Consider the restriction map (−)A`: C2

A(M) → C2A(A`). By Corollary 1.30, we have C2

A(A`) =A`, so for any ϕ ∈ C2

A(M), we have ϕ|A`= aA`

for some a ∈ A. Each element n ∈ N generates a leftA-module A〈n〉 ' A`/an where an is some left ideal of A (namely, the ideal of all those a ∈ A such thatan = 0). It follows that the pair (A`, N) satisfies the hypotheses of Lemma 1.35, so (−)|A`

is injective,This means ϕ = aM , so we’re done.

Definition 1.37. Let A be a ring. A left (resp. right) A-module is said to be free if there exists an

isomorphism M '⊕

i∈I A` (resp. M '⊕

i∈I Ar). When I is finite, we write A⊕n`def=⊕

i∈I A` where|I| = n.

Corollary 1.38. Let A be a ring and M be a left A-module which admits a direct summand isomorphic toA`. Then AM = C2

A(M) and

Z(EndA(M)) = Z(AM ),

i.e. an element of the center of EndA(M) is equal to multiplication by some element of Z(AM ). This holdsin particular if M is a free left A-module. Furthermore, if A is commutative, Z(EndA(M)) = AM .

17

Chapter 1

Proof. We have

Z(EndA(M)) = CEndZ(M)(EndA(M)) ∩ EndA(M)

= CEndZ(M)(CA(M)) ∩ EndA(M)

= C2A(M) ∩ EndA(M)

= AM ∩ EndA(M)

= Z(AM ).

The last equality is clear since saying that aM is an A-linear map is exactly saying that it commutes withall other bM for b ∈ A, i.e. lies in Z(AM ).

Remark 1.39. We have Z(A)M ⊆ Z(AM ), but we don’t have equality in general. The point is, twoendomorphisms aM and bM may commute without a and b commuting in A.

Definition 1.40. Let A be a ring and M a left A-module. We say that M is finitely generated or is afinite A-module if there exists a surjective map A⊕n` →M of left A-modules for some n ≥ 1.

Corollary 1.41. Let A be a PID (note that principal ideal domains are assumed commutative by definition!)and M be a finitely generated left A-module. Then AM = C2

A(M) and Z(EndA(M)) ' A/AnnA(M)(recall that AnnA(M) = ker(a 7→ aM )).

Proof. By the classification of finitely generated modules over a PID, we have an isomorphism M '⊕ni=1A/ai where we can choose the ideals such that a1 ⊆ a2 ⊆ · · · ⊆ an. It follows that AnnA(M) = a1,

so we can interpretM as an A/a1-module. Without loss of generality, assume a1 = 0. We can then applyProposition 1.36 and Corollary 1.38.

Corollary 1.42. Let V be a finite-dimensional vector space over a field K and ϕ1, ϕ2 ∈ EndK(V ). Thefollowing are equivalent :

(i) There exists a polynomial p ∈ K[T ] such that p(ϕ1) = ϕ2

(ii) For all ψ ∈ EndK(V ) which commutes with ϕ1, ψ commutes with ϕ2.

Proof. Turn V into aK[T ]-module Vϕ1 by setting pvdef= p(ϕ1)(v). The ring EndK[T ](V ) equals the set of

K-linear endomorphisms of V which commute with polynomials in ϕ1 ; this is equivalent to commutingwith ϕ1. By Corollary 1.41, we have K[T ]V /AnnK[T ](V ) ' Z(EndK[T ](V )), so after translating bothsides of this equality into the statements (i) and (ii), we are done.

1.3 Tensor products and Hom

For the rest of this chapter, A denotes a ring.

Definition 1.43. Let M be a right A-module and N be a left A-module. If L is an abelian group, a mapϕ : M ×N → L is called A-balanced if for any m1,m2 ∈M,n1, n2 ∈ N and a ∈ A, the following holds :

ϕ(m1 +m2, n1) = ϕ(m1, n1) + ϕ(m2, n1)

ϕ(m1, n1 + n2) = ϕ(m1, n1) + ϕ(m1, n2)

ϕ(m1a, n1) = ϕ(m1, an1).

In other words, it is a Z-bilinear map for which the action of a ∈ A can be done either on M or on Nwithout changing the result of the application of ϕ.

18

The tensor product of M and N over A is the abelian group M ⊗A N defined as the quotient of thefree abelian group Z⊕(M×N) modulo the subgroup generated by the relations

(m1 +m2, n1)− (m1, n1)− (m2, n1)

(m1, n1 + n2)− (m1, n1)− (m1, n2)

(m1a, n1)− (m1, an1).

The coset of the Z-module basis element (m,n) ∈M ×N in M ⊗AN is denoted by m⊗ n. It comes witha canonical A-balanced map ι : M ×N →M ⊗N given by (m,n) 7→ m⊗ n.

Proposition 1.44. (Universal property of the tensor product) LetM be a right A-module, N a left A-moduleand L an abelian group. There is a canonical bijection

ϕ : M ×N → L | ϕ is A-balanced ←→ HomZ(M ⊗A N,L)

sending ϕ to ϕ in such a way that the following diagram commutes :

M ×N M ⊗A N

Lϕ

ι

ϕ

Proof. Obvious from our definition of M ⊗A N . Details are left to the reader.

Remark 1.45. Since we have shown thatM⊗AN satisfies the above universal property, the pair (M⊗AN, ι)is unique up to a unique isomorphism making a commutative triangle with M ×N as above.

More generally, if we assume additionally that we have rings B,C , thatM is a left B-module, N a rightC-module and L a (B,C)-bimodule, then M ⊗A N is a (B,C)-bimodule via

∀b ∈ B, c ∈ C,m ∈M,n ∈ N, b(m⊗ n)cdef= bm⊗ nc

and the set of A-balanced maps ϕ : M ×N → L satisfying

ϕ(bm, nc) = bϕ(m,n)c

are in bijection with Hom(B,C)(M ⊗A N,L). Even more generally, the module structures on M and Nwhich are compatible with their respective A-module structures induce module structures on M ⊗A N (ifthe module structure on M is on the right or the structure on N is on the left, just replace it by the corre-sponding structure on the opposite ring and use the same construction).

Furthermore, the construction (M,N) 7→M ⊗A N gives a bifunctor ⊗ : Mod-A×A-Mod→ Ab. Ifany ofM or N admits module structures compatible with their A-module structure, ⊗ will also be functorialin the category of pairs of A-modules admitting those structures. Note that ⊗ is an additive functor in bothof its arguments, meaning that if f, f ′ ∈ EndA(M) and g, g′ ∈ EndA(N), then

(f + f ′)⊗ g = f ⊗ g + f ′ ⊗ g, f ⊗ (g + g′) = f ⊗ g + f ⊗ g′.

Definition 1.46. Let A,B be rings andM,N be (A,B)-modules. The abelian group Hom(A,B)(M,N) au-tomatically becomes an (End(A,B)(N),End(A,B)(M))-bimodule via post-/pre-composition : for (ϕB, ϕA) ∈EndB(N)× EndA(M), f ∈ Hom(A,B)(M,N)

ϕBfϕAdef= ϕB f ϕA.

One easily checks that this is still a morphism of (A,B)-bimodules : for (a, b) ∈ A×B and m ∈M ,

(ϕBfϕA)(amb) = (ϕBf)(aϕA(m)b) = ϕB(a(fϕA)(m)b) = a(ϕBfϕA)(m)b.

19

Chapter 1

Remark 1.47. WhenM,N are two left A-modules, we can define module structures on HomA(M,N) usingmodule structures on M and N compatible with their A-module structures. For instance, if M is also aleft C-module, N a left B-module and the structures are compatible, we can turn HomA(M,N) into a(B,C)-bimodule via

∀m ∈M, b ∈ B, c ∈ C, (bfc)(m)def= bf(cm).

This is possible because the compatibility condition is equivalent to asking that the endomorphisms cMand bN are A-linear, giving rise to morphisms of rings C → EndA(M) and B → EndA(N). The(EndA(N),EndA(M))-bimodule structure on HomA(M,N) can therefore be restricted to a (B,C)-bimodulestructure.

We can also apply this technique if one of the module structures onM or N is a right module structureinstead of a left one, as long as it is compatible with the left A-module structures on M or N ; it sufficesto replace these right module structures by left module structures on the opposite ring. This is possible,in particular, when M is an (A,B)-bimodule and N a (A,C)-bimodule, making HomA(M,N) a (C,B)-bimodule via

∀m ∈M, b ∈ B, c ∈ C, (cfb)(m)def= f(mb)c.

Furthermore, the construction (M,N) 7→ HomA(M,N) gives a bifunctor ⊗ : A-Modopp×A-Mod→ Ab.If any ofM orN admits module structures compatible with their A-module structure, HomA(−,−) will alsobe functorial in the category of pairs of A-modules admitting those structures. By linearity of composition,HomA(−,−) is also additive in both of its arguments.

Lemma 1.48. Let M,N, Mii∈I , Nii∈I be left A-modules. We have natural isomorphisms

HomA(⊕i∈I

Mi, N) '∏i∈I

HomA(Mi, N), HomA(M,∏i∈I

Ni) '∏i∈I

HomA(M,Ni).

Proof. The first statement is equivalent to saying that a morphism of left A-modules ϕ : M → N isdetermined by its restrictions ϕ|Mi : Mi → N since ϕ =

⊕i∈I ϕ|Mi . For the second one, letting

πi : N → Ni be the canonical projection, this is the fact that a morphism ϕ : M → N is determined byits components πi ϕ : M → Ni.

Corollary 1.49. Let M,N be two left A-modules and assume N =⊕

i∈I Ni for a family Nii∈I ofsubmodules. If M is finitely generated, then the isomorphism

HomA(M,∏i∈I

Ni) '∏i∈I

HomA(M,Ni)

restricts to the following isomorphism on the two following subsets :

HomA(M,⊕i∈I

Ni) '⊕i∈I

HomA(M,Ni).

Proof. The isomorphism of Lemma 1.48 puts in correspondence a morphism ϕ : M →⊕

i∈I Ni withits projections πi ϕi∈I . Since M is finitely generated, such a morphism ϕ is determined by theimage of a subset m1, · · · ,mk. Because ϕ(mi) =

⊕i∈I ni, only finitely many indices i ∈ I contain

non-zero components of the elements ϕ(m1), · · · , ϕ(mk). It follows that only finitely many of the mapsπi ϕ are non-zero. Conversely, if only πi1 ϕ, · · · , πis ϕ are non-zero, then the corresponding mapϕ : M →

∏i∈I Ni satisfies ϕ(M) ⊆

∑sj=1Nij , so ϕ(M) lands inside N in a well-defined manner.

20

Theorem 1.50. (Tensor-hom adjunction) Let A,B be rings, M a right A-bimodule, N a (A,B)-bimoduleand P a right B-bimodule. (It follows that M ⊗A N is a right B-module and HomB(N,P ) is a rightA-module.) There is a natural isomorphism of abelian groups

HomB(M ⊗A N,P ) ' HomA(M,HomB(N,P )).

Moreover, this is an isomorphism of right EndA(M)-modules, End(A,B)(N)-modules and EndB(P )-modules. This implies that if C is a ring and any of M , N or P admit a C-module structure which iscompatible with the module structures we already gave on that module, then both sides admit that corre-sponding C-module structure and the natural isomorphism above is an isomorphism of C-modules.

Proof. Working out the details is easy but long and pointless. Seeing the main idea is the most useful part.A morphism of right B-modules ϕ : M ⊗AN → P corresponds to an A-balanced map ϕ : M ×N → Pfor which the restriction ϕm : N → P given by ϕm(n) = ϕ(m,n) is B-linear. The fact that it isA-balanced implies that for a ∈ A,

ϕam(n) = ϕm(an) = (ϕma)(n)

by definition of the A-module structure on HomB(N,P ), which means that the construction m 7→ ϕm isA-linear. This process can be reversed, thus gives the isomorphism. Naturality is obvious. The fact thatthis is an isomorphism of EndA(M)-modules (resp. End(A,B)(N), EndB(P )) follows from the naturalityof this isomorphism in all three arguments. Restricting this property to any subring of one of these threerings gives the statement about compatible module structures.

Theorem 1.51. Let M be a right A-module and N be a left A-module.

(i) The functor of left A-modules to abelian groups given by N 7→M ⊗A N is right-exact.

(ii) The functor of right A-modules to abelian groups M 7→M ⊗A N is right-exact.

Proof. Let P be a left A-module. Recall that HomA(N,P ) is an (A,A)-bimodulesince both N and P are left A-modules. Since M is a right A-module, the functorHomA(M,HomA(−, P )) = HomZ(M,−) HomA(−, P ) is a composition of left-exact functors,namely HomA(−, P ) : A-Mod → A-Mod-A and HomA(M,−)|A-Mod-A : A-Mod-A → A-Mod-A(because both M and HomA(N,P ) are right A-modules). The first functor sends a right-exact sequenceto a left-exact one, and the second functor preserves left-exact sequences. Therefore, the composition isa contravariant functor mapping a right-exact sequence to a left-exact one and it is naturally isomorphicto HomA(M ⊗A −, P ) by Theorem 1.50. Since the latter is exact for all left A-modules P , it follows thatthe functor M ⊗A − is right-exact by Proposition 1.16 since the functor HomA(−, P ) is contravariant.This proves part (i).

For part (ii), note that the functor HomA(−,HomA(N,P )) is contravariant left-exact for all left A-modules N ,P and it is naturally isomorphic to HomA(− ⊗A N,P ). Since the latter is left-exact for allP , the functor −⊗A N is right-exact.

Definition 1.52. Let M0, · · · ,Mn be abelian groups and A1, · · · , An be rings with the following relations :

(i) M0 is a right A1-module

(ii) Mn is a left An-module

(iii) For each 1 ≤ i ≤ n− 1, Mi is a (Ai, Ai+1)-bimodule.

21

Chapter 1

Then we can form the abelian group

M0 ⊗A1 M1 ⊗A2 M2 ⊗A2 · · · ⊗An Mn

as follows. Consider the free abelian group Z⊕∏n

i=0Mi on the set n + 1-tuples (m0, · · · ,mn) ∈∏ni=0Mi.

Quotient out by the submodule generated by the relations

(m0, · · · ,mi, · · · ,mn) + (m0, · · · ,m′i, · · · ,mn)− (m0, · · · ,mi +m′i, · · · ,mn)

(m0, · · · ,miai,mi+1, · · · ,mn)− (m0, · · · ,mi, aimi+1, · · · ,mn).

This gives us the abelian group. It is a bimodule for each ring Ai by allowing multiplication on the Mi−1-factor on the right and on theMi-factor on the left. If A0 (resp. An+1) is a ring andM0 is a left A0-module(resp. if Mn is a right An+1-module), then the corresponding tensor product also admits this structure.

Proposition 1.53. (Associativity of the tensor product) Let A0, A1, A2, A3 be rings,M0 a (A0, A1)-bimodule,M1 a (A1, A2)-bimodule and M2 a (A2, A3)-bimodule. We have two isomorphism of abelian groups whichis simultaneously an isomorphism of A0, A1, A2 and A3-bimodules :

M0 ⊗A1 (M1 ⊗A2 M2) 'M0 ⊗A1 M1 ⊗A2 M2 ' (M0 ⊗A1 M1)⊗A2 M2.

Proof. It suffices to see that the correspondences

m0 ⊗ (m1 ⊗m2)←→ m0 ⊗m1 ⊗m2 ←→ (m0 ⊗m1)⊗m2

produce well-defined maps in each direction and they are inverse to each other, as can be seen whenapplied on the generators of the respective modules given above.

Proposition 1.54. (Tensor product with respect to opp) Given a left (resp. right) A-module M , recallRemark 1.6 for the definition of the opposite module structureMopp on its opposite ring Aopp. LetM0 be an(A0, A1)-bimodule and M1 be an (A1, A2)-bimodule. We have an isomorphism of (Aopp

2 , Aopp0 )-bimodules

(M0 ⊗A1 M1)opp 'Mopp1 ⊗Aopp

1Mopp

0 .

In particular, if A0, A1, A2 are commutative rings, the tensor product is commutative up to the naturalisomorphism given in the proof.

Proof. Givenm ∈M0, writemopp ∈Mopp0 for the same element but which gets affected by multiplication

from the opposite side in the reversed order. It suffices to map (m0 ⊗ m1)opp to mopp1 ⊗ mopp

0 . The(Aopp

2 , Aopp0 )-linearity becomes obvious since for a2 ∈ A2, m0 ∈ M0 and m1 ∈ M1, denoting the

opposite multiplication with the symbol ∗, we have

a2 ∗ (m0 ⊗m1)opp = (m0 ⊗m1a2)opp 7→ (m1a2)opp ⊗mopp0 = a2 ∗mopp

1 ⊗mopp0 .

The argument is similar for a0 ∈ A0.

1.4 Isotypical modules

Isotypical modules are to be seen as generalizations of free modules over a ring. We will see their importancein Theorem 1.59, where we establish a correspondence between the free modules over EndA(M) = CA(M)and the isotypical modules of type M .

Definition 1.55. LetM,N be two left A-modules. We say thatM is isotypical of type N ifM '⊕

i∈I Ni

where Ni ' N . When this is the case, we write M ' N⊕I .

22

Proposition 1.56. Let M be an isotypical A-module of type N .

(i) The restriction map (−)|N : C2A(M)→ C2

A(N) is an isomorphism.

(ii) The C2A(M)-module M is also isotypical of type N when N is seen as a C2

A(M)-module via itsC2A(N)-module structure and the isomorphism (−)|N : C2

A(M) ' C2A(N) given in part (i).

Proof. By Proposition 1.31 (ii) and Lemma 1.35 (ii), we know that (−)|N is injective and that the directsummands isomorphic to N in (ii) are isomorphic as C2

A(M)-modules. It only remains to show that(−)|N is surjective.

Let ϕ ∈ C2A(N), and let ψ

def= ϕ⊕I ∈ EndZ(M), so that whenever m ∈ Ni = N , we have ψ(m) =

ϕ(m) ⊆ Ni. It is clear that ψ|N = ϕ, so we only need to show that ψ ∈ C2A(M). Let α ∈ CA(M)

and m ∈ Ni for some i ∈ I . Denote the projection of M onto Ni by πi. Note that πiα is an A-linearendomorphism of Ni, thus commutes with ϕ. Therefore,

(αψ)(m) =∑i∈I

(πiαψ)(m) =∑i∈I

(πiα)(ϕ(m)) =∑i∈I

ϕ((πiα)(m)) = ψ

(∑i∈I

(πiα)(m)

)= (ψα)(m).

By linearity, we obtain αψ = ψα, which completes the proof.

Remark 1.57. We had already proved Proposition 1.56 in the case where F = A`, so this is a generalizationfrom the case of free A-modules to the case of isotypical A-modules.

Theorem 1.58. Let M be a left A-module. Since the A-module structure and CA(M)-module structureson M are compatible by definition, when V is a left A-module and W is a right CA(M)-module, we candefine

S(W )def= W ⊗CA(M) M, T (V )

def= HomA(M,V ).

By the constructions of Section 1.3, S(W ) is canonically a left A-module and T (V ) is canonically a rightCA(M)-module, which gives a canonical isomorphism

HomA(S(W ), V ) ' HomCA(M)(W,T (V )).

The functors S and T are therefore adjoint functors. We can describe the unit and counit of this adjunctionexplicitly as follows :

εV : S(T (V )) = HomA(M,V )⊗CA(M) M → V, εV (ϕ⊗m) 7→ ϕ(m)

andηW : W → T (S(W )) = HomA(M,W ⊗CA(M) M), ηW (w)(m) = w ⊗m.

The morphism εV is a morphism of A-modules and ηW is a morphism of CA(M)-modules.

Proof. We can apply Theorem 1.50 directly. One needs to see M and W as right Aopp-modules insteadof left A-modules so that the statement applies accordingly to the notation given there, but this makesliterally no difference in the result.

Theorem 1.59. Let M be a finitely generated left A-module, V a left A-module isotypical of type M andW a free right CA(M)-module.

(i) The left A-module S(W ) is isotypical of type M . More explicitly, for any CA(M)-basis wii∈I of

W , the submodules Sidef= 〈wi〉CA(M) ⊗M satisfy S(W ) =

⊕i∈I Si.

23

Chapter 1

(ii) The morphism ηW : W → HomA(M,W ⊗CA(M) M) is an isomorphism of CA(M)-modules.

(iii) If W ′ is another right CA(M)-module, the morphism of abelian groups

S : HomCA(M)(W′,W )→ HomA(S(W ′), S(W ))

is an isomorphism.

(iv) Write V =⊕

i∈I Vi where ιi : M → V is the isomorphism onto Vi. Then T (V ) = HomA(M,V ) is afree right CA(M)-module with basis ιii∈I .

(v) The morphism εV : S(T (V )) = HomA(M,V )⊗CA(M) M → V is an isomorphism of A-modules.

(vi) If V ′ is another left A-module, the morphism of abelian groups

T : HomA(V, V ′)→ HomCA(M)(T (V ), T (V ′))

is an isomorphism.

Proof. (i) Writing W = CA(M)⊕I , we obtain

S(W ) = W ⊗CA(M) M ' (CA(M)⊕I)⊗CA(M) M ' (CA(M)⊗CA(M) M)⊕I 'M⊕I .

A choice of basis corresponds to an isomorphism W ' CA(M)⊕I , which explains the secondstatement.

(ii) The morphism ηW commutes with the following isomorphisms since CA(M) = HomA(M,M) :

W ' HomA(M,M)⊕I ' HomA(M,M⊕I) ' HomA(M,W ⊗CA(M) M) = S(T (W )).

(iii) Because S and HomCA(M)(W′,−) commute with direct sums, it suffices to show that

HomCA(M)(W′,HomA(M,M)) = HomCA(M)(W

′, CA(M)) ' HomA(W ′ ⊗CA(M) M,M),

which is precisely the statement of the tensor-hom adjunction.

(iv) Without loss of generality, write V = M⊕I . It follows that

T (V ) = HomA(M,M⊕I) ' HomA(M,M)⊕I ' CA(M)⊕I

is a free right CA(M)-module.

(v) Letting V = M⊕I , the morphism εV commutes with the isomorphism

HomA(M,V )⊗CA(M) M ' CA(M)⊕I ⊗CA(M) M ' CA(M)⊕I ⊗CA(M) M 'M⊕I = V.

(vi) Because T and HomA(−, V ′) commutes with direct sums, it suffces to show that

HomA(M,V ′) ' HomCA(M)(HomA(M,M),HomA(M,V ′)) =' HomCA(M)(CA(M),HomA(M,V ′)),

which is clear ; the isomorphism from right to left is given by evaluation of a CA(M)-linear mapϕ : CA(M)→ HomA(M,V ′) at 1 ∈ CA(M), i.e. ϕ 7→ ϕ(1).

Corollary 1.60. Let M be a finitely generated left A-module, V a left A-module isotypical of type M andW a free right CA(M)-module.

24

(i) The map f 7→ S(f) induces an isomorphism of rings

EndCA(M)(W ) ' EndA(S(W )) = EndA(W ⊗CA(M) M).

(ii) The map g 7→ T (g) induces an isomorphism of rings

EndA(V ) ' EndCA(M)(T (V )) = EndCA(M)(HomA(M,V )).

(iii) There is a one-to-one correspondence between the A-submodules V ′ ≤ V which are direct summandsof V isotypical of typeM and the free right CA(M)-submodulesW ′ ≤W which are direct summandsof W via the formulas

V ′ 7→ T (V ′), W ′ 7→ S(W ′).

Proof. Part (i) is obtained by setting W ′ = W in Theorem 1.59 (iii) and part (ii) is obtained by settingV ′ = V in Theorem 1.59 (vi). This argument makes them isomorphisms of abelian groups ; functorialityof this isomorphism in the general case makes it an isomorphism of rings. Part (iii) follows from parts (i)and (ii) since the ring isomorphisms given there map projections to projections.

25

Chapter 2

Finiteness conditions on rings and modules

In this chapter, A denotes a ring. The three conditions we will study are the properties of being artinian,noetherian and of finite length ; these apply to both modules and rings.

2.1 Artinian, noetherian and finite length modules

Definition 2.1. Let M be a left A-module.

(i) We say that M is artinian if it satisfies one of the following equivalent conditions :

• Every non-empty subset of submodules of M possesses a minimal element

• Every countable descending chainM1 ⊇M2 ⊇ · · · ⊇Mn ⊇ · · · of submodules ofM stabilizes,which means there exists n ≥ 1 such that for all n′ ≥ n, Mn′ = Mn.

(ii) We say that M is noetherian if it satisfies one of the following equivalent conditions :

• Every non-empty subset of submodules of M possesses a maximal element

• Every countable ascending chain M1 ⊆ M2 ⊆ · · · ⊆ Mn ⊆ · · · of submodules of M stabilizes,which means there exists n ≥ 1 such that for all n′ ≥ n, Mn′ = Mn.

The equivalence between those two definitions is clear by using Zorn’s Lemma. Note that M is an artinian(resp. a noetherian) left A-module if and only if it is artinian (resp. noetherian) as a left AM -module.

Example 2.2. (i) If K is a field and V is a finite-dimensional vector space over K , then V is artinianand noetherian because given a chain of subspaces, the dimension of each member of the chain tellsus when the chain stops.

(ii) Let Mii∈I be an infinite family of non-zero left A-modules and consider Mdef=⊕

i∈I . Then Mis neither artinian or noetherian. It suffices to give a total order on I and consider the chain of

submodules M ′idef=∑

j≤iMj (resp. M ′′idef=∑

j≥iMj ).

Proposition 2.3. Let M be a noetherian left A-module. If S ⊆ M is such that A〈S〉 = M , there existsn ≥ 1 and s1, · · · , sn ∈ S such that A〈s1, · · · , sn〉 = M . In particular, every noetherian left A-module isfinitely generated.

Proof. Without loss of generality, assume S is infinite. Let MT T⊆S be the set of submodules given by

MTdef= A〈T 〉 where T is assumed finite. This collection is not empty, so let MT0 be a maximal element.

26

For any s ∈ S, MT0∪s ⊆ MT0 ⊆ MT0 + A〈s〉 = MT0∪s, which means s ∈ MT0 . Therefore S ⊆ MT0 ,which implies MT0 = M .

Since S = M generates M , this implies M is finitely generated.

Proposition 2.4. Let M be a left A-module. The following are equivalent :

(i) M is noetherian

(ii) Every A-submodule M ′ ≤M is finitely generated.

Since the condition (ii) is inherited from M to any of its submodules, it follows that every submodule of anoetherian A-module is noetherian.

Proof. ( (i) ⇒ (ii) ) By definition, a chain of submodules of M ′ is also a chain of submodules of M , soM ′ is noetherian, hence finitely generated by Proposition 2.3.

( (ii) ⇒ (i) ) Suppose we are given a chain M1 ⊆ M2 ⊆ · · · ⊆ Mn ⊆ · · · of submodules of M . Let

M ′def=⋃n≥1Mn. Since M ′ is finitely generated, its generators all lie in Mn for some n ≥ 1, meaning

that Mn = Mn+1 = · · · = M ′. It follows that M is noetherian.

Proposition 2.5. Consider the following exact sequence of left A-modules :

0 N M M/N 0ιN πN

(i) M is artinian if and only if N and M/N are artinian.

(ii) M is noetherian if and only if N and M/N are noetherian.

Proof. We begin with (i).

(⇒) If Nii∈N is a decreasing chain of submodules of N , then ιN (Ni)i∈I is a decreasing chain ofsubmodules of M , so it stabilizes ; since ιN is injective, our original chain stabilizes, so N is artinian.Similarly, if Pii∈N is a decreasing chain of submodules of M/N , then π−1

N (Pi)i∈I is a decreasingchain of submodules of M , hence stabilizes ; since πN is surjective, Pi = πN (π−1

N (Pi)), so our originalchain stabilizes and M/N is artinian.

(⇐) Let Mii∈I be a decreasing chain of submodules of M/N . The projections πN (Mi)i∈I forma chain in the artinian module M/N , thus stabilizes. Therefore, there exists n1 ≥ 1 such thatMi +N = Mn1 +N for all i ≥ n1. The restrictions ι−1

N (Mi)i∈N form a chain in the artinian moduleN , thus also stabilizes, which means there exists n2 ≥ 1 with Mi ∩N = Mn2 ∩N for all i ≥ n2. Letting

ndef= maxn1, n2, we see that Mi + N = Mn + N and Mi ∩N = Mn ∩N . It follows that Mi = Mn

since Mi ⊆ Mn + N but m ∈ Mi ∩ (Mn + N \Mn) is impossible since then m ∈ Mi ∩ N ; this givesMi ⊆Mn, and the symmetry of the argument gives equality. In other words, M is artinian.

The proof in the noetherian case is the same ; just replace the words “artinian” by “noetherian” and thewords “decreasing chain” by “increasing chain”.

Corollary 2.6. A finite direct sum of artinian (resp. noetherian) left A-modules is artinian (resp. noetherian).

27

Chapter 2

Proof. By induction on the number of modules considered. IfM1, · · · ,Mn are artinian (resp. noetherian),

letMdef=⊕n

i=1Mi andM ′def=⊕n−1

i=1 . By assumption, M ′ andMn are artinian (resp. noetherian), henceso is M by considering the split exact sequence involving those three.

Definition 2.7. Let M be a left A-module.

(i) The module M is called simple if it is nonzero and its set of submodules equals 0,M.

(ii) Given a left A-module M and a left ideal a E A, we let

aMdef=

n∑i=1

aimi

∣∣∣∣∣n ≥ 1, ai ∈ a, mi ∈M

.

The annihilator of a module M is the kernel of the map AM → EndA(M), i.e.

AnnA(N)def= a ∈ A | aM = 0.

If m ∈ M , we also define the annihilator of m as the annihilator of A〈m〉, namely the set of a ∈ Asuch that am = 0. A left A-module M for which AnnA(M) = 0 is said to be faithful.

(iii) An element a ∈ A is called nilpotent if an = 0 for some n ≥ 1 ; the set of nilpotent elements inA is denoted by Nil(A). Note that in contrast with the case of commutative rings, Nil(A) is not anideal since it is not even a subgroup of A in general! This can be seen by the example of the ringMat2×2(A) and the equation [

0 10 0

]+

[0 01 0

]=

[0 11 0

],

where the two matrices we are adding square to zero but their sum squares to the identity matrix.

(iv) A left ideal a E A is called a nilideal if a ⊆ Nil(A).

(v) The (left) Jacobson radical is the intersection of all left maximal ideals of A. We denote it by Jac(A).

Proposition 2.8. Let A be a ring. The following conditions on a ∈ A are equivalent :

(i) a ∈ Jac(A)

(ii) 1− xa admits a left inverse for all x ∈ A

(iii) For every simple left A-module M , we have a ∈ AnnA(M), i.e. am = 0 for all m ∈M .

Proof. ( (i) ⇒ (ii) ) Let a ∈ Jac(A) and x ∈ A. Repeating the argument of Krull’s theorem in thenon-commutative case, if 1 − xa does not have a left inverse, the left ideal A〈1− xa〉 E A is proper,hence Zorn’s Lemma implies that A〈1− xa〉 is contained in some left maximal ideal m E A. But then1− xa, xa ∈ m, hence 1 = (1− xa) + xa ∈ m, a contradiction.

( (ii) ⇒ (iii) ) Suppose M is a simple left A-module and there exists m ∈ M with am 6= 0. Since Mis simple, aM being a left A-submodule of M , we have aM = M . In particular, there exists x ∈ Awith x(am) = m, e.g. (1−xa)m = 0. Since 1−xa admits a left inverse z, we havem = z((1−xa)m) = 0.

( (iii) ⇒ (i) ) Let m E A be a left maximal ideal. Then A/m is a simple left A-module, hence a ∈AnnA(A/m) = m. Taking the intersection over all left maximal ideals m, we obtain a ∈ Jac(A).

Theorem 2.9. (Nakayama’s Lemma) Let M be a finitely generated left A-module. If Jac(A)M = M , thenM = 0.

28

Proof. Suppose M 6= 0 and let m1, · · · ,mn be a set of generators where n > 0 is chosen minimal.Write

m1 = a1m1 + · · ·+ anmn, ai ∈ Jac(A).

We can re-write this as(1− a1)m1 = a2m2 + · · ·+ anmn.

Since (1− a1) admits a left-inverse, we see that m1 ∈ 〈m2, · · · ,mn〉A, hence m2, · · · ,mn is also a setof generators for M . This contradicts the minimality of n, therefore we must have n = 0, i.e. M = 0.

Proposition 2.10. Let A be a ring. If a E A is a nilideal, then a ⊆ Jac(A).

Proof. Let a E A be a nilideal and pick a ∈ a. For x ∈ A, since xa ∈ a is nilpotent, we have (xa)n = 0for some n ≥ 1. Therefore (

n−1∑i=0

(xa)i

)(1− xa) = 1− (xa)n = 1,

showing that 1− xa has a left-inverse for all x ∈ A. This means a ∈ Jac(A) by Proposition 2.8.

Proposition 2.11. Let K be a field and A a K-algebra. If A is a finite-dimensional K-vector space, thenthere exists n ≥ 1 such that Jac(A)n = 0.

Proof. Since the sequence Jac(A) ⊇ Jac(A)2 ⊇ · · · is a decreasing sequence of finite-dimensional K-vector subspaces of A, the sequence eventually becomes stationary, so there exists n ≥ 1 such thatJac(A)n = Jac(A)n+1 = Jac(A)Jac(A)n. By Nakayama’s Lemma, Jac(A)n = 0.

The next part of this preliminary section is needed to prove the existence of a Jordan-Hölder series ofa left A-module of finite length, which is a necessary tool to work with simple subquotients of a module offinite length.

Lemma 2.12. Let M a left A-module and F,E,U ⊆ M be A-submodules. Suppose F ⊆ E. Then(E ∩ U) + F = E ∩ (U + F ).

Proof. We have (E ∩ U) + F = (E ∩ U) + (E ∩ F ) = E ∩ (U + F ).

We see thatF ⊆ (E ∩ U) + F = E ∩ (U + F ) ⊆ E,

so that in some way we inserted U in between E and F . We call the module (E ∩ U) + F = E ∩ (U + F )the insertion of U between E and F , and we denote it by InsE,F (U).

Theorem 2.13. (Zassenhaus’ Lemma, a.k.a. the Butterfly Lemma) Let A be a ring, F , E, F ′ and E′ beA-submodules of the left A-module M with the property that F ⊆ E and F ′ ⊆ E′. We have a naturalisomorphism of left A-modules

((E ∩ E′) + F )/((E ∩ F ′) + F ) ' ((E′ ∩ E) + F ′)/((E′ ∩ F ) + F ′),

which tells us thatInsE,F

(E′)/InsE,F

(F ′)' InsE′,F ′ (E) /InsE′,F ′ (F ) .

So the insertions of E′ and F ′ between E and F have, up to isomorphism, the same quotient as theinsertions of E and F between E′ and F ′.

29

Chapter 2

E

InsE,F (E′)

InsE,F (F ′)

F

(E ∩ F ′) + (E′ ∩ F )

E ∩ E′

E′

InsE′,F ′ (E)

InsE′,F ′ (F )

F ′

Proof. By symmetry with respect to the pairs E ↔ E′ and F ↔ F ′, it suffices to prove that

InsE,F(E′)/InsE,F

(F ′)' (E ∩ E′)/((E ∩ F ′) + (E′ ∩ F )).

Consider the natural maps

E ∩ E′ (E ∩ E′) + F ((E ∩ E′) + F )/((E ∩ F ′) + F ) = InsE,F (E′) /InsE,F (F ′) .

We see that the composition is surjective because E ∩E′ gets mapped and the elements of F are zero inthe last quotient. The kernel of this map is

(E ∩ E′) ∩ ((E ∩ F ′) + F ) = E ∩ F ′ + E′ ∩ F,

so by the isomorphism theorems, we are done.

Definition 2.14. Let M be a left A-module. A finite series for M is a collection (M0, · · · ,Mn) of A-submodules of M such that M0 = 0, Mn = M and Mi ⊆Mi+1 for i = 0, · · · , n− 1 (equality Mi = Mi+1

is allowed) ; we denote a finite series forM byM•. The length ofM• = (M0, · · · ,Mn) is defined to be theinteger n.

Two finite series M• and N• of A-submodules of M , of length n and r respectively, are called equiva-lent if r = n and there exists a permutation σ ∈ Sn (the group of permutations of n elements) such thatMi/Mi−1 ' Nσ(i)/Nσ(i)−1. The finite series P• of M of length s is called a refinement of the series M• ifM0, . . . ,Mn ⊆ P0, . . . , Ps.

A finite seriesM• is called a Jordan-Hölder series forM if all succesive quotientsMi/Mi−1 are simple.In other words, the series does not allow proper refinements, hence is a maximal element of the set of finiteseries ofM , partially ordered under the relation of refinement. The moduleM is said to be of finite lengthif it admits a Jordan-Hölder ( JH) series.

Theorem 2.15. LetM be a left A-module. Then any two finite series of submodules ofM allow refinementswhich are equivalent.

Proof. Let M• and N• be two series of submodules of M of length r and s respectively. For i = 1, . . . , r

and j = 0, . . . , s, let Pi,jdef= InsMi−1,Mi (Nj). Then Pi,0 = Mi−1 and Pi,s = Mi, namely

0 = P1,0 ⊆ · · · ⊆ P1,s = M1 = P2,0 ⊆ · · · ⊆ P2,s ⊆ · · · ⊆ Pr,0 ⊆ · · · ⊆ Pr,s = M.

This refinement arises from inserting the series N• between each of the consecutive modules in M•.

Analogously, define Qi,jdef= InsNj−1,Nj (Mi) for j = 1, . . . , s and i = 0, . . . , r, which refines N• by

30

insertingM• between each of the consecutive modules in N•. We order the submodules Qi,j analogously,namely

0 = Q0,1 ⊆ · · · ⊆ Qr,1 = N1 = Q0,2 ⊆ · · · ⊆ Qr,2 ⊆ · · · ⊆ Q0,s ⊆ · · · ⊆ Qr,s = M.

We now show that P•,• and Q•,• are equivalent. By Zassenhaus’ Lemma, if i = 1, . . . , r and j = 1, . . . , s,we have

Pi,j/Pi,j−1 = InsMi−1,Mi (Nj) /InsMi−1,Mi (Nj−1)

' InsNj−1,Nj (Mi) /InsNj−1,Nj (Mi−1)

= Qi,j/Qi−1,j .

Corollary 2.16. Let M be a left A-module of finite length. Any finite series of submodules of M can berefined to a Jordan-Hölder series. Any two Jordan-Hölder series ofM are equivalent, and in particular musthave the same length.

Proof. Let M• be a series of submodules of M . Fix a Jordan-Hölder series N• of M . By Theorem 2.15,there exists refinements M ′• and N ′• of M• and N• respectively, such that M ′• and N ′• are equivalent.Since N• is a Jordan-Hölder series, N ′• is formed of N• by repeating certain terms, so that in N ′•,all successive quotients are simple or 0. Since N ′• andM

′• are equivalent, the same property holds forM ′•.

Deleting the superfluous terms (i.e. removing the successive quotients which are zero and keep only onecopy of each submodule) in N ′• and M

′•, we obtain Jordan-Hölder series M ′′• and N ′′• with the following

properties :

- M ′′• refines M•

- N ′′• = N•

- M ′′• is equivalent to N ′′• .

This means that the series M• allows the Jordan-Hölder series M ′′• as a refinement.

If M• was already a Jordan-Hölder series, then in the previous construction M• = M ′′• and M ′′• isequivalent to N ′′• where N ′′• = N•, hence M• and N• are equivalent.

Definition 2.17. Let M be a left A-module of finite length. The length of M is the length of any Jordan-Hölder series (which is well-defined by Corollary 2.16) and is denoted by `A (M) or ` (M) if A is understood.

Corollary 2.18. Let M a left A-module of finite length.

(i) If N ≤M is a submodule, N• is a Jordan-Hölder series for N and (M/N)• is a Jordan-Hölder seriesfor M/N , denoting by πN : M →M/N the canonical projection, the series

0 = N0 ( · · · ( Nk = N = π−1N ((M/N)0) ( π−1

N ((M/N)1) ( · · · ⊆ π−1N ((M/N)`) = M

is a Jordan-Hölder series for M . In particular, `A (M) = `A (N) + `A (M/N) (we say that length isadditive on short exact sequences).

(ii) IfM = N ⊕P for two submodules N,P ≤M and N•, P• are two Jordan-Hölder series for N and Prespectively, then

0 = N0 ( · · · ( Nk = N = N ⊕ P0 ( N ⊕ P1 ( · · · ( N ⊕ P` = N ⊕ P = M

is a Jordan-Hölder series for M .

31

Chapter 2

Proof. For (i), the successive quotients of submodules are all simple by the isomorphism theorems forA-modules. Part (ii) is a straightforward corollary of part (a).

Corollary 2.19. Let M a left A-module of finite length. Let M• be a Jordan-Hölder series of length n forM . If N1 ≤ N2 ≤ M are A-submodules such that N2/N1 is simple, there exists i ∈ 1, · · · , n such thatMi/Mi−1 ' N2/N1.

Proof. Refine the finite series 0 = N0 ⊆ N1 ⊆ N2 ⊆ N3 = M to a Jordan-Hölder series N• forM ; because N2/N1 is simple, N• can only be refined by adding extra submodules contained in N1

or containing N2. Since M• and N• are equivalent by Corollary 2.16, the simple quotient N2/N1 isisomorphic to one of the simple quotients Mi+1/Mi, as desired.

Corollary 2.20. Let M be a left A-module. Then M is of finite length if and only if it is artinian andnoetherian.

Proof. (⇒) Suppose we are given an increasing chain Mii∈N of submodules of M ; without loss ofgenerality, assume M0 = 0. It follows that for any n ≥ 1, the finite series (M0, · · · ,Mn,M) can berefined to a Jordan-Hölder series, so it can contain at most `A (M) strict inclusions. This implies that forn large enough, Mi = Mn for all i ≥ n, hence M is noetherian. The same argument proves that M isartinian by reversing the roles of M and 0 in this proof.

(⇐) Set M0def= 0. Since M is artinian, the set of its non-zero submodules admits a minimal element,

namely M1 ; the minimality implies that M1 is simple. The A-module M/M1 is also artinian, so bychoosing a minimal nonzero submodule of it, we obtain a submodule M1 ≤M2 ≤M such that M2/M1

is simple. Assuming (M0,M1, · · · ,Mn) have been constructed such that the successive quotients aresimple, we can construct Mn+1 as long as Mn 6= M . This recursive construction cannot continueindefinitely because M is noetherian. Therefore Mn = M eventually, which means we have found aJordan-Hölder series for M .

Proposition 2.21. (Existence of the Fitting decomposition) Let M be an A-module and ϕ ∈ EndA(M).

(i) If imϕ = im (ϕ2), then M = kerϕ+ imϕ.

(ii) If kerϕ = ker(ϕ2), we have kerϕ ∩ imϕ = 0.

(iii) If M is artinian, there exists n ≥ 0 such that M = ker(ϕn) + im (ϕn).

(iv) If M is noetherian, there exists n ≥ 0 such that ker(ϕn) ∩ im (ϕn) = 0.

(v) If M is of finite length, there exists n ≥ 0 such that M = ker(ϕn)⊕ im (ϕn).

The decomposition of a finite length A-module M in (v) is called its Fitting decomposition with respectto ϕ.

Proof. (i) Pick m ∈ M . There exists m′ ∈ M such that ϕ(m) = ϕ2(m′), which means m− ϕ(m′) ∈kerϕ. Therefore, m = (m− ϕ(m′)) + ϕ(m′) ∈ kerϕ+ imϕ.

(ii) If m ∈ kerϕ ∩ imϕ, pick m′ ∈ M such that m = ϕ(m′). This means m′ ∈ ker(ϕ2) = kerϕ,therefore m = ϕ(m′) = 0.

(iii) Consider the decreasing chain of submodules im (ϕn)n≥1. Since M is artinian, it stabilizes, sothere exists n ≥ 1 such that im (ϕn) = im ((ϕn)2). Therefore M = ker(ϕn) + im (ϕn) by part (i).

(iv) Consider the increasing chain of submodules ker(ϕn)n≥1. Since M is noetherian, it stabilizes,so there exists n ≥ 0 such that ker(ϕn) = ker((ϕn)2). Therefore ker(ϕn) ∩ im (ϕn) = 0.

32

(v) Combine parts (iii) and (iv).

Corollary 2.22. Let M be a left A-module and ϕ ∈ EndA(M).

(i) If M is artinian, ϕ is an isomorphism if and only if it is injective.

(ii) If M is noetherian, ϕ is an isomorphism if and only if it is surjective.

Proof. For (i), this follows from the expression M = ker(ϕn) + im (ϕn) = im (ϕn). For (ii), this followsfrom ker(ϕn) = ker(ϕn) ∩ im (ϕn) = 0.

Definition 2.23. A left A-module is called indecomposable if it is non-zero and does not admit non-trivialdirect summands, i.e. cannot be written as M = M1 ⊕M2 where M1,M2 are non-zero.

Theorem 2.24. Let M be a left A-module which is nonzero, indecomposable and of finite length. Theset of endomorphisms of EndA(M) which are not isomorphisms form a two-sided ideal contained inNil(EndA(M)) ; in particular, an endomorphism of M is either nilpotent or an isomorphism.

Proof. Consider ϕ ∈ EndA(M). The module M admits the Fitting decompositionM = ker(ϕn) ⊕ im (ϕn), but since M is indecomposable, one of the two summands have to bezero. If ker(ϕn) = 0, ϕ is an isomorphism ; if im (ϕn) = 0, ϕ is nilpotent.

If ϕ1, ϕ2 ∈ EndA(M) are such that ϕ1 ϕ2 is an isomorphism, then ϕ1 and ϕ2 are both isomorphisms(this follows trivially from looking at the set of units in EndA(M)). So if either one of ϕ1 or ϕ2 isnot an isomorphism, then their composition isn’t either, which gives the statement about the nilpotentendomorphisms of M forming a two-sided ideal in EndA(M) when it comes to multiplication. Foraddition, suppose ϕ1, ϕ2 nilpotent and ψ = ϕ1 + ϕ2 invertible, so that idM = ψ−1ϕ1 + ψ−1ϕ2. Choosen1, n2 ≥ 1 such that (ψ−1ϕi)

ni = 0. The endomorphisms ψ−1ϕ1 and ψ−1ϕ2 = idM −ψ−1ϕ1 commute,so since they are both nilpotent, by the binomial theorem,

idM = (ψ−1ϕ1 + ψ−1ϕ2)n1+n2 =

n1+n2∑j=0

(n1 + n2

j

)(ψ−1ϕ1)j(ψ−1ϕ2)n1+n2−j = 0,

a contradiction to M 6= 0.

Theorem 2.25. Let M be a left A-module of finite length.

(i) There exists a direct sum decomposition M =⊕n

i=1Mi where each Mi is indecomposable.

(ii) If M =⊕m

j=1M′j is another direct sum decomposition of M in indecomposable submodules, then

m = n and there exists a permutation σ ∈ Sn (the group of permutations of 1, · · · , n) such thatMσ(i) 'M ′i for 1 ≤ i ≤ n.

Proof. Part (i) follows by induction on `A (M) ; if M has length 1, it is simple, thus indecomposable and“M = M” is the only possible decomposition of M in indecomposable submodules. If `A (M) > 1 andM is indecomposable, we repeat the case where `A (M) = 1. If it is not, then M is a direct sum of twonon-zero submodules which have length strictly smaller than the length of M by Corollary 2.18 (i). Theinduction hypothesis completes the proof of this part.

33

Chapter 2

Onto part (ii). Letn⊕i=1

Mi = M =

m⊕j=1

M ′j

be two direct sum decompositions of M into indecomposables. Let πini=1 and π′jmj=1 be the twofamilies of projections corresponding to those decompositions. Without loss of generality, assumen ≤ m. We will prove by induction on 0 ≤ r ≤ n the following result : there exists an injectionσr : 1, · · · , r → 1, · · · ,m and an automorphism ϕr ∈ EndA(M) such that ϕr(M ′i) = Mσr(i) for all1 ≤ i ≤ r.

For r = 0, the statement is trivial. Suppose 0 < r ≤ n and assume the statement has been provenfor r − 1. This means there exists σr−1 : 1, · · · , r − 1 → 1, · · · ,m and an automorphismϕr−1 ∈ EndA(M) such that ϕr−1(M ′i) = Mσr−1(i). By replacing each M ′i by ϕr−1(M ′i) = Mσr−1(i), wecan assume without loss of generality that M ′i = Mi for 1 ≤ i ≤ r − 1.

The restriction to M ′r of the projection π′r =∑n

i=1 π′rπi is the identity of M ′r . Since M

′r is indecompos-

able, the endomorphisms π′rπi cannot all be nilpotent, otherwise idM ′r would be nilpotent, a contradiction.Pick k ∈ 1, · · · , n such that π′rπk|M ′r is an isomorphism. We must have k ≥ r since for k < r, we haveπk = π′kπk, which means π′rπk = π′rπ

′kπk = 0. We set

σr|1,··· ,r−1 = σr−1, σr(r)def= k, ϕr

def= idM − π′r + πkπ

′r.

If m ∈M satisfies ϕr(m) = 0, we have

0 = π′rϕr(m) = (π′r − π′2r + π′rπkπ′r)(m) = (π′rπk)(π

′r(m)) =⇒ π′r(m) = 0,

hencem = ϕr(m) + π′r(m)− (πkπ

′r)(m) = 0 + 0− πk(0) = 0,

which means ϕr is injective, hence an automorphism of M . Clearly, ϕr(m) = m when m ∈∑m

j=1j 6=r

M ′j

and ϕr(M ′r) ⊆Mk. This means that Mk is the direct sum of ϕr(M ′r) and Mk ∩

(∑mj=1j 6=r

M ′j

), but since

Mk is indecomposable and ϕr(M ′r) 6= 0, this means Mk = ϕr(M′r), which proves our claim.

Finally, the assumption n < m is contradictory since it implies

n⊕i=1

Mi = M =m⊕j=1

M ′j '

n⊕j=1

ϕ−1n (Mσn(j))

⊕ m⊕j=n+1

M ′j

,

and by computing lengths, we deduce that⊕m

j=n+1M′j has length zero.

2.2 Artinian and noetherian rings

Definition 2.26. Let A be a ring. We say that A is

(i) left-artinian (resp. left-noetherian) if the left A-module A` is left-artinian (resp. left-noetherian).Equivalently, A is left-artinian (resp. left-noetherian) if every descending (resp. ascending) chain ofleft ideals of A stabilizes.

(ii) right-artinian (resp. right-noetherian) if the right A-module A` is right-artinian (resp. right-noetherian).

34

Equivalently, A is right-artinian (resp. right-noetherian) if every descending (resp. ascending) chain ofright ideals of A stabilizes.

(iii) When A is such that A` (resp. Ar) is of finite length, the left-length (resp. right-length) of A is thelength of A` (resp. Ar).

When A is commutative, left-artinian and right-artinian is the same notion (similarly for noetherian), sowe call such rings artinian (resp. noetherian). If A is commutative and A`, Ar are of finite length, theleft-length equals the right-length, so we simply call it the length of A.

Example 2.27. (i) Let K be a field and A a K-algebra which is of finite dimension over K . Then A isleft-artinian, right-artinian, left-noetherian and right-noetherian since left (resp. right) ideals of A areK-vector subspaces of A, so finite series have length bounded by dimK A.

(ii) A principal ideal domain is a commutative noetherian ring since every ideal is principal, hence finitelygenerated. For example, the ring of integers Z is noetherian. However, Z is not artinian since for any

n ≥ 1, the ideal adef= nZ is such that the chain of ideals ann≥1 is strictly increasing.

(iii) Let Mii∈I be an infinite family of non-zero left A-modules and Mdef=⊕

i∈IMi. Then EndA(M)is neither left-artinian or left-noetherian. To see this, for each i ∈ I , let ai E EndA(M) be the leftideal defined by

aidef= ϕ ∈ EndA(M) | ϕ(Mi) = 0.

By looking at the projections πi of M to Mi, we see that the collection of left ideals aii∈I are in adirect sum inside of EndA(M), which gives the claim (c.f. Example 2.2 (ii)).

(iv) We will see later (c.f. Corollary 4.37) that a left-artinian ring is left-noetherian. However, there existsleft-artinian modules which are not left-noetherian ; in fact, this is true even in the commutative case,i.e. there exists an abelian group (in other words, a Z-module) which is artinian but not noetherian.A classical example is that of the Z-subalgebra of Q given by Z[1

p ] where p ∈ Z is a prime number,which we then mod out by Z :

Z[1p ]/Z =

apn + Z | a ∈ Z, n ≥ 1, (a, pn) = 1

.

To see this, it suffices to see that the only proper subgroups of Z[1p ]/Z are those generated by 1

pn +Z for

some n ≥ 1. Write Gndef=⟨

1pn

⟩Z. Note that by definition, Z[1

p ] =⋃n≥1Gn. If H ≤ Z[1

p ] is a proper

subgroup and x ∈ Gn ∩H for some n ≥ 1, then x = apn where (a, pn) = 1„ so there exists b, c ∈ Z

such that ab + pnc = 1, which means that bx ≡ 1pn (mod Z), hence bx ∈ H . Therefore Gn ⊆ H .

Letting nH be the largest integer satisfying GnH ⊆ H , we see by this argument that H = GnH . Thisalso implies that Z[1

p ] is not finitely generated, hence not noetherian ; for if a1pn1 , · · · ,

akpnk is a finite

subset, their Z-span is contained in⋃ki=1Gni , a proper subgroup. However, any descending chain of

subgroups of Z[1p ]/Z corresponds to a descending sequence of non-negative integers, thus stabilizes,

which means Z[1p ]/Z is artinian.

(v) A finite product of artinian (resp. noetherian) rings is artinian (resp. noetherian). To see this, pick

A1, · · · , An artinian (resp. noetherian) rings, so that Adef= A1 × · · · × An is a left A-module. A

left ideal a E A has to be of the form a1 × · · · × an where ai E Ai by considering the idempotents(0, · · · , 0, 1, 0, · · · , 0) ∈ A (with a 1 at the ith coordinate). Therefore, a decreasing (resp. increasing)sequence eventually stabilizes because it stabilizes in each component. More generally, if Mi is anartinian (resp. noetherian) Ai-module for i = 1, · · · , n, then M1 × · · · ×Mn is an artinian (resp.noetherian) A1 × · · · × An-module by the same argument. Furthermore, any quotient of an artinian(resp. noetherian) ring by a two-sided ideal is again an artinian (resp. noetherian) ring.

35

Chapter 2

(vi) Subrings of artinian (resp. noetherian) rings are not necessarily artinian (resp. noetherian). The

best example would be the polynomial ring in infinitely many variables over a field K , say Adef=

K[xnn∈N], seen as a subring of its quotient field (which is artinian and noetherian). Consideringthe sequence of ideals

andef= (x1)nA , bn

def= (x1, · · · , xn)A

we see that an is an infinite strictly descending chain and bn is an infinite strictly ascending chain, soA is neither artinian nor noetherian.

Proposition 2.28. A finitely generated left A-module over an artinian (resp. noetherian) ring is artinian(resp. noetherian).

Proof. This follows from Proposition 2.5 since M is the quotient of the left A-module A⊕n for somen ≥ 1.

Corollary 2.29. Let A be a left-artinian ring and B be a left-finite A-algebra (i.e. an A-algebra which makesB a finitely generated left A-module). Then B is a left-artinian ring. (This generalizes Example 2.27 (i).)

Proof. If bnn∈N is a descending chain of left ideals in B, it is a descending chain of A-submodules ofB`, which is artinian. Therefore, this chain stabilizes, i.e. B is left-artinian.

Proposition 2.30. LetM be a faithful left A-module and mii∈I be a family of generators ofM/. The mapA→M⊕I given by a 7→ (ami)i∈I is injective and A-linear, hence A` can be seen as a left A-submodule ofM⊕I .

Proof. It is obvious that this map is a morphism of left A-modules. Suppose ami = 0 for all i ∈ I . Ifϕii∈I ⊆ EndA(M) and m =

∑∗i∈I ϕi(mi) ∈M , then

a

( ∗∑i∈I

ϕi(mi)

)=

∗∑i∈I

ϕi(ami) = 0,

which means a ∈ AnnA(M) = 0 since EndA(M)〈mii∈I〉 = M/, henceM is a faithful left A-module.

Proposition 2.31. Let M be an artinian (resp. noetherian) left A-module. If M/ is finitely generated, thering AM is artinian (resp. noetherian).

Proof. The left A-module M is also a faithful left A/AnnA(M)-module which is left-artinian (resp. left-noetherian) and AM ' A/AnnA(M), so without loss of generality, assumeM is a faithful left A-module.By Proposition 2.30, A` is isomorphic to an A-submodule of M⊕n for some n ≥ 1, which means it is aleft-artinian (resp. left-noetherian) A-module, i.e. A is a left-artinian (resp. left-noetherian) ring.

Corollary 2.32. Assume A is a commutative ring and let M be a noetherian A-module. Then AM isnoetherian.

Proof. Since M is finitely generated, so is M/ since AM ⊆ EndA(M) by the commutativity assumption.The result follows from Proposition 2.31.

36

2.3 Simple modules

Proposition 2.33. LetM be a nonzero left A-module. ThenM is simple if and only if for all m ∈M \0,Am = M . In particular, a simple module is a cyclic module, which means it is generated by a singleelement.

Proof. This follows from the fact that Am is a nonzero submodule of M for any m ∈M \ 0.

Proposition 2.34. Let A be a ring and a E A be a left ideal of A. The left A-module A`/a is cyclic, andit is simple if and only if a is a left maximal ideal of A, which means that if b is another left ideal of Asatisfying a ⊆ b ⊆ A`, then b = a or b = A`.

Proof. The left A-module A`/a is generated by 1 + a, thus is cyclic. The left A-submodules of A/a arein one-to-one correspondence with the left ideals of A containing a, thus the second statement holds.

Theorem 2.35. Let M be a simple left A-module. Given m ∈ M \ 0, the map A` → M given bya 7→ am induces an isomorphism A`/m 'M where m is a left maximal ideal of M . There is a one-to-onecorrespondence between the set of left maximal ideals of A and the set of isomorphism classes of simpleleft A-modules given by m 7→ [A`/m] (the brackets stand for the equivalence class of the left A-module).

Proof. If two simple left A-modules M1,M2 are isomorphic, they have the same annihilator, hence areboth isomorphic to A`/m where m = AnnA(M1) = AnnA(M2) is a left maximal ideal. Conversely,A`/m is simple for any left maximal ideal m, which gives the result.

Example 2.36. (i) Let A be a ring. The simple submodules of A` are the nonzero left ideals of A whichare minimal with respect to inclusion in the set of nonzero left ideals of A ; we call them the minimalleft ideals of A. An obvious example of when A has minimal left ideals is when A is left-artinian, inwhich case we can use the descending chain condition and Zorn’s lemma to prove the existence of aminimal left ideal.

Such ideals do not always exist, as shown by the case of a (commutative) local ring (A,m) whichis also an integral domain since if a E A and A is not a field (in the case of a field, there are nonon-zero ideals), we have 0 6= ma ( a (am 6= 0 because A is a domain, and a = ma implies a = 0 byNakayama’s Lemma). Another trivial example is when A = Z since 0 6= a E Z implies 0 6= a2 ( a.In fact, any UFD or Dedekind domain also has no minimal ideals as a module over itself for the samereason.

(ii) Let K be a field, V a K-vector space, and A ⊆ EndK(V ) be the subring generated as a vectorspace by idV and the K-endomorphisms of V which are of finite rank. It follows that V is a simpleA-module since if W ≤ V is an A-submodule, then W contains any finite-dimensional subspace ofV , thus is equal to V . In particular, V is simple as an EndK(V )-module, which is equal to V / whenwe see V as a K-module.

Let µ ∈ HomK(V,K) be a non-zero linear form on V . For v ∈ V , denote by ϕv ∈ A the endomor-phism given by ϕv(w) = µ(w)v. It is clear that for any ψ ∈ EndK(V ), we have

(ψ ϕv)(w) = ψ(µ(w)v) = µ(w)ψ(v) = ϕψ(v)(w) =⇒ ψϕv = ϕψ(v)

hence the map ϕ : V → EndK(V ) sending v to ϕv is a left EndK(V )-linear map ; if we restrict itscodomain to A, it is also a left A-linear map. Since its kernel is a proper left EndK(V )-submodule of

37

Chapter 2

V , it is zero (which can also easily be seen directly since µ 6= 0), which means V maps isomorphicallyonto a minimal left ideal of EndK(V ).

(iii) Let A be a ring. For A` to be a simple A-module, it is necessary and sufficient that 0 be the onlyproper left ideal of A, meaning that A is a division ring. In this case, any simple left A-module isisomorphic to A`. In particular, the left A-module A` is simple if and only if the right A-module Aris simple.

(iv) Let A be a principal ideal domain. The simple A-modules are of the form A/p where p = (p)A ∈ Ais generated by a prime element of A. For n > 1, the A-modules A/pn are indecomposable, but notsimple ; to see this, assume (A, p) is local (after localization at p), so that picking x ∈ pk \ pk+1 for1 ≤ k < n, we obtain (x)A ' pk/pn ' A/pn−k 6' A/pn. In particular, if A/pn ' a ⊕ b where a, bare A-submodules (i.e. ideals of A), then a and b are annihilated by pn−1, a contradiction.

Definition 2.37. Let M be a left A-module. A maximal submodule of M is a left A-submodule N of Msuch that M/N is simple. In other words, N is maximal in the set of proper submodules of M partiallyordered under inclusion.

Remark 2.38. When M is a noetherian left A-module, maximal submodules always exist. We generalizethis in the next proposition.

Proposition 2.39. Let M be a left A-module and L ≤ M a proper submodule. If M is finitely generated,then there exists a maximal submodule of M containing L.

Proof. It suffices to see that we can apply Zorn’s lemma, i.e. that every chain has an upper bound. If

L ⊆ N1 ⊆ · · · ⊆ Nr ⊆ · · · is a chain of proper submodules of M , then Ndef=⋃r≥1Nr is a proper

submodule of M . Otherwise, each generator lies in Nr for some r ≥ 1, meaning that M is contained inNr, a contradiction.

Corollary 2.40. LetM be a finitely generated left A-module. There exists a simple left A-moduleM ′ such

that adef= AnnA(M ′) satisfies aM 6= M .

Proof. Let N be a maximal submodule of M and let adef= AnnA(M/N). Then aM ⊆ N (M .

Proposition 2.41. Let A be a ring, b a minimal left ideal of A (c.f. Example 2.36) and M a simple faithfulleft A-module. Then M ' b as A-modules. In particular, all minimal left ideals of a ring A are isomorphicas A-modules.

Proof. Let b ∈ b \ 0. Since M is a faithful A-module, there exists m ∈ M \ 0 such that bm 6= 0.

Define mdef= AnnA(m). Since M is simple and b /∈ m, we have m + b = A. Since b is a minimal ideal,

b ∩ m = 0 ; otherwise, b ∩ m = b implies b ⊆ m = AnnA(m), which is impossible since bm 6= 0.Therefore A` ' m ∩ b, hence A/m ' b.

Theorem 2.42. Let M be a faithful left A-module admitting a Jordan-Hölder series (M0, · · · ,Mn) (c.f.Definition 2.14). IfM/ is finitely generated, every simple left A-module is isomorphic to one of the successivequotients Mi+1/Mi for some 0 ≤ i < n.

Proof. Every simple left A-module is a quotient of the left A-module A`. We have already seen that A`is isomorphic to a submodule of M⊕n in Proposition 2.30 because M is assumed faithful. It follows thatevery simple left A-module is isomorphic to one of the successive quotients of a Jordan-Hölder series forM⊕n, but we can construct a Jordan-Hölder series for M⊕n using the Jordan-Hölder series for M (c.f.

38

Corollary 2.18 (ii)), so that the set of isomorphism classes of simple A-modules coming from successivequotients of M equals that of M⊕n, completing the proof.

2.4 Completely reducible modules

Theorem 2.43. Let M be a left A-module which is the sum of a family Mii∈I of simple A-submodules.For every A-submodule L ≤M , there exists a subset J ⊆ I such that M = L⊕

⊕j∈JMj .

Proof. Let Ω be the set of all J ⊆ I such that the modules Mjj∈J and L are in a direct sum. Weclaim that Zorn’s Lemma applies to Ω ⊆ P(I) since the condition of being in a direct sum only concernsfinitely many elements of J at a time, hence if Jαα∈Λ is a family of subsets of Ω which are totallyordered under inclusion, the union

⋃α∈Λ Jα also lies in Ω. So assume J is a maximal element of Ω. We

claim that M = L ⊕⊕

j∈JMj . Let Ndef= L ⊕

⊕j∈JMj ( M . Since J is maximal in Ω, if Mj′ is a

simple submodule ofM , then N ∩Mj′ 6= 0, henceMj′ ∩N = Mj′ becauseMj′ is simple. This impliesMj′ ⊆ N ; since Mj′ was arbitrary, this implies M ⊆ N , hence M = N .

Theorem 2.44. Let M be a left A-module. The following are equivalent :

(i) M is the sum of a family of simple A-submodules

(ii) M is the direct sum of a family of simple A-submodules

(iii) Every A-submodule of M is a direct summand.

A left A-module M satisfying one of the equivalent conditions above is called completely reducible (orsemisimple in most of the literature).

Proof. ( (i)⇒ (ii) ) Choose L = 0 in Theorem 2.43.

( (ii) ⇒ (iii) ) This follows by letting the submodule in (iii) be the submodule L in Theorem 2.43.

( (iii) ⇒ (i) ) Let L be a cyclic submodule of M . By Proposition 2.39, there exists a maximal submoduleL′ of L, so that L/L′ is simple. Let N ′ ≤ M be such that L′ ⊕ N ′ = M . The submodule L ∩ N ′ isnon-zero, otherwise L⊕N ′ = M , which means that L/L′ ' M/M = 0, a contradiction. It follows thatL∩N ′ is a non-zero submodule of L satisfying (L∩N ′)∩L′ = 0 and (L∩N ′) +L′ = L by maximalityof L′ in L, hence L = L′ ⊕ (L ∩N ′) and L/L′ ' L ∩N ′ is simple. Therefore, any non-zero submoduleof M contains a simple submodule.

Consider the sum S of all simple submodules of M and let P be a submodule such that M = S ⊕ P .If P is nonzero, it contains a simple submodule of M , so that P ∩ S 6= 0, a contradiction. ThereforeP = 0, which completes the proof.

Remark 2.45. A left A-module M is completely reducible if and only if it is a completely reducible AM -module. By Theorem 2.44, any left A-module M which is the sum of completely reducible submodules isitself completely reducible.

Example 2.46. (i) Simple left A-modules are completely reducible.

(ii) If A is a field, every cyclic A-module is simple, so every A-module is semisimple. This is the fact thatevery vector space admits a basis.

39

Chapter 2

(iii) The Z-module Z is not semisimple since simple Z-modules are of the form Z/pZ where p is a primenumber and the only finite subgroup of Z is 0.

Corollary 2.47. Let M be a left A-module which is the sum of the simple submodules Mii∈I . For anyA-submodule N , there exists J ⊆ I such that N '

∑j∈JMj and M/N '

∑j∈I\JMj .

Proof. By Theorem 2.43, we see that there exists J ⊆ I such that M = N ⊕∑

j∈JMj , so that N 'M/

(∑j∈JMj

)is the quotient of M by one of its submodules. Therefore, it suffices to deal with the

case of the quotient. But by the same expression, M/N '∑

j∈JMj , so we are done.

Corollary 2.48. If M is a completely reducible left A-module and N ≤ M is a submodule, then N andM/N are completely reducible.

Proof. Since M is completely reducible, it is the sum of its simple submodules. The result follows byCorollary 2.47 combined with Theorem 2.44.

Corollary 2.49. Let M be a left A-module which is the sum of the family of simple submodules Mii∈I .Then any simple submodule of M is isomorphic to one of the Mi’s.

Proof. If N ≤M , it is isomorphic to the sum of submodules of the formMj where j ∈ J ⊆ I for a givensubset J . For any j ∈ J , we have 0 Mj ≤

∑i∈JMi ' N , which means Mj = N when N is simple.

2.5 Isotypical components of completely reducible modules

Theorem 2.50. Let M be a completely reducible left A-module. If N0 is a simple A-submodule of M , let

Nii∈I be the set of simple submodules of M isomorphic to N0. Their sum N0 def=∑

i∈I Ni is isotypicalof type N0, so is a completely reducible submodule of M ; it is uniquely characterized as the sum of allsimple A-submodules of M isomorphic to N0, so we call it the isotypical component of M of type N0.When Pii∈I is a family of simple submodules of M satisfying

(i) Each simple A-submodule of M is isomorphic to one of the Pi

(ii) No two distinct Pi are isomorphic

then letting

P 0i

def=∑N≤MN'Pi

N

gives the isotypical component P 0i of M of type Pi, and M =

⊕i∈I P

0i is the direct sum of its isotypical

components.

Proof. It is clear that M is the sum of its isotypical components and that its isotypical components areindeed isotypical by Theorem 2.44. The isotypical components are in a direct sum, for if i ∈ I , a non-zero element in P 0

i ∩∑

j∈I\i P0j implies that both submodules intersected contain a simple submodule

isomorphic to Pi, but the sum only contains simple submodules isomorphic to Pj for j 6= i by assumption(i) and Corollary 2.47, a contradiction by assumption (ii).

Corollary 2.51. Let M be a completely reducible left A-module with isotypical components Mii∈I , each

40

Mi being isotypical of type Pi. If N ≤ M is a submodule, then the isotypical component of N of type Pi,denoted by Ni, is equal to N ∩Mi. Furthermore, N '

⊕i∈I N ∩Mi.

Proof. We already know that N is completely reducible, hence it is the direct sum of its isotypicalcomponents. Since a simple submodule of N is a simple submodule of M , the isotypical component Ni

of N of type Pi must be contained in Mi, which means N =⊕

i∈I Ni =⊕

i∈I N ∩Mi.

Remark 2.52. If M is an arbitrary left A-module, we can still define the isotypical components as the sumof the simple A-submodules of a given isomorphism type and the sum

∑i∈IMi will still be direct. The

equality M =⊕

i∈IMi will hold precisely when M is completely reducible.

Proposition 2.53. Let M =⊕

i∈IMi and N =⊕

i∈I Ni be two completely reducible left A-moduleswritten in their decompositions in isotypical components of type Pi and let f : M → N be a morphism ofA-modules. Then f(Mi) ⊆ Ni. In particular,

(i) We have HomA(M,N) '∏i∈I HomA(Mi, Ni) via the map f 7→ (f |Mi)i∈I .

(ii) If M = N , we have EndA(M) =∏i∈I EndA(Mi).

Proof. This follows from the fact that f(Mi) is a quotient of Mi, thus also isotypical of type Pi (c.f.Corollary 2.47) ; this implies that f(Mi) ⊆ Ni since Ni is the sum of all simple submodules of N ofisomorphism type Pi, and f(Mi) is such a sum. The parts (i) and (ii) are trivial corollaries.

Corollary 2.54. Let M be a completely reducible left A-module and N ≤ M an A-submodule. Thefollowing are equivalent :

(i) N ≤M/ is a submodule, which means it is an EndA(M)-submodule of M

(ii) N is the (direct) sum of isotypical components of M .

Proof. ( (ii) ⇒ (i) ) By Proposition 2.53, if f ∈ EndA(M), then f |N : N → M maps each isotypicalcomponent into itself, which means N is stable under EndA(M).

( (i) ⇒ (ii) ) Let N1, N2 ≤ M be two simple A-submodules of M which are isomorphic. Since N1

is a direct summand of M , fix a complement N ′1 so that M = N1 ⊕ N ′1 and let π1 : M → M be

the corresponding projection. If ϕ : N1 → N2 is an isomorphism, then ψ12def= ϕ π1 ∈ EndA(M)

and ψ12(N1) = N2. Therefore, if N is stable under A-endomorphisms of M and contains one simplesubmodule of isomorphism type Pi, it contains the entire isotypical component Mi of M . Since N =⊕

i∈I N ∩Mi, this means either N ∩Mi = 0 or N ∩Mi = Mi, as desired.

Proposition 2.55. Let M be a completely reducible left A-module with Nii∈I and N ′jj∈J two familiesof simple submodules of M such that

⊕i∈I Ni = M =

⊕j∈J N

′j . Then |I| = |J |.

Proof. By writing M as a direct sum of its isotypical components, we can assume without loss ofgenerality that M is isotypical. If I is finite, then the A-length of M satisfies `A (M) = |I|, whence J isalso finite since any finite collection J ′ ⊆ J satisfies |J ′| ≤ |I| from the fact that

∑j∈J ′ N

′j ≤ M , which

implies |J | ≤ |I|. It follows that |J | = `A (M) = |I|.

Suppose I and J infinite. For each i ∈ I , fix mi ∈ Ni \0 so that Ni = A〈mi〉. Let Ji ⊆ J be defined asfollows ; write mi =

∑∗j∈J m

′ij with m

′ij ∈ N ′j and let Ji be the set of those j ∈ J for which m′ij 6= 0. We

have J =⋃i∈I Ji because otherwise, there exists j ∈ J such that m′ij = 0 for all i ∈ I , meaning that N ′j

41

Chapter 2

is not necessary in the sum∑

j∈J N′j to obtain M , a contradiction to the fact that our sums are direct.

This means that |J | ≤ |I × N| = |I||N| = |I| because I is infinite, and by symmetry of the hypotheses,we obtain |I| = |J |.

Definition 2.56. Let M be a completely reducible left A-module and P a simple A-module. We denotethe isotypical component of M of type P by MP (this includes the case where MP = 0, i.e. when M hasno submodule isomorphic to P ). We generalize the notion of left-length of an completely reducible moduleto include the case where M might not be a module of finite length in the sense of Jordan-Hölder series :the length of M , denoted by `A (M), is the cardinality of I where M =

⊕i∈I Ni is a decomposition of M

as a direct sum of simple A-modules. If M has isotypical component MP , we denote the length of MP by[M : P ]A. Therefore, if J(M) denotes an index set containing an isomorphism class representative of eachsimple left A-submodule of a moduleM , then `A (M) =

∑P∈J(M)[M : P ]A. In particular, `A (M) <∞ if

and only if it admits a Jordan-Hölder series, in which case M is a direct sum of simple left A-modules.

If M is a left A-module, we say that M is isotypically simple if M is isotypical of type P for somesimple left A-module P (whose isomorphism type is then uniquely determined). The isotypical componentsof a completely reducible left A-module are therefore its maximal isotypically simple submodules.

Corollary 2.57. Let M,N be two completely reducible left A-modules. Then M and N are isomorphic ifand only if for any simple left A-module P , we have [M : P ]A = [N : P ]A.

Proof. Decompose M and N into isotypical components. By Proposition 2.53, assume without loss ofgenerality that M and N are isotypical of type P . Then M ' P⊕I and N ' P⊕J , so the result followsfrom Proposition 2.55.

Corollary 2.58. Let M be a completely reducible left A-module. Then `A (M) <∞ if and only if M is afinitely generated A-module.

Proof. Let P be a simple left A-module. The module M is the direct sum of its isotypical componentsand the length of M is the sum of the [M : P ]A where P is simple, hence without loss of generality,assume M is isotypical of type P . Then AnnA(M) = m is a left maximal ideal where A/m ' P andM is finitely generated if and only if we have a surjective morphism of A-modules A⊕r` → M , which bythe latter is equivalent to a surjective morphism (A`/m)⊕r → M . This implies `A (M) ≤ r < ∞, andconversely, `A (M) <∞ implies M ' (A`/m)⊕ `A(M) is finitely generated.

Another way to prove this is as follows : write M =⊕

i∈IMi as a direct sum decomposition into simpleA-submodules. If I is finite, since each Mi is cyclic, M is finitely generated. Conversely, if M is finitelygenerated, a finite subset of M can only span a finite amount of the Mi’s, thus I must be finite.

2.6 Centralizer and bicentralizer of a completely reducible module

Proposition 2.59. Let M be a completely reducible left A-module.

(i) The module M is simple (resp. isotypically simple) if and only if it is a simple (resp. isotypicallysimple) C2

A(M)-module.

(ii) The A-submodules of M are precisely the C2A(M)-submodules of M//.

(iii) If M is an isotypically simple left A-module of type P where P is a simple A-submodule of M , thenthe restriction map ϕ 7→ ϕ|P induces an isomorphism C2

A(M) ' C2A(P ).

(iv) Let MP P∈I be the isotypical components of M where I indexes a family of simple submodules of

42

M as in Theorem 2.50. The map

C2A(M)→

∏P∈I

C2A(MP ), ϕ 7→ (ϕ|MP

)P∈I

is an isomorphism.

Proof. Since AM ⊆ C2A(M), we already know that a C2

A(M)-submodule of M is an A-submodule.Conversely, if N ≤ M is an A-submodule, then N is a direct summand, which means it is a C2

A(M)-submodule by Proposition 1.31. Therefore, A-submodules and C2

A(M)-submodules agree, so the notionsof simple (resp. isotypically simple) agree in both cases (namely, over A and over C2

A(M)), which provespart (i) and (ii). Part (iii) follows from Proposition 1.56. It remains to prove part (iv).

It is clear that the given map is injective since each MP is a C2A(M)-submodule, so that ϕ =

⊕P∈I ϕ|P .

For surjectivity, let ϕP ∈ C2A(MP ). There exists a unique ϕ ∈ EndZ(M) such that ϕ|P = ϕP , so it

suffices to show that ϕ ∈ C2A(M). For each ψ ∈ CA(M) = EndA(M), recalling Proposition 2.53, we see

that ψ(MP ) ⊆MP , which implies that ψ|MP∈ CA(MP ). Since ϕP ∈ C2

A(MP ), it commutes with ψ|MP,

hence ϕ commutes with ψ, which completes the proof.

Remark 2.60. Equip M with the discrete topology and EndZ(M) with the topology of pointwise conver-gence (or equivalently, the subspace topology coming from the inclusion EndZ(M) ⊆MM = HomSet(M,M)where each copy ofM has the discrete topology). The topology we just put EndZ(M) makes it into a Haus-dorff topological ring, meaning that addition and multiplication (given by composition of endomorphisms)is continuous ; this is obvious by using the definition of continuity via convergent sequences, and the Haus-dorff property is obtained as follows : given ϕ,ψ ∈ EndZ(M) distinct, they differ at at least one m ∈ M(unless M = 0, in which case EndZ(M) = 0), so since M is Hausdorff, we can pick two neighborhoods Uϕof ϕ(m) and Uψ of ψ(m). We then set

Vϕdef= ϕ′ ∈ EndZ(M) | ϕ′(m) ∈ Uϕ

and define Vψ similarly. The Vϕ and Vψ are disjoint neighborhoods of ϕ and ψ in EndZ(M).

Furthermore, if S ⊆ EndZ(M), the centralizer CEndZ(M)(S) (c.f. Definition 1.20) is closed in EndZ(M),for if ϕn ∈ CEndZ(M)(S) converges to ϕ ∈ EndZ(M) and ψ ∈ S, we see that for all m ∈ M , there exists alarge enough integer nm such that

ϕ(ψ(m)) = ϕnm(ψ(m)) = ψ(ϕnm(m)) = ψ(ϕ(m)) =⇒ ϕ ∈ CEndZ(M)(S).

It follows that CA(M) and C2A(M) are closed subrings of EndZ(M), and we will now prove that AM is

dense in C2A(M) when M is completely reducible ; this is called the density theorem. For two subsets

S1, S2 ⊆ EndZ(M), the equality S1 = S2 expresses the following : pick ϕ2 ∈ S2 arbitrary. A neighborhoodbase of ϕ2 can be taken as

Um1,··· ,mn

def= ψ ∈ EndZ(M) | ∀1 ≤ j ≤ n, ψ(mj) = ϕ2(mj) =

n⋂j=1

Umj .

Then S1 = S2 if and only if for every m1, · · · ,mn ∈ M and ϕ2 ∈ S2, there exists ϕ1 ∈ S1 such thatϕ1(mj) = ϕ2(mj) for 1 ≤ j ≤ n. The equality AM = C2

A(M) now boils down to the statement of thedensity theorem given below.

Lemma 2.61. Let M be a completely reducible left A-module, ψ ∈ C2A(M) and m ∈ M . There exists

a ∈ A such that ψ(m) = am.

43

Chapter 2

Proof. This is because the A-submodule A〈m〉 is also a C2A(M)-submodule by Proposition 2.59, which

means that ψ(m) ∈ A〈m〉 can be written in the form am for some a ∈ A.

Theorem 2.62. (Density theorem) Let M be a completely reducible left A-module, ψ ∈ C2A(M) and

m1, · · · ,mn ∈M . There exists a ∈ A such that ψ(mj) = amj for all 1 ≤ j ≤ n.

Proof. Consider the bicentralizer of M⊕n. If P ≤ M is a simple submodule, then we have a canonicalisomorphism (M⊕n)P ' (MP )⊕n, which induces a canonical isomorphism

C2A(M) '

∏P

C2A(P ) ' C2

A(M⊕n)

by Proposition 2.59, which is essentially given by ψ 7→ ψ⊕ndef= (ψ, · · · , ψ). Applying Lemma 2.61 to

M⊕n, ψ⊕n and (m1, · · · ,mn), we see that there exists a ∈ A such that ψ(mj) = amj , as desired.

Corollary 2.63. Let M be a completely reducible left A-module. When M/ is a finitely generatedEndA(M)-module, we have the equality AM = C2

A(M).

Proof. Let m1, · · · ,mn be a set of generators for M/. By the density theorem, for each ϕ ∈ C2A(M),

there exists a ∈ A such that ϕ|m1,··· ,mn = aM |m1,··· ,mn. Since both ϕ and aM are endomorphisms ofM/ by definition of CA(M) and C2

A(M), this means that ϕ = aM , which implies AM = C2A(M).

Corollary 2.64. LetM1, · · · ,Mn be pairwise non-isomorphic simple left A-modules such that eachMi/ is

finitely generated. If a1, · · · , an ∈ A, there exists a ∈ A such that aMi = (ai)Mi for 1 ≤ i ≤ n.

Proof. It suffices to see that the isotypical components of Mdef=⊕n

i=1Mi are the Mi’s and that M/ isfinitely generated (because EndA(M) '

∏ni=1 EndA(Mi) by Proposition 2.53), which implies

AM = C2A(M) '

n∏i=1

C2A(Mi) =

n∏i=1

AMi

by Proposition 2.59 and the previous corollary.

2.7 Centralizer of a simple module

Theorem 2.65. (Schur’s Lemma) Let M,N be two left A-modules and f : M → N be a morphism notidentically zero.

(i) If M is simple, f is injective.

(ii) If N is simple, f is surjective.

(iii) If both M and N are simple, f is an isomorphism.

As a result, when M is a simple A-module, EndA(M) is a skew field.

Proof. The first three parts follow from the fact that ker f ≤M and im f ≤ N . Setting M = N , we seethat every non-zero endomorphism ofM is an isomorphism, meaning that for each f ∈ EndA(M)\0,f−1 ∈ EndA(M) satisfies ff−1 = f−1f = idM , completing the proof.

44

It follows that whenM is simple,M/ is a left CA(M)-vector space. Note that C2A(M) = CCA(M)(M

/) =EndCA(M)(M

/) is precisely the set of CA(M)-linear endomorphisms of this vector space.

Lemma 2.66. Let K be an algebraically closed field and D be a skew field which is also a K-algebra offinite dimension. Then D = K .

Proof. Let x ∈ D. Denote by K(x) the smallest skew field of D generated by K and x, or in otherwords, the set of all y ∈ D which can be written as a product of polynomials in x and their inverses. ThisK-algebra is by definition a skew field which is commutative, and therefore is a finite field extension ofK since D is finite over K . The fact that K is algebraically closed implies that every algebraic extensionof K is equal to K , therefore K(x) = K , meaning that x ∈ K .

Corollary 2.67. (Burnside’s Theorem) LetK be an algebraically closed field, A aK-algebra andM a simpleleft A-module which is finite-dimensional over K (via its A-module structure). Then AM = C2

A(M) =EndK(M) and CA(M) = KM = K is the ring of homotheties of M seen as a K-module.

Proof. It suffices to see that CA(M) = KM since the fact that K is a field implies KM = K (becauseM 6= 0 and AnnA(M) E K) and the other statements in the theorem follow from Corollary 2.63. Theinclusions

KM ⊆ CA(M) = EndA(M) ⊆ EndK(M)

finish the proof by Lemma 2.66 since EndA(M) is a skew field and a K-algebra of finite rank (becausedimK EndK(M) = (dimKM)2 <∞ by assumption on M ).

Corollary 2.68. Let K be an algebraically closed field, A a K-algebra and M1, · · · ,Mn be pairwise non-isomorphic simple left A-modules which are finite-dimensional overK (via their A-module structure). Givenci ∈ EndK(Mi) for 1 ≤ i ≤ n, there exists a ∈ A such that aMi = ci for all 1 ≤ i ≤ n. In other words, themap A→

∏ni=1 EndK(Mi) given by restriction (a 7→ (aM1 , · · · , aMn)) is surjective.

Proof. By Proposition 2.53, we know that Mdef=⊕n

i=1Mi is finitely generated over

CA(M) =

n∏i=1

CA(Mi) =

n∏i=1

KMi = K⊕n

because it is finite-dimensional over K ; the result follows from Corollary 2.64.

45

Chapter 3

Simple and semisimple rings

In this chapter, A denotes a ring and D denotes a division ring.

3.1 Left & right vector spaces over a skew field

We treat the particular case of D-modules when D is a skew field (the letter D stands for “division ring”).

Definition 3.1. Let D be a division ring. A left D-vector space (resp. right D-vector space) is a left D-module (resp. right D-module). In this section, we denote a left D-vector space by the letter V . Submodulesare called vector subspaces and elements of V are called vectors.

A basis for a left D-vector space V is a subset S ⊆ V such that V =⊕

s∈S D〈s〉. A subset L ⊆ Vis called linearly independent if it is a basis of D〈L〉. A subset S ⊆ V is said to span V or to be agenerating subset of V if D〈S〉 = V .

Lemma 3.2. Every cyclic left D-vector space is simple and there is only one isomorphism class of simpleleft D-vector spaces ; they are all isomorphic to D`.

Proof. Let V be a cyclic D-module and v ∈ V be a generator. If 0 W ≤ V is a non-zero vectorsubspace, by hypothesis we can write an element w ∈ W \ 0 in the form dv for some d ∈ D. Thenv = d−1(dv) = d−1w ∈W , meaning that W ⊇ V , i.e. W = V .

If V is simple, then the map ϕv : D` → V defined by ϕv(d) = dv is left D-linear and non-zero whenv ∈ V \ 0 because ϕv(d′d) = d′dv = d′ϕv(d), meaning that it is an isomorphism D` ' V .

Theorem 3.3. Every left D-vector space V has a basis. More precisely, if L ⊆ S ⊆ V are two subsetswhere L is linearly independent (i.e. L is a basis of the vector space D〈L〉) and S is a generating subset ofV , there exists a basis B for V such that L ⊆ B ⊆ S.

Proof. Consider the vector subspaceWdef= D〈L〉 and the family of simple D-submodules D〈s〉s∈S . By

Lemma 3.2, we see that V is the sum of its simple submodules, hence is semisimple by Theorem 2.44.By Theorem 2.43 applied to the family D〈s〉s∈S and the vector subspace W , we see that there exists a

subset L′ ⊆ S such that V = W ⊕⊕

`′∈L′ D〈`′〉. This means that Bdef= L ∪ L′ is a basis for V since L

is a basis for W . By definition, L ⊆ B ⊆ S, so we are done. To prove that V admits a basis, set L = ∅and S = V .

46

Definition 3.4. Let V be a left D-vector subspace. Since V is isotypically simple of type D`, the dimension

of V is defined as its length as a left D-module (c.f. Definition 2.56), namely dimD Vdef= `D (V ). The

dimension of V equals the cardinality of any basis for V by Theorem 3.3. The codimension of W ≤ V iswritten codimD(W,V ) and is defined as dimD(V/W ). If f : V → W is a D-linear map, the rank of f is

given by rk(f)def= dimD im f .

Corollary 3.5. Let V be a left D-vector space.

(i) The vector space V is finite-dimensional if and only if it is finitely generated.

(ii) Two left D-vector spaces are isomorphic if and only if they have the same dimension.

(iii) A linearly independent subset L ⊆ V is a basis if and only if it is maximal among the set of linearlyindependent subsets of V .

(iv) A generating subset S ⊆ V is a basis if and only if it is minimal among the set of generating subsetsof V .

(v) Dimension is additive : if V =⊕

i∈I Vi, then dimV =∑

i∈I Vi.

(vi) Every vector subspace W ≤ V admits a complement, namely a subspace W ′ ≤ V such that V =W ⊕W ′. In particular every short exact sequence of vector spaces splits.

Proof. For part (i), this follows from Theorem 3.3 (if it is finitely generated, the finite set of generatorscontains a basis). Part (ii) follows from Corollary 2.57 and parts (iii) and (iv) follow from Theorem 3.3.Part (v) follows since a basis B for V can be given by taking a union of bases Bi for each Vi, so thatB =

⋃i∈I Bi satisfies

dimD V = |B| =∑i∈I|Bi| =

∑i∈I

dimD Vi.

Part (vi) follows from the fact that V is completely reducible.

Proposition 3.6. Let A be a ring and ϕ : A→ D be a morphism of rings where D is a division ring. If Eis a free left A-module, then any two bases for E have the same cardinality.

Proof. Turn D into a (D,A)-bimodule by dadef= dϕ(a). Consider the left D-vector space V

def= D⊗A E.

If B is a basis for E, then 1⊗ B def= 1⊗ b | b ∈ B is a basis for V of the same cardinality. Therefore,

any two bases B,B′ of E induce two bases 1⊗B, 1⊗B′ of V which are of the same cardinality, implying|B| = |B′|.

Corollary 3.7. Let A be a commutative ring. If E is a free A-module, any two bases for E have the samecardinality.

Proof. By Krull’s theorem, A admits a maximal ideal m, so we can apply Proposition 3.6 to the projectionπm : A→ A/m where A/m is a field.

Proposition 3.8. Let0 V1 V2 · · · Vn 0

be an exact sequence of finite-dimensional left D-vector spaces. Then

n∑i=1

(−1)i dimD Vi = 0.

47

Chapter 3

Proof. By induction on n. If n = 1 or n = 2, there is nothing to prove. Consider the new sequencewhose beginning is replaced by

0 V2/V1 V3 · · · Vn 0.

It is still exact, hence −dimD(V2/V1) +∑n

i=3(−1)i−1 dimD Vi = 0. Since dimD(V2/V1) = dimD V2 −dimD V1 by the case n = 2, substituting and multiplying by −1 gives

∑ni=1(−1)i dimD Vi = 0.

Corollary 3.9. Let W1,W2 be two vector subspaces of a left D-vector space V . We have the formula

dimD(W1 +W2) + dimD(W1 ∩W2) = dimDW1 + dimDW2.

Proof. It suffices to apply Proposition 3.8 to the exact sequence

0 W1 ∩W2 W1 ⊕W2 W1 +W2 0.

where the second map is given by the difference between the two inclusions ιWi : Wi → V .

Corollary 3.10. Let V be a finite-dimensional left D-vector space and Wii∈I a family of vector subspacessatisfying

∑i∈IWi = V . We have the inequality

dimD V ≤∑i∈I

dimDWi.

Proof. Since V is finite-dimensional, we can assume without loss of generality that I is finite. Writethe family as W1, · · · ,Wn. By induction on n and replacing the family W1, · · · ,Wn−1 by

∑n−1i=1 Wi, it

suffices to restrict to the case n = 2, which follows from Corollary 3.9.

Theorem 3.11. (Rank-nullity formula) Let f : V → W be a D-linear map between left D-vector spaces.Then

rk(f) + dim ker f = dimV, rk(f) + dim coker f = dimW.

As a corollary, rk(f) ≤ mindimD V,dimDW.

Proof. Write V = ker f ⊕W ′ and W = im f ⊕W ′, so that

im f ' V/ ker f 'W ′, coker f 'W/im f,

hence by taking dimensions, we obtain both equalities.

Proposition 3.12. Let V be a finite-dimensional left D-vector space and f ∈ EndD(V ). The following areequivalent :

(i) The endomorphism f is bijective

(ii) The endomorphism f is injective

(iii) The endomorphism f is surjective

(iv) The endomorphism f admits a left-inverse

(v) The endomorphism f admits a right-inverse

(vi) The endomorphism f is an isomorphism.

48

(vii) rk(f) = dimD V .

Proof. Obvious from the rank-nullity formula.

Example 3.13. Let n,m, p ≥ 1 be three integers and f : D⊕m` → D⊕n` and g : D⊕n` → D⊕p` be twoD-linear maps which are determined by their matrix forms, constructed as follows : if v1, · · · , vm is thestandard basis of D⊕m` and w1, · · · , wn is the standard basis of D⊕n` , defining coefficients bij ∈ D viathe equations f(vj) =

∑ni=1 aijwi, for any d1, · · · , dm ∈ D, we have

f

m∑j=1

djvj

=

m∑j=1

djf(vj) =

m∑j=1

djf(vj) =

m∑j=1

dj

(n∑i=1

bijwi

)=

n∑i=1

m∑j=1

djbij

wi.

Repeating this computation in an analogous manner, we deduce that if (bij) is the matrix form of f and (aij)is the matrix form of g, then the matrix form of g f corresponds to the following matrix multiplication :a11 · · · a1n

.... . .

...ap1 · · · apn

b11 · · · b1m

.... . .

...bn1 · · · bnm

= [aij ][bij ] =

[n∑k=1

bikakj

].

Note that is not usual matrix multiplication, but matrix multiplication where multiplication of coeffi-cients is performed in Dopp rather than in D. It follows from this example that the isomorphism of ringsEndD(D`) ' Dopp of Proposition 1.28 extends to the isomorphism EndD(D⊕n` ) ' Matn×n(Dopp).

3.2 Centralizer of a completely reducible module

We go back to the situation where A is a ring.

Proposition 3.14. Let P be a simple A-module and M an isotypically simple module of type P . Considerthe rings EndA(P ) and EndA(M) (recall that EndA(P ) is a division ring since P is a simple A-module).

(i) The A-module HomA(P,M) can be given the structure of a (EndA(M),EndA(P ))-bimodule (c.f.Remark 1.47), i.e. it is a left EndA(M)-module, a right EndA(P )-vector space and both modulestructures are compatible.

(ii) The canonical morphism EndA(M)→ EndEndA(P )(HomA(P,M)) induced by the EndA(M)-modulestructure on HomA(P,M) (compatible with the EndA(P )-module structure, hence multiplication byan element of EndA(M) is EndA(P )-linear) is an isomorphism of rings. The EndA(M)-moduleHomA(P,M) is simple and its centralizer satisfies

CEndA(M)(HomA(P,M)) ' EndA(P )opp.

(iii) The canonical map HomA(P,M) ⊗EndA(P ) P → M given by ϕ ⊗ p 7→ ϕ(p) is an isomorphism ofleft A-modules and of left EndA(M)-modules.

(iv) The EndA(M)-module M/ is isotypical of type HomA(P,M).

(v) We have the formulas

`A (M) = dimEndA(P ) HomA(P,M), `EndA(M) (M/) = dimEndA(P ) P.

49

Chapter 3

(vi) The left A-submodules of M and the EndA(P )-vector subspaces of HomA(P,M) are in one-to-onecorrespondence via the formulas

(M ′ ≤M) 7→ HomA(P,M ′) ⊆ HomA(P,M), (V ≤ HomA(P,M)) 7→ V ⊗EndA(P ) P ⊆M,

the inclusions being produced by the canonical maps : the inclusion HomA(P,M ′) ⊆ HomA(P,M)is given by post-composition with the inclusion mapM ′ ⊆M and the inclusion V ⊗EndA(P ) M ⊆Mis given by recalling that V ⊆ HomA(P,M), so we can map a product ϕ⊗ p ∈ V ⊗EndA(P ) P to thecomposition ϕ(p), which is injective by part (iii).

Proof. Part (i) follows from Proposition 2.59. Since P is a cyclic A-module, the mapEndA(M) → EndEndA(P )(HomA(P,M)) is an isomorphism by letting the pair (P,M) play therole of the pair (M,V ) in Theorem 1.59 (vi). Since EndA(P ) is a skew field, HomA(P,M) is a simpleEndEndA(P )(HomA(P,M))-module (this is the fact that given two vectors in a vector space V , thereis an endomorphism of V mapping one vector to the other), hence a simple EndA(M)-module by part (i).

We can use the fact that HomA(P,M) is a free EndA(P )-module to apply Corollary 1.38 and deducethat

EndA(M) ' EndEndA(P )(HomA(P,M)) = CEndA(P )(HomA(P,M))

=⇒ CEndA(M)(HomA(P,M)) ' C2EndA(P )(HomA(P,M)) = EndA(P )HomA(P,M) = EndA(P )opp.

The A-linearity and EndA(M)-linearity of the map given in part (iii) is obvious ; it is an isomorphismby Theorem 1.59 (v). If we fix an isomorphism of A-modules M ' P⊕I , we obtain an isomorphism ofEndA(P )-modules

HomA(P,M) ' HomA(P, P⊕I) ' EndA(P )⊕I ,

which implies `A (M) = dimEndA(P ) HomA(P,M).

Considering P as a right EndA(P )opp-module and HomA(P,M) as a left EndA(P )opp-module,we can identify HomA(P,M) ⊗EndA(P ) P with P ⊗EndA(P )opp HomA(P,M) ; this identification isEndA(M)-linear. Note that the action of EndA(P )opp on HomA(P,M) corresponds to the action ofCEndA(M)(HomA(P,M)) by part (ii) and P is a free EndA(P )opp-module since EndA(P )opp is a divi-sion ring. By applying Theorem 1.59 (i) to this situation, the role of the triple (A,M,W ) there beingreplaced by that of (EndA(M),HomA(P,M), P ) here, we see that

P ⊗EndA(P )opp HomA(P,M) ' HomA(P,M)⊗EndA(P ) P 'M/

is an isotypical EndA(M)-module of type HomA(P,M). Part (vi) is a corollary of Theorem 1.59 becauseevery subspace of HomA(P,M) is free and every submodule of M is isotypical of type P .

Corollary 3.15. Let P be a simple left A-module ; recall that Ddef= CA(P ) is a division ring by Schur’s

Lemma and that P / is a left D-vector space. Let V be a right CA(P )-vector space (so that V ⊗D P / is aleft A-module by multiplication on elements of P ) and S ⊆ EndD(V ) be a subset. Given ϕ ∈ EndD(V ),

we can extend this to an endomorphism ϕ⊗ idS ∈ EndA(V ⊗D P /) by (ϕ⊗ idS)(v ⊗ p) def= ϕ(v)⊗ p. For

an A-submodule N ≤ V ⊗D P /, the following are equivalent :

(i) The submodule N is left invariant by every ϕ ∈ S, i.e. (ϕ⊗ idP )(N) ⊆ N

(ii) The submodule N has the form W ⊗D P / where W ≤ V is a D-vector subspace left invariant byevery ϕ ∈ S, i.e. ϕ(W ) ⊆W .

50

Proof. Find a set I such that V ' d⊕I and let Mdef= P⊕I . Note that we have an identification

V ' D⊕I = EndA(P )⊕I ⊆ HomA(P, P⊕I) = HomA(P,M).

given by summing the maps (ϕi)i∈I ∈ EndA(P )⊕I to get a map⊕

i∈I ϕi : P → P⊕I ; this inclusion isalso right D-linear, so V can be seen as a D-vector subspace of HomA(P,M). It follows that the functor(−)⊗D P applied to this inclusion preserves its injectivity (because the corresponding exact sequence ofD-modules is split), so we get an inclusion

V ⊗D P ⊆ HomA(P,M)⊗D P 'M

by Proposition 3.14 (iii). By Proposition 3.14 (vi), the left A-submodules ofM are in an inclusion-preservingbijection with the right D-vector subspaces of HomA(P,M), so the left A-submodules of V ⊗D P arein bijection with the right D-vector subspaces of V . By writing N = W ⊗D P , we see that N is stableunder the endomorphisms ϕ ⊗ idP for all ϕ ∈ S if and only if W is stable under the endomorphismsϕ ∈ S, which proves the equivalence.

Corollary 3.16. Let E be a division ring and S be a family of ring endomorphisms of E. Note that foreach ϕ ∈ S, ϕ(E) ⊆ E is isomorphic to E (via the map ϕ : E → ϕ(E)), so it is a division subring of E.

Let Ddef= ES be the set of e ∈ E which are left invariant by each ϕ ∈ S (i.e. ϕ(e) = e for all ϕ ∈ S). Note

that D is a division ring (because E is). For a right D-vector space V , we turn V ⊗D E into a left E-vectorspace. For an E-vector subspace U ≤ V ⊗D E, the following are equivalent :

(i) The subspace U is stable under every endomorphism of the form idV ⊗ ϕ for ϕ ∈ S

(ii) There exists a D-vector subspace W ≤ V such that U = W ⊗D E.

Proof. Consider the Z-subalgebra A of EndZ(E) generated by S and (Er)E , the ring of homothetiesof the right E-module E ; note that in A, the multiplication of two right homotheties is not given bymultiplication in E but by composition in EndZ(E), so E is a left A-module (even though we are actingby homotheties on the right) ; if eEr denotes multiplication by e ∈ E on the right, the left-A-linearity ofmultiplication by eEr ∈ A is reflected in the equation

(eEr e′Er)(e′′) = eEr(e′Er

(e′′)).

The left A-module E is simple since (Er)E ⊆ A and the right E-module Er is simple. We also see thatCA(E) ⊆ CE(Er) since if ψ ∈ EndZ(E) commutes with all elements of A, it commutes in particularwith all elements of (Er)E , i.e. is a right E-linear map. But we have seen that as sets, CE(Er) = EE`

(c.f. Proposition 1.28 ; we said “as sets” because we need to add “opp” to get an isomorphism of rings), soCA(E) ⊆ (E`)E . This means that if ψ ∈ CA(E), then there exists eψ ∈ E such that ψ(e′) = eψe

′ andfor all ϕ ∈ S,

eψ = eψϕ(1) = ψ(ϕ(1)) = ϕ(ψ(1)) = ϕ(eψ) =⇒ eψ ∈ D.

Conversely, if eψ ∈ D, then ψ ∈ CA(E), so it follows that CA(E) = (E`)D is the ring of homothetiesof E seen as a left D-module and can therefore be identified with D. The result now follows fromCorollary 3.15.

Proposition 3.17. Let M be a completely reducible with decomposition into isotypical components writtenas M =

⊕P MP . Then M/ =

⊕P (MP )/ is also a decomposition of M/ into isotypical components,

each (MP )/ being isotypically simple of type HomA(P,MP ) by Proposition 3.14 (iv). In particular, if M iscompletely reducible, then M/ is also completely reducible.

51

Chapter 3

Proof. First of all, recall that by Proposition 2.53 (ii),

CA(M) = EndA(M) =∏P

EndA(MP ) =∏P

CA(MP )

where the identification is given by ϕ 7→ (ϕ|MP)P . Since each MP is isotypical, we can apply Proposi-

tion 3.14 (iv) and deduce that (MP )/ is also an isotypical CA(MP )-module, hence an isotypical CA(M)-submodule of M . Since MP is stable under C2

A(M) = CA(M/) by Proposition 1.31, it is a sum ofisotypical components of M/ by Corollary 2.54, but since (MP )/ is itself isotypical, (MP )/ is an isotypi-cal component of M/.

3.3 Semisimple rings

Proposition 3.18. Let A be a ring. The following are equivalent :

(i) Every left A-module is completely reducible

(ii) The A-module A` is completely reducible.

A ring A with these properties is called semisimple.

Proof. The implication ( (ii) ⇒ (i) ) is obvious. For the converse, note that every left A-module isisomorphic to the quotient of an isotypical module of type A` by choosing a set of generators, so theresult follows from the fact that the direct sum and quotient of completely reducible modules is completelyreducible.

Corollary 3.19. If ϕ : A→ B is a surjective morphism of rings and A is semisimple, then B is semisimple.In particular, if a E A is a two-sided ideal and A is semisimple, then A/a is semisimple.

Proof. The morphism turns B into an A-module which is completely reducible, so in particular B` iscompletely reducible, which means B is semisimple.

Remark 3.20. Let A be a semisimple ring.

(i) Since A` is a cyclic A-module (being generated by 1 ∈ A) and semisimple, it is of finite length byCorollary 2.58, hence A` is left-artinian and left-noetherian.

(ii) The opposite ring Aopp is also semisimple, thus there is no necessity of defining the notions of “left-semisimple” and “right-semisimple”. To see this, note that (Aopp)` ' Ar = (A`)

/ by Proposition 1.28.But then the fact that A` is semisimple implies that (A`)

/ is semisimple by Proposition 3.17.

Proposition 3.21. Let M be a left A-module such that M/ is finitely generated. The following are equiva-lent :

(i) The module M is completely reducible.

(ii) The ring of homotheties AM is semisimple.

Proof. The implication ( (ii)⇒ (i) ) is trivial sinceM is a left AM -module. Conversely, ifM is completelyreducible, (AM )` is isomorphic to a submodule ofM⊕n by Proposition 2.30, hence AM is semisimple.

Corollary 3.22. Let D be a division ring and V a finite-dimensional left D-vector space. Let A be a subringof the ring of endomorphisms EndD(V ). Then the following are equivalent :

52

(i) The A-module V is completely reducible.

(ii) The ring A is semisimple.

In particular, this is true if D = K is a field and A is a K-subalgebra of EndK(V ).

Proof. Note that D ⊆ EndA(V ) since A ⊆ EndD(V ) ; this implies that V / is finitely generated since itis a finite-dimensional D-vector space. The EndD(V )-module V is faithful, so A = AV is semisimple ifand only if V is a completely reducible A-module by Proposition 3.21.

Proposition 3.23. A ring A is semisimple if and only if it is isomorphic to the endomorphism ring of acompletely reducible B-module M of finite length for some ring B.

Proof. (⇒) If M is a completely reducible B-module, so is M/ by Proposition 3.17. Since M// isfinitely generated (because M is by its complete reducibility and BM ⊆ C2

B(M)), we deduce thatA ' EndB(M) = CB(M)M/ is semisimple by Proposition 3.21.

(⇐) If A is semisimple, then by Remark 3.20, so is

Aopp ' (AAr)opp ' CA(A`) = EndA(A`).

by Proposition 1.28. The left A-module A` is semisimple, so this means A is isomorphic to the endomor-phism ring of the semisimple left A-module A`.

Proposition 3.24. Let A be a semisimple ring. The isotypical components of A`, the isotypical componentsof Ar and the minimal two-sided ideals of A all coincide (by “minimal”, we mean “minimal with respect tobeing non-zero under the partial order given by inclusion”).

Proof. The two-sided ideals of A are the left A-submodules of A` stable under multiplication on the rightby A, i.e. table under AAr , or equivalently, under (AAr)opp ' CA(A`) by Proposition 1.28. It followsfrom Corollary 2.54 that the two-sided ideals of A are precisely the sum of isotypical components of A`,so the minimal two-sided ideals of A are the isotypical components of A`. By repeating the process withAr, we come to the same conclusion.

Proposition 3.25. Let A be a semisimple ring. Every simple left A-module is isomorphic to a minimal leftideal of A.

Proof. Let M be a simple left A-module. There exists a left maximal ideal m such that M ' A`/m, butsince A` is a semisimple left A-module, A` = m ⊕ n for some left ideal n ≤ A`. The left A-module nis also completely reducible, so the maximality of m implies the minimality of n. Apply the argumentsymmetrically for the case of right A-modules.

Proposition 3.26. Let A be a semisimple ring. For every left ideal a E A, there exists an idempotent ea ∈ asuch that a = A(ea) = aea and 1 = ea + ea′ where a′ is a supplementary left ideal to a, i.e. A = a⊕ a′.

Proof. Let a′ ≤ A` be a supplementary left ideal of a in A, so that A = a ⊕ a′. By Corollary 1.18, thefamily of projections πa, πa′ is a family of orthogonal idempotents, but since πa, πa′ ∈ EndA(A`) =

CA(A`) = (AAr)opp, we see that edef= πa(1), e′

def= πa′(1) satisfy

A(e) = a, A

(e′)

= a′, e2 = e, (e′)2 = e′, ee′ = e′e = 0.

53

Chapter 3

Letting eadef= e and ea′

def= e′, we see that a = A(ea) = aea.

Proposition 3.27. Let A be a semisimple ring. The following are equivalent :

(i) There is only one isomorphism class of simple left A-modules

(ii) There is only one isomorphism class of simple right A-modules

(iii) The ring A has a unique left maximal ideal

(iv) The ring A has a unique right maximal ideal

(v) The left A-module A` is isotypically simple

(vi) The right A-module Ar is isotypically simple

(vii) The only two-sided ideals of A are A and 0.

A non-zero ring satisfying any of the equivalent properties above is called simple. If B is a commutativering, a B-algebra A is called simple if A is simple as a ring.

Proof. By Proposition 3.24, we see that (v), (vi) and (vii) are equivalent. It is clear that (i) and (iii) (resp. (ii)and (iv)) are equivalent since for two left ideals m1,m2 E A, we have A/m1 ' A/m2 as left A-modules ifand only if m1 = m2 (compare annihilators of those modules). It is also clear that (i) (resp. (ii)) implies (v)(resp. (vi)). The converse is also true by Corollary 2.47 since any simple left (resp. right) A-module is aquotient of the isotypical module A⊕I` (resp. A⊕Ir ), so they all belong to the same isomorphism class.

Example 3.28. • Any skew field D is simple since D` is a simple left D-module.

• Any simple commutative ring is a field since it has no non-zero proper two-sided ideals.

• Simple rings are semisimple by definition since A` is completely reducible and isotypical ; the factthat it is completely reducible is equivalent to the semisimplicity of A.

• If A is a simple ring, so is Aopp since the condition (vii) is symmetric with respect to multiplication.

Proposition 3.29. Let M be a left A-module such that M/ is finitely generated. The following are equiva-lent :

(i) AM is a simple ring

(ii) M is isotypically simple

When either of the assumptions hold, ifM is isotypical of type P where P is simple, we have P ' AM/mM

where mM is the unique left maximal ideal of AM , or equivalently, the unique left maximal ideal of Acontaining AnnA(M).

Proof. We already know that AM is semisimple if and only if M is completely reducible by Proposi-tion 3.21. If AM is simple, there is only one class of simple left A-modules, so the complete reducibilityof M implies that it is isotypically simple. Conversely, as in the proof of Proposition 3.21, we see that(AM )` is isomorphic to a submodule of M⊕n for some n ≥ 1, so (AM )` is isotypically simple, henceAM is simple.

54

Theorem 3.30. Let A be a semisimple ring. The minimal two-sided ideals A1, · · · , An admit a unit elementand form simple subrings of A such that as a ring, A =

∏ni=1Ai. Conversely, if A1, · · · , An are simple

rings, then∏ni=1Ai is a semisimple ring whose minimal two-sided ideals are the following subsets, each

identified with Ai :0× · · · × 0×Ai × 0× · · · × 0, i = 1, · · · , n.

Proof. Consider the semisimple ring A. The completely reducible left A-module A` is a direct sum ofits isotypical components A1, · · · , An (there are finitely many of them since A` is of finite length byRemark 3.20), which are minimal two-sided ideals of A by Proposition 3.24. Under the decompositionA` =

⊕ni=1Ai, write 1 ∈ A as 1 =

⊕ni=1 ei where ei ∈ Ai. Note that the fact that the ideals Ai are

two-sided implies that AiAj ⊆ Ai∩Aj = 0 for 1 ≤ i, j ≤ n with i 6= j. The action of 1 on Ai is thereforethe same as that of ei, so ei is a unit element for the subring Ai. By definition, Ai is an isotypical leftA-module, hence also isotypical as a module over the ring A/AnnA(Ai) ' Ai, which means it is simple.

Conversely, given simple rings A1, · · · , An, the ring Adef=∏ni=1Ai is clearly semisimple since A` =⊕n

i=1Ai is completely reducible when seen as a direct sum of the completely reducible submodules Aibecause the A-module Ai is annihilated by the Aj for i 6= j.

Definition 3.31. The minimal two-sided ideals of a semisimple ring are called its simple components.

Corollary 3.32. Every quotient of a semisimple ring A by a two-sided ideal a is a semisimple ring isomor-phic to a subring of A, for which we have an isomorphism of rings A ' (A/a)× a and a is a subring of Awith πa(1) as its identity element.

Proof. The A/a-submodules of an A/a-module M correspond to its A-submodules (since their annihi-lators all contain a). Since A is semisimple, M is a completely reducible A-module, hence a completelyreducible A/a-module, which means A/a is semisimple. Since a is a left A-submodule of A, it admits acomplement a′ such that A` = a⊕ a′, which implies that A ' a×A/a. The rest is obvious.

Proposition 3.33. Let A1, · · · , An be simple rings and Adef=∏ni=1Ai. The ring A admits precisely n

distinct isomorphism classes of simple left A-modules. Given a simple A-module M , there exists a uniquei ∈ 1, · · · , n such that

AnnA(M) = A(A1, · · · ,mi, · · · , An)A = A1 × · · · ×Ai−1 × 0×Ai+1 × · · · ×An(where mi is the unique left maximal ideal of Ai) and M is a simple Ai-module under the identificationA/AnnA(M) ' Ai.

Proof. Let M be a simple left A-module. For 1 ≤ i ≤ n, consider the submodule

AiMdef=

∗∑j

ajmj

∣∣∣∣∣∣ aj ∈ Ai, mj ∈M

.

Then AiM is an A-submodule of M since AiAj = 0 for i 6= j. Therefore, there exists precisely one isuch that AiM 6= 0 (in which case AiM = M ), because the contrary implies

M = AiM = Ai(AjM) = 0.

Assume AiM = M 6= 0. By letting mi be the unique left maximal ideal of Ai and identifying thequotients Ai/mi ' A/AnnA(M), the Ai-module M is simple precisely because it is a simple A-module.It follows that each simple A-module is a simple Ai-module for a unique 1 ≤ i ≤ n, so there are preciselyn isomorphism classes of simple A-modules, each one being represented by the simple Ai-module Ai/mi

where mi is the unique left maximal ideal of Ai.

55

Chapter 3

Theorem 3.34. Let D be a division ring.

(i) Let V be a finite-dimensional left D-vector space of dimension r > 0. The ring Adef= EndD(V ) is

simple and the isotypically simple left A-module A` is of type V and has length r, i.e. `A (A`) = r.Considered as an A-module, V is simple.

(ii) Let A be a simple ring such that `A (A`) = s and M be a simple left A-module. We obtain a

division ring Ddef= CA(M) = EndA(M) and the equality dimDM = s. Furthermore, the canonical

morphism of rings ϕ : A → C2A(M) = EndD(M) defined by a 7→ aM is an isomorphism, so that

A ' EndD(M).

In other words, a finite-dimensional vector left D-vector space V of dimension dimD V corresponds to itsendomorphism ring A of length `A (A`) via A = EndD(V ) and D = EndA(V ), for which dimD V =`A (A`). An element ϕ ∈ EndZ(V ) commutes with every element of AV (resp. DV ) if and only if it is inDV (resp. AV ), so this gives the identities

Z(A) = Z(EndD(V )) = Z(D) = Z(EndA(V )).

Proof. (i) If v1 ∈ V \ 0, one can extend v1 to a D-basis v1, v2, · · · , vr for V by Theorem 3.3,so that as an A-module, A〈v1〉 = V . This shows that V is a simple A-module. It is a faithfulA-module since an endomorphism of V which vanishes on all of V is zero by definition. Considerthe A-linear map A → V ⊕r defined for ϕ ∈ A = EndD(V ) by ϕ 7→ (ϕ(v1), · · · , ϕ(vr)). Thismap is bijective ; it is the correspondence between an endomorphism of V and the collectionof its r columns when written in matrix form. Since V is a simple A-module, this shows that`A (A`) = `A (V ⊕r) = r. In particular, A is isotypically simple of type V .

(ii) We know that AA`= C2

A(A`) by Corollary 1.38. Since A` is isotypically simple of type M , weobtain a canonical isomorphism

A ' AA`= C2

A(A`) ' C2A(M⊕s) ' C2

A(M).

This also implies that AM = C2A(M) and that M is a faithful A-module. Since C2

A(M) =CCA(M)(M

/) = EndD(M), we just identified A with EndD(M), the endomorphism ring of theleft D-vector space M . Since A is of finite length, it is left-artinian, so the dimension dimDM isfinite ; otherwise, the ring EndD(M) = A would not be left-artinian by Example 2.27. By part (i),we see that dimDM = `A (A`) = r.

The last equality follows from Corollary 1.38 since V is a free (and thus faithful) D-module.

Corollary 3.35. Let D1, D2 be division rings and Mi be a finite-dimensional vector space over Di fori = 1, 2. If the rings EndD1(M1) and EndD2(M2) are isomorphic, then D1 and D2 are isomorphic anddimD1 M1 = dimD2 M2.

Proof. The isomorphism between EndD1(M1) and EndD2(M2) show that these rings have equal lengthsas left modules over themselves, thus the equality in dimensions. Since Di is the endomorphism ring ofany simple EndDi(Mi)-module and that the rings EndDi(Mi) are isomorphic, so are D1 and D2 whenthe modules M1 and M2 are identified along the isomorphism of rings EndD1(M1) ' EndD2(M2).

Corollary 3.36. Let A be a ring and M be an isotypically simple A-module of finite length r. The ring

Bdef= EndA(M) is simple and `B (B`) = r.

56

Proof. By Proposition 3.14, fixing a simple A-module P such that M ' P⊕r, we see that B =EndA(M) ' EndA(P ) is isomorphic to the endomorphism ring of a left EndA(P )-vector space ofdimension r. We can then apply Theorem 3.34 to complete the proof.

Corollary 3.37. Let M be a left A-module over the semisimple ring A. Then M/ is finitely generated andAM = C2

A(M). (In fact, an upper bound for the number of generators may be computed as `A (A`).

Proof. The second assertion follows from the first via Corollary 2.63 since the semisimplicity of Aimplies that M is completely reducible.

Since M has finitely many isotypical components, we can restrict to the case that M is isotypicallysimple of type P where P is a simple left A-module. Seeing P as a simple module over one of thesimple components Ai of A instead of all of A (this is because only one of the Ai’s is not contained in

the annihilator of P ), we see that Ddef= EndA(P ) is a division ring. By Proposition 3.14 (v), note that

B = EndA(M) satisfies `B (M/) = dimD P . The latter is an upper bound for the minimal numberof generators of M/, so it suffices to show that dimD P < ∞ ; this is clear by Theorem 3.34 sinceEndD(P ) = Ai is simple of finite length and dimD P = `Ai ((Ai)`) <∞.

As for the upper bound, note that `A (A`) =∑n

i=1 `Ai ((Ai)`) where A =∏ni=1Ai is the decomposition

of A as the product of its simple components.

Theorem 3.38. Let A be a semisimple ring with simple components A1, · · · , An, so that A =∏ni=1Ai.

Then Z(A) =∏ni=1 Z(Ai) and each Z(Ai) is a field. More precisely, if A = EndD(V ) where D is a division

ring and V a left D-vector space of finite dimension, then Z(A) = Z(D)V consists of those endomorphismsof V given by multiplication by central elements of D. In particular, the following holds :

(i) The ring A is a simple ring if and only if Z(A) is a field

(ii) If A is a semisimple ring, then Z(A) is semisimple.

Proof. The identity Z(A) =∏ni=1 Z(Ai) follows by definition of multiplication. The fact that Z(Ai) is

a field follows by Corollary 1.38 identifying Ai with the endomorphism ring of a simple Ai-module P so

that Didef= EndAi(Pi) is a division ring satisfying Ai = EndDi(Pi) and

Z(Ai) = Z(EndDi(Pi)) = Z(Di)Pi ' Z(Di) ⊆ Di.

As a division subring of the division ring Di which is commutative, Z(Ai) is a field. It is then clear thatZ(A) is a field if and only if n = 1 since a direct product of fields is not a field.

Definition 3.39. Let K be a field.

(i) A central K-algebra is a K-algebra A with structure map ϕ : K → A satisfying Z(A) = ϕ(K).

(ii) A central simple algebra over K (written CSA for short) is a central K-algebra A which is a simplering.

Corollary 3.40. Let K be a field.

(i) If A is a finite-dimensional CSA over K , then there exists a division ring D which is a CSA over Kand for which A ' Matn×n(D) for some integer n ≥ 1.

(ii) If A is a CSA which is finite-dimensional over K and we assume K is algebraically closed, thenA ' EndK(K⊕n) ' Matn×n(K) for some integer n ≥ 1.

57

Chapter 3

Proof. Let P be a simple A-module. By Theorem 3.34, we see that letting Ddef= EndA(P ), we have

A ' EndD(P ). Letting n = dimD P , this proves that A ' Matn×n(Dopp), so since Dopp is a divisionring and Z(Dopp) = Z(D) = Z(EndA(P )) = Z(A) = K , the division ring D is also a CSA. Thisproves part (i), and part (ii) follows since a division ring which is a CSA over K is isomorphic to K as aK-algebra.

58

Chapter 4

Radicals

In this chapter, A denotes a ring and M is a left A-module.

4.1 Radical of a module

Definition 4.1. Let U ≤ A be a subgroup of the additive group (A,+) and V ≤ M be a subgroup of theadditive group (M,+). Their product is denoted by UV and is the abelian subgroup of M generated bythe products uv where u ∈ U and v ∈ V . In symbols,

UVdef= 〈uv |u ∈ U, v ∈ V 〉Z.

The map U × V → M defined by (u, v) 7→ uv is Z-bilinear, so it factors through a map of abelian groupsU ⊗Z V and UV is the image of the corresponding map U ⊗Z V →M .

More generally, if U1, · · · , Un ≤ A are subgroups of (A,+), we define

U1 · · ·UnVdef= 〈u1 · · ·unv |ui ∈ Ui, v ∈ V 〉Z.

Again, this is the image of the map U1 × · · · × Un × V →M defined by (u1, · · · , un, v) 7→ u1 · · ·unv, andalso the image of the corresponding additive map U1 ⊗Z · · · ⊗Z Un ⊗Z V →M .

Taking the particular case where M = A`, we can define the products U1U2 for two additive subgroupsU1, U2 ≤ A by the same formulas. This definition of the product is associative, namely, for three additivesubgroups U1, U2, U3 ≤ A, we have

U1(U2U3) = (U1U2)U3 = U1U2U3.

which is clear since all three are generated by the same subsets of A by associativity of multiplication in A.

Remark 4.2. (i) If Uii∈I is a family of additive subgroups of A and Vjj∈J is a family of additivesubgroups of M , letting U =

∑i∈I Ui and V =

∑j∈J Vj , we obtain

UV =∑

i∈I,j∈JUiVj .

This is clear since tensor products commute with direct sums.

(ii) If U1, U2 ≤ A are two additive subgroups and f : A → B is a morphism of rings to another ring B,then f(U1U2) = f(U1)f(U2) (since they are both generated by the same elements).

59

Chapter 4

(iii) If a E A is a left ideal, then for every additive subgroup V ≤ M , aV is an A-submodule of M . Inparticular, for any additive subgroup U ≤ A, aU is a left ideal of A. Analogously, if b is a right idealin A, Ub is a right ideal of A, so ab is a two-sided ideal in A.

(iv) Let a1, a2 be two left ideals of A. The operation (a1, a2) 7→ a1a2 is an associative operation on the

set of left ideals of A. We define the nth power of a left ideal by the recursive definition a0 def= A

and andef= an−1a. For every n ≥ 0, we have an+1 ⊆ an. We define a similar notion for right ideals.

Multiplying a right ideal b with a left ideal a to form the subgroup ba ≤ A, we obtain the inclusionba ⊆ b ∩ a. More generally, if a2, · · · , an−1 are two-sided ideals of A, a1 is a right ideal of A and anis a left ideal of A, then

a1 · · · an ⊆n⋂i=1

ai.

This is clear since a1 · · · ai−1aiai+1 · · · an ⊆ A · · ·AaiA · · ·A = ai. If additionally, a1 is a left ideal,then a1 · · · an is a left ideal ; similarly, if an is also a right ideal, then a1 · · · an is a right ideal.

Definition 4.3. Let A be a ring.

(i) An element x ∈ A is said to be nilpotent if there exists n ≥ 1 such that xn = 0. The set of nilpotentelements of A is denoted by Nil(A). (When A is commutative, Nil(A) is a subgroup by the binomialtheorem.)

(ii) If x, y ∈ A are two elements such that xy = yx = 1, we say that x, y are units of A. The set of unitsof A is denoted by A× and it forms a group under multiplication. If xy = 1 for some x, y ∈ A butyx is not necessarily equal to 1, we say that x is right-invertible with right inverse y ; similarly, y isleft-invertible with left inverse x.

(iii) A left (resp. right) ideal a of A is said to be nilpotent if there exists n ≥ 1 such that an = 0.

(iv) A left (resp. right) ideal a contained in Nil(A) is called a nilideal.

Remark 4.4. (i) When x ∈ Nil(A), 1− x ∈ A×. To see this, assume n ≥ 0 is such that xn = 0 and usethe geometric progression formula :(

n−1∑i=0

xi

)(1− x) = (1− x)

(n−1∑i=0

xi

)= 1− xn = 1.

(ii) The equation an = 0 is equivalent to the following property : any product of n elements of a is equalto zero (because those products generate an).

(iii) A nilpotent ideal a is evidently a nilideal, but the converse is not true in general. (It would be niceto have easy examples, but Bourbaki (c.f. Algèbre, Chapitre 8, §6, n 1, Remarques) has a rather hardone.)

Definition 4.5. Recall that a proper submodule N ≤ M is said to be maximal if M/N is simple (c.f.Definition 2.37). The radical of a left A-moduleM is denoted by rad(M) and is defined as the intersectionof all maximal submodules of M . If rad(M) = 0, we say that M is radical-free. Equivalently, rad(M) isthe intersection of all kernels of morphisms of A-modules f : M → P where P is a simple left A-module,or in other words, rad(M) is the set of all m ∈ M such that whenever f : M → P is a morphism ofA-modules and P is simple, then f(m) = 0.

It is possible that M admits no maximal submodule in which case we set rad(M)def= M (this is the

same definition as above but intersecting over the empty family of maximal submodules of M ).

60

Remark 4.6. The module A/a is radical-free if and only if a is the intersection of left maximal ideals.

Proposition 4.7. If M is a non-zero finitely generated left A-module, then rad(M) 6= M .

Proof. This follows from the existence of maximal submodules, proven in Proposition 2.39.

Proposition 4.8. If f : M → N is a morphism of left A-modules, then f(rad(M)) ⊆ rad(N).

Proof. Let P be a simple A-module and g : N → P a morphism of left A-modules. The compositiong f : M → P vanishes on rad(M), hence g vanishes on f(rad(M)) ; this means f(rad(M)) ⊆ ker g.Taking the intersection over all such g gives the result.

Proposition 4.9. Let N be an A-submodule of the left A-module M .

(i) We have rad(N) ⊆ rad(M)

(ii) We have rad(M/N) ⊇ (rad(M) + N)/N . In particular, if N ⊆ rad(M), we have rad(M/N) =rad(M)/N .

(iii) If Mii∈I is a family of left A-modules, we have rad(∏i∈IMi) ⊆

∏i∈I rad(Mi).

(iv) If Mii∈I is a family of left A-modules, we have rad(⊕

i∈IMi) =⊕

i∈I rad(Mi).

Proof. Part (i) and the first statement of part (ii) are corollaries of Proposition 4.8. For the secondstatement of part (ii), the stronger statement of equality instead of inclusion follows from the fact thatN is contained in every maximal submodule of M , hence the maximal submodules of M and those ofM/N are in bijective correspondence.

Let Ni be a maximal submodule of Mi. Then

N(i)def= Ni ×

∏j∈Ij 6=i

Mj

is a maximal submodule of Mdef=∏i∈IMi, hence rad(

∏i∈IMi) ⊆ N(i) for all maximal submodules

Ni of Mi, meaning that rad(∏i∈IMi) ⊆

∏i∈I rad(Mi), which is part (iii).

Applying the same argument as in part (iii) to the direct sum, we obtain rad(⊕

i∈IMi) ⊆⊕

i∈I rad(Mi).However, since Mi ⊆

⊕j∈IMj , in this case we can also obtain rad(Mi) ⊆ rad(

⊕j∈IMj) by part (i).

Summing all these submodules gives⊕

i∈I rad(Mi) ⊆ rad(⊕

i∈IMi), thus implying the equality inpart (iv).

Corollary 4.10. Let M be a finitely generated left A-module and N ≤ M an A-submodule satisfyingN + rad(M) = M . Then N = M .

Proof. By Proposition 4.9 (ii), we see that rad(M/N) ⊇ (rad(M) + N)/N = M/N , which impliesequality throughout. By Proposition 4.7, this implies that M/N = 0.

Corollary 4.11. The submodule rad(M) is the smallest submodule N such that M/N is radical-free.

Proof. If M/N is radical-free, by Proposition 4.9, we see that 0 = rad(M/N) ⊇ (rad(M) + N)/N ,

61

Chapter 4

hence N = rad(M) +N , meaning rad(M) ⊆ N . By the second statement of Proposition 4.9 part (ii),

rad(M/rad(M)) = rad(M)/rad(M) = 0.

Proposition 4.12. LetM be a finitely generated left A-module with chosen set of generators m1, · · · ,mr.For m ∈M , the following are equivalent :

(i) m ∈ rad(M)

(ii) For all a1, · · · , ar ∈ A, the elements m1 + a1m, · · · ,mr + arm also form a set of generators forM .

Proof. In any case, set Ndef= A〈m1 + a1m, · · · ,mr + arm〉. If m ∈ rad(M), we see that N +rad(M) =

M , hence N = M , which proves that (i) implies (ii). For the converse, suppose that m ∈M \ rad(M), sothat there exists a maximal submodule N ofM not containing m. We know that A〈m〉+N = M , so foreach index 1 ≤ i ≤ r, since mi ∈ M = A〈m〉 + N , we find ai ∈ A such that mi + aim ∈ N , meaningthat A〈m1 + a1m, · · · ,mr + arm〉 ⊆ N . In particular, A〈m1 + a1m, · · · ,mr + arm〉 6= M .

Proposition 4.13. Let M be a finitely generated left A-module. The following are equivalent :

(i) M is radical-free, i.e. rad(M) = 0

(ii) M is isomorphic to a submodule of a product of simple left A-modules.

As a corollary, every finitely generated and completely reducible left A-module is radical-free.

Proof. ( (i)⇒ (ii) ) Let Mii∈I be a family of maximal submodules of M whose intersection is zero andconsider the family of maps πi : M → M/Mi ; note that M/Mi is simple by definition of maximal-ity. Their product gives a morphism π : M →

∏i∈IM/Mi which has kernel equal to

⋂i∈IMi = 0.

Therefore, M ' π(M) and the latter is a submodule of a product of simple left A-modules.( (ii) ⇒ (i) ) Let Pii∈I be a family of simple left A-modules and assume M ⊆

∏i∈I Pi. Note that by

definition, a simple module is radical-free. It follows that

rad(M) ⊆ rad

(∏i∈I

Pi

)⊆∏i∈I

rad(Pi) = 0.

Since a completely reducible module M can be written as M '⊕

i∈I Pi ⊆∏i∈I Pi, we see that

rad(M) = 0.

4.2 Radical of a ring

Let A be a ring. Recall Definition 2.7 where we defined the Jacobson radical of A (or simply the radical of

A) as the radical of the left A-module A`, i.e. Jac(A)def= rad(A`) is the intersection of all the left maximal

ideals of A (this is the non-commutative generalization of the Jacobson radical of a commutative ring). Asin the case of left A-modules, the ring A is said to be radical-free if Jac(A) = 0.

Remark 4.14. In theory, we should have defined the left Jacobson radical and the right Jacobson radical ofA, as we quickly mentioned in Definition 2.7. However, we will see that if we defined these two notions,they will always agree, so we only have to speak of the Jacobson radical of a ring. Because of this, we will

62

call it the Jacobson radical and when necessary, the notion which could have been called the right Jacobsonradical would be equal to the left Jacobson radical of Aopp, so we will denote it by Jac(Aopp) instead ofintroducing a new notation.

Proposition 4.15. Let A be a ring.

(i) If M is a left A-module, the annihilator

AnnA(M)def= a ∈ A | aM = 0

is a two-sided ideal of A.

(ii) The radical Jac(A) is the intersection of all the annihilators of simple left A-modules. By part (i), it isa two-sided ideal of A.

(iii) The radical Jac(A) is the smallest element in the set of annihilators of all completely reducible leftA-modules.

Proof. (i) This is obvious since if a ∈ AnnA(M), b ∈ A and m ∈ M , we have bm ∈ M , hence(ab)m = a(bm) = 0 and (ba)m = b(am) = b0 = 0.

(ii) It suffices to see that if P is a simple A-module, then the morphism ϕ : A` → P is entirely

determined by pdef= ϕ(1) since ϕ(a) = aϕ(1) = ap. By the simplicity of P , p will always generate

P as long as p 6= 0. It also follows that AnnA(P ) = a ∈ A | ap = 0 = kerϕ is a left maximalideal since if p 6= 0, A`/AnnA(P ) = A`/ kerϕ ' P . Conversely, if m is a left maximal ideal of A,the morphism ϕ : A` → A`/m is a morphism of left A-modules from A` to a simple left A-module

Pdef= A`/m with kernel equal to m. This establishes the correspondence between the kernels of

morphisms to simple modules and annihilators of simple modules, so their intersections are thesame.

(iii) If P is a simple left A-module, the kernel of the morphism ϕ : A` → P defined by ϕ(1)def= p ∈

P \0 is independent of the choice of p since it is equal to AnnA(P ). IfM is completely reduciblewith isotypical component MP of type P , then

AnnA(M) =⋂

P simpleMP 6=0

AnnA(MP ) =⋂

P simpleMP 6=0

AnnA(P ).

Letting m E A denote an arbitrary left maximal ideal of A and M0def=⊕

mEAA`/m, we see that

Jac(A) =⋂mEA

m =⋂mEA

AnnA(A`/m) = AnnA(M0)

contains the annihilator of every completely reducible left A-module, which proves the claim sinceM0 is itself completely reducible by construction.

Corollary 4.16. et A be a ring and M a left A-module. (Recall Theorem 2.9.)

(i) We have Jac(A)M ⊆ rad(M).

(ii) (Nakayama’s Lemma, second version) IfM is finitely generated and N ≤M is an A-submodule ofMsuch that N + Jac(A)M = M , then N = M .

63

Chapter 4

Proof. (i) Every simple A-module is annihilated by Jac(A), so if f : M → P is a morphism ofA-module where P is a simple left A-module, we see that f(Jac(A)M) = Jac(A)f(M) ⊆Jac(A)P = 0, which means Jac(A)M ⊆ ker f . Taking the intersection over all such f givesthe result.

(ii) This follows from Theorem 2.9 by considering the left A-module M/N , since Jac(A)(M/N) =M/N implies M/N = 0, i.e. N = M .

Theorem 4.17. (Snake’s lemma). Let

M1 N1 P1 0

0 M2 N2 P2

f1

α

g1

β γ

f2 g2

be a commutative diagram of left A-modules with exact rows. Then this diagram can be extended

kerα kerβ ker γ

M1 N1 P1 0

0 M2 N2 P2

cokerα cokerβ coker γ

f0 g0

δ

α β γ

f3 g3

and there exists a connecting homomorphism δ : ker γ → cokerα (“connecting” is nothing more but itsname, i.e. does not mean δ has any extra properties) such that the following sequence is exact :

kerα kerβ ker γ cokerα cokerβ coker γ.f0 g0 δ f3 g3

Furthermore, if f1 : M1 → N1 is injective, so is f0 : kerα → kerβ, and if g2 : N2 → P2 is surjective, so isg3 : cokerβ → coker γ.

Proof. We define δ as follows. Let p1 ∈ ker γ. Then there exists n1 ∈ N1 with g1(n1) = p1. Sinceg2(β(n1)) = γ(g1(n1)) = γ(p1) = 0, we have β(n1) ∈ ker g2 = im f2, so there exists m2 ∈ A2 withf2(m2) = β(n1). We let δ(p1) be the equivalence class of m2 in cokerα = M2/imα.The choice of n1 and m2 in the above construction is not unique, but the equivalence class of f2(m2) incokerα is. To see this, suppose n1, n

′1 are such that g1(n1) = p1 = g1(n′1). Then n1 − n′1 ∈ ker g1 =

im f1, so there exists m1 ∈ A1 with f1(m1) = n1 − n′1. Pick m2,m′2 such that f2(m2) = β(n1) and

f2(m′2) = β(n′1). Then

f2(m2 −m′2) = f2(m2)− f2(m′2) = β(n1)− β(n′1) = β(n1 − n′1) = β(f1(m1)) = f2(α(m1)).

Since f2 is injective, this means m2 ≡ m′2 (mod imα), showing that δ(p1) is well-defined.Onto exactness at ker γ. Assume p1 = g1(n1) with n1 ∈ kerβ. Then f2(m2) = β(n1) = 0, but f2 isinjective, so m2 = 0, which means δ(p1) = 0. Therefore δ g0 = 0. Now suppose δ(c1) = 0, and chooseb1 and a2 which defined δ(c1). This means a2 = α(a1) for some a1 ∈ A1, and since

β(n1) = f2(m2) = f2(α(m1)) = β(f1(m1)) =⇒ n1 − f1(m1) ∈ kerβ,

we see that g1(n1 − f1(m1)) = g1(n1) = p1, so ker δ = im g0.

64

Finally, exactness at cokerα. The composition f3δ is clearly zero ; picking c1 ∈ ker γ and correspondingn1,m2, we have

f3(δ(p1)) = f3(m2) = f2(m2) = β(n1) = 0.

Now assume m2 ∈ cokerα satisfies f3(m2) = 0. This means f2(m2) = 0, so there exists n1 ∈ B1 withβ(n1) = f2(m2). It follows that δ(g1(n1)) = m2 by definition of δ, so ker f3 = im δ.For left-exactness of the kernel, if M1 → N1 is injective, note that the maps kerα→M1 → N1 are bothinjective, hence f0 is injective. Similarly one proves that g3 is surjective if N2 → P2 is.

Corollary 4.18. Let A be a ring, M a left A-module and b E A a right ideal. The right ideal b and theleft A-moduleM can be multiplied to form the subgroup bM ofM , in which case we can take the quotientM/bM as abelian groups (bM is not a left A-submodule of M in general). We have an isomorphism ofabelian groups

(Ar/b)⊗AM 'M/bM.

Furthermore, if b is a two-sided ideal, then we have an isomorphism of left A-modules A/b⊗AM 'M/bM .

Proof. Consider the following commutative diagram of abelian groups :

kerα kerβ ker γ

b⊗AM Ar ⊗AM (Ar/b)⊗AM 0

0 bM M M/bM 0

cokerα cokerβ coker γ

δ

α β γ

The vertical maps α, β and γ are induced by the multiplication map A×M →M of the left A-moduleM . The second row is exact by construction and the first row is right-exact by the right-exactness of thetensor product (c.f. Theorem 1.51). It is clear that β is an isomorphism and α is surjective, so ker γ = 0 byexactness of the kernel-cokernel exact sequence from the Snake Lemma. Surjectivity of γ is also obvioussince the right square between the two sequences commutes, the map M → M/bM is surjective and βis an isomorphism.

Corollary 4.19. Let M be a left A-module.

(i) If M is finitely generated and b E A is a right ideal contained in Jac(A), then (Ar/b) ⊗A M = 0implies M = 0.

(ii) Let N be a finitely generated left A-module and b E A a right ideal contained in Jac(A). Iff : M → N is a morphism of A-modules (M is not assumed finitely generated) and the mapid ⊗ f : (Ar/b)⊗AM → (Ar/b)⊗A N is surjective, then f is surjective.

Proof. (i) Since Ar/b ⊗A M ' M/bM by Corollary 4.18, the equation (Ar/b) ⊗A M = 0 impliesJac(A)M = M , hence M = 0 by Theorem 2.9 (iii) since M is finitely generated.

(ii) It is clear that im (idAr/b ⊗ f) = Ar/b⊗ im f , so the surjectivity of im (idAr/b ⊗ f) implies

Ar/b⊗ im f = im (idAr/b ⊗ f) = Ar/b⊗N =⇒ Ar ⊗ (N/im f) = 0

65

Chapter 4

since we can tensor the following exact sequence with Ar/b :

0 im f N N/im f 0.

By part (i), since N is finitely generated, so does N/im f , which implies N/im f = 0, i.e. f issurjective.

Proposition 4.20. Let A be a ring and a E A be a two-sided ideal.

(i) We have the inclusion of two-sided ideals Jac(A/a) ⊇ (Jac(A) + a)/a in A/a.

(ii) If a ⊆ Jac(A), we have Jac(A/a) = Jac(A)/a.

(iii) The two-sided ideal Jac(A) is the smallest two-sided ideal a such that A/a is radical-free.

Proof. Parts (i) and (ii) are straightforward corollaries of Proposition 4.9. By part (i), if Jac(A/a) = 0, thenJac(A) + a = a, which means Jac(A) ⊆ a ; together with part (ii) which shows that Jac(A/Jac(A)) = 0,this proves part (iii).

Theorem 4.21. Let A be a ring and x ∈ A. The following are equivalent :

(i) x ∈ Jac(A)

(ii) For all a ∈ A, 1− ax is left-invertible, i.e. there exists ya,x ∈ A such that ya,x(1− ax) = 1.

Proof. This is actually a corollary of Proposition 4.12 applied to M = A` and m1 = 1 since Jac(A) =rad(A`), A` = A〈1〉 and the left-invertibility condition is equivalent to A〈1− ax〉 = A`.

Lemma 4.22. Let x ∈ A. If x is left-invertible and right-invertible, then x is invertible (i.e. we can pick aleft-inverse for x which is also a right-inverse). Furthermore, when this is the case, the inverse is unique ; wedenote the inverse by x−1.

Proof. Suppose y is a left-inverse and z is a right-inverse. Then

z = 1z = (yx)z = y(xz) = y1 = y.

Applying this to the case where both y and z are inverses for x, we obtain unicity of the inverse.

Corollary 4.23. Let A be a ring.

(i) The two-sided ideal Jac(A) is the largest two-sided ideal such that 1− x is invertible for all x ∈ A.

(ii) We have Jac(A) = Jac(Aopp) (c.f. Remark 4.14).

(iii) An element x ∈ A belongs to Jac(A) if and only if 1− ax is invertible for all a ∈ A.

(iv) Every nilideal (i.e. a left or right ideal consisting of nilpotent elements) is contained in Jac(A).

(v) If Aii∈I is a family of rings, then rad(∏i∈I Ai) =

∏i∈I Jac(Ai).

Proof. Let a E A be a two-sided ideal such that 1 − x is invertible for all x ∈ A. Then for all a ∈ A,ax ∈ a, hence 1 − ax is also invertible, and in particular left-invertible, which shows that a ⊆ Jac(A).

66

To finish the proof, it suffices to show that x ∈ Jac(A) implies that 1− x is invertible.

By assumption, there exists y1,x ∈ A such that y1,x(1 − x) = 1, i.e. 1 − y1,x = −y1,xx. It

follows that zxdef= 1 − y1,x = −y1,xx ∈ Jac(A). Therefore, we can pick y1,zx ∈ A such that

y1,zxy1,x = y1,zx(1 − zx) = 1. It follows that y1,x is both left-invertible and right-invertible, hence isinvertible and y−1

1,x = 1− x by Lemma 4.22.

The statement of part (i) being symmetric with respect to multiplication, we see that Jac(A) andJac(Aopp) are both characterized by part (i) and thus must be equal, which proves part (ii). Part (iii)follows from part (ii) and Theorem 4.21.

Let n be a nilideal and x ∈ n. Since ax ∈ n is nilpotent, 1−ax is invertible by Remark 4.4, which impliesax = 1− (1−ax) ∈ Jac(A) by part (iii). Finally, part (v) also follows from part (iii) since if ai, xi ∈ A, theI-tuple (1− aixi)i∈I = 1− (ai)i∈I(xi)i∈I is invertible if and only if each 1− aixi ∈ Ai is invertible.

Theorem 4.24. Let A be a ring and a E A a left ideal. The following are equivalent :

(i) a ⊆ Jac(A)

(ii) If M is a finitely generated left A-module, the equation aM = M implies M = 0.

Proof. By Proposition 4.7 and Theorem 2.9, if a ⊆ Jac(A) and M is a non-zero finitely generated leftA-module, we have rad(M) 6= M , hence aM ⊆ Jac(A)M ⊆ rad(M) 6= M . Conversely, if aM 6= Mfor every non-zero finitely generated left A-module, then aP 6= P for every simple left A-module. By thesimplicity of P , this means aP = 0, i.e. a ⊆ AnnA(P ), so a ⊆ Jac(A) by Proposition 4.15.

Example 4.25. (i) Let A be an integral domain and let Bdef= A[[x1, · · · , xn]] be the commutative ring

of formal power series in the n variables x1, · · · , xn. This is an integral domain, which can beproved by induction on n since B = A[[x1, · · · , xn−1]][[xn]] and if f, g ∈ A[[x]] satisfy fg = 0, thenwriting f =

∑n≥k fnx

n and g =∑

n≥` gnxn with fk, g` 6= 0, it follows that fkg` = 0, which is a

contradiction since A is assumed to be an integral domain. The group of units B× is the set of thosef ∈ B for which f(0, · · · , 0) ∈ A×. When A is a field, the set of f ∈ B for which f(0, · · · , 0) = 0form the unique maximal ideal of B, so that (B, Jac(B)) is a local ring. See [Commutative Algebra,Proposition 4.38] for explanations.

(ii) When (A,m) is a commutative local integral domain which is not a field (so that m = Jac(A) 6= 0),the field of quotients Q(A) is radical-free, so it is possible for a subring of a radical-free ring to havea non-zero radical (in constrast with the case of modules where N ≤M implies rad(N) ⊆ rad(M)).

(iii) Let A be an integral domain and I a set. The polynomial ring Bdef= A[xii∈I ] is radical-free. To see

this, pick f ∈ B \ 0 ; if g ∈ B has positive degree, then so does 1− fg, which cannot be invertible,hence f ∈ B \ Jac(B), which implies Jac(B) = 0.

(iv) Consider the subring A[x1, · · · , xn] ⊆ A[[x1, · · · , xn]]. We know that A[[x1, · · · , xn]] is a local ring,hence its radical ideal is the one described in part (i). Note that even though Jac(A[x1, · · · , xn]) = 0,we have

Jac(A[[x1, · · · , xn]]) ∩A[x1, · · · , xn] = (x1, · · · , xn)A[x1,··· ,xn] 6= 0.

In other words, it is not because a subring A of a ring B satisfies Jac(A) = 0 that A ∩ Jac(B) = 0.

(iv) Let K be a field, I a set and A ⊆ KI be a K-subalgebra. Then Jac(A) = 0. To see this, for everyi ∈ I , the map KI → K defined by f 7→ fi, when restricted to A, has as a kernel the maximal ideal

67

Chapter 4

mi E A of functions f : I → K vanishing at i ∈ I . The intersection of all those ideals is the set offunctions vanishing everywhere, thus equal to zero.

Proposition 4.26. Let A be a PID.

(i) A is radical-free if and only if it is a field or Spec (A) is infinite.

(ii) Since PIDs are UFDs, let x ∈ A be written as x = upi11 · · · pinn where p1, · · · , pn ∈ A are primeelements, i1, · · · , in ≥ 1 are integers and u ∈ A×. Then A/A(x) is radical-free if and only if x issquare-free, i.e. i1 = · · · = in = 1.

Proof. (i) Fields are clearly radical-free, so assume A is not a field, hence it admits at least one primeelement. The maximal ideals of A are of the form (p)A where p ∈ A is a prime element. Theelement x ∈ A belongs to Jac(A) =

⋂p (p)A if and only if it is divisible by each prime element p.

If x ∈ Jac(A) \ 0, this means x is divisible by all primes in A, so there must be finitely manyof them. Conversely, if Spec (A) = (p1)A , · · · , (pn)A is finite, then Jac(A) = (p1 · · · pn)A 6= 0(because PIDs are integral domains).

(ii) Write x =∏nj=1 p

ijj (we can assume u = 1 since (x)A = (ux)A). By the Chinese Remainder

Theorem (c.f. [Commutative Algebra, Theorem 2.17]), we have

A/ (x)A =

n∏j=1

A/ (pj)ijA .

We know by Corollary 4.23 (v) that this implies Jac(A/ (x)A) =∏nj=1 Jac(A/ (pj)

ijA), so we can

restrict to the case where x = pi is a prime power. By part (i), since Spec (A) = (p)A is finite, itis radical-free if and only if it is a field, i.e. i = 1.

Remark 4.27. Letting A = Z, we see that a radical-free ring can admit quotients with non-zero radical (inthis case, they are Z/nZ where n is a non-zero non-square-free integer) since Jac(Z/nZ) = mZ/nZ wherem is the square-free part of n, i.e. if n =

∏nj=1 p

ijj with ij ≥ 1, thenm =

∏nj=1 pj . Furthermore, Jac(Z) = 0

implies Jac(Z)M = 0 for all Z-modules M , even though rad(M) 6= 0 for some abelian groups (such as theexamples of Z/nZ above since their Jacobson radical is the same as their radical as a Z-module).

4.3 Radical of artinian rings/modules

Theorem 4.28. Let A be a left-artinian ring. Then Jac(A) is the largest two-sided nilpotent ideal of A.

Proof. Every two-sided nilpotent ideal of A is contained in Jac(A) by Corollary 4.23 (iv), so it suffices toshow that Jac(A) is nilpotent. The sequence Jac(A)nn∈N is decreasing, so since A is artinian, thereexists n0 ≥ 1 such that for all n ≥ 1, we have Jac(A)n0+n = Jac(A)n0 . Suppose Jac(A)n0 6= 0. Wededuce the existence of left ideals a E A such that Jac(A)n0a 6= 0 (since Jac(A) is such an ideal). Thefact that A is artinian implies that A` is an artinian left A-module, which means that every decreasingsequence of left ideals of A stabilizes. By Zorn’s Lemma, this proves the existence of a left ideal a0

minimal with respect to the condition that Jac(A)n0a0 6= 0. Since

Jac(A)n0(Jac(A)a0) = Jac(A)n0+1a0 = Jac(A)n0a0 6= 0,

we deduce that Jac(A)a0 ⊆ a0 is also a left ideal of A satisfying the required restriction, implying thatJac(A)a0 = a0. To obtain a contradiction, it suffices to show that a0 is finitely generated by Nakayama’sLemma (c.f. Theorem 2.9 (iii)) since it would imply a0 = 0 (and Jac(A)n0a0 6= 0). By assumption, there

68

exists a ∈ a0 such that Jac(A)n0A(a) 6= 0, so since A(a) ⊆ a0, this implies A(a) = a0 is generated by

one element.

Theorem 4.29. Let M be a left A-module. The following are equivalent :

(i) M is completely reducible and of finite length

(ii) M is completely reducible and finitely generated

(iii) M is artinian and radical-free.

Proof. The equivalence of (i) and (ii) is given by Corollary 2.58. We already know that a completelyreducible module of finite length is artinian by Corollary 2.20 and radical-free by Proposition 4.13.

Conversely, suppose M artinian, radical-free and non-zero (the result is trivial if M = 0). Consider theset of all finite intersections of maximal submodules of M . Since M 6= 0 and rad(M) = 0, this set isnon-empty, i.e. M admits at least one maximal submodule. Since it is artinian, every decreasing chain ofsuch finite intersections eventually stabilizes, so we can apply Zorn’s Lemma and find a minimal elementN0. If N is a maximal submodule of M , we have N ∩ N0 ⊆ N0, so by minimality of N0, this impliesN ∩ N0 = N0, i.e. N0 ⊆ N . So N0 equals the intersection of all maximal submodules of M since itis itself a finite intersection of maximal submodules of M , therefore N0 = rad(M) = 0. Pick maximalsubmodulesM1, · · · ,Mr ofM such that

⋂ri=1Mi = 0. The product of the projection mapsM →M/Mi

gives a morphism of A-modules f : M →∏ri=1M/Mi where each M/Mi is a simple left A-module, so

since∏ri=1M/Mi is a completely reducible left A-module of finite length, so does M ' f(M).

Corollary 4.30. The quotient of an artinian left A-module M by its radical rad(M), namely M/rad(M),is a completely reducible left A-module of finite length.

Proof. Indeed, this quotient is artinian by Proposition 2.5 and radical-free by Corollary 4.11, so we canapply Theorem 4.29.

Corollary 4.31. Let A be a ring. Then A is semisimple if and only if it is left-artinian and radical-free. (Inparticular, it is also right-artinian since Aopp is then semisimple.)

Proof. This follows from Corollary 4.30 applied to A` and the fact that when A is semisimple, A` is offinite length by Corollary 2.58 since it is completely reducible by definition.

Corollary 4.32. The quotient of a left-artinian ring A by its radical, i.e. A/Jac(A), is semisimple.

Proof. This follows straightforwardly from Corollary 4.31.

Corollary 4.33. A non-zero ring A is simple if and only if it is artinian and 0, A is its set of two-sidedideals. In other words, if a ring has 0, A as its set of two-sided ideals, then the following are equivalent :

(i) A is simple

(ii) A is semisimple

(iii) A is left-artinian.

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Chapter 4

Proof. By definition (c.f. Proposition 3.27), a non-zero ring A is simple if and only if it is semisimple andhas 0, A as its set of two-sided ideals. By Theorem 2.9, the condition on the set of two-sided ideals ofA implies that Jac(A) = 0 (because Jac(A) = A implies A = 0 and Jac(A) is a two-sided ideal), so Ais simple if and only if it is semisimple, which is equivalent to A being left-artinian by Corollary 4.31.

4.4 Modules over an artinian ring

Theorem 4.34. Let A be a left-artinian ring and M a left A-module. The following are equivalent :

(i) M is completely reducible

(ii) Jac(A)M = 0

(iii) AM is a semisimple ring.

As a corollary, a left-artinian ring has only finitely many isomorphism classes of simple left A-modules whichcan be put in a one-to-one correspondence with the simple components of the semisimple ring A/Jac(A).

Proof. Of course, (iii) implies (i) since every module over a semisimple ring is completely reducibleby definition (c.f. Proposition 3.18). To see that (i) implies (ii), it suffices to see that AnnA(M) is theintersection of the annihilator of a family of simple left A-modules and Jac(A)M is the intersection ofthe annihilators of all simple left A-modules, implying that Jac(A) ⊆ AnnA(M), e.g. Jac(A)M = 0.Finally, the equation Jac(A)M = 0 implies that AM = A/AnnA(M) is a quotient ring of A/Jac(A), sosince this ring is artinian and radical-free, it is semisimple by Corollary 4.31.

The last statement follows from Proposition 3.33 since for any simple left A-moduleM , we have Jac(A) ⊆AnnA(M), so the A-module structure ofM is the same as its A/Jac(A)-module structure, and so A andA/Jac(A) have the same isomorphism classes of simple modules.

Theorem 4.35. Let A be a ring possessing a two-sided nilpotent ideal n such that A/n is semisimple (for ex-

ample, if A is left-artinian, we can take ndef= Jac(A), c.f. Theorem 4.34 where we take M

def= (A/Jac(A))`).

For an arbitrary left A-module M , the following are equivalent :

(i) M is of finite length

(ii) M is artinian

(iii) M is noetherian.

Proof. ( (i)⇒ (ii),(iii) ) This follows by Corollary 2.20.

( (ii),(iii) ⇒ (i) ) To show that M is of finite length, since nr = 0 for some r ≥ 1, it suffices to consider

the modules Midef= niM/ni+1M for 0 ≤ i < r. Since Mi is an A/n-module (because n ⊆ AnnA(M))

and `A (Mi) = `A/n (Mi), it suffices to show that the A/n-module Mi has finite length. But A/n issemisimple, so Mi is a completely reducible A/n-module. Whether M is noetherian or artinian, thisimplies that Mi is a finite direct sum of simple A/n-modules, and therefore has finite length.

Corollary 4.36. Let A be a left-artinian ring andM a finitely generated left A-module. ThenM is of finitelength.

70

Proof. Since M is a quotient of the artinian left A-module A⊕n` , it is artinian, hence of finite length byTheorem 4.35.

Corollary 4.37. Let A be a left-artinian ring. Then A` has finite length, hence A is left-noetherian.

Proof. This follows from Theorem 4.35 and Corollary 4.36.

Corollary 4.38. In a commutative artinian ring, we have Jac(A) = Nil(A).

Proof. Since A is commutative, Nil(A) is an ideal of A. Because A is noetherian by Corollary 4.37, thisideal is finitely generated. Write Nil(A) = A(a1, · · · , an). Fix N such that aNi = 0 for all 1 ≤ i ≤ n. Forany b1, · · · , bn ∈ A, by the multinomial formula, we have

(b1a1 + · · ·+ bnan)M =∑

j1+···+jn=Mj1,··· ,jn≥0

(M

j1, · · · , jn

) n∏i=1

(biai)ji .

PickingM large enough so that each n-tuple non-negative integers (j1, · · · , jn) satisfying j1 + · · ·+ jn =M is such that at least one of the coefficients is larger than N , this sum always vanishes, showingthat Nil(A)M = 0, i.e. Nil(A) is a nilpotent ideal. Since Jac(A) is a nilpotent ideal of A, we haveJac(A) ⊆ Nil(A). By Theorem 4.28, we have equality.

Proposition 4.39. Let A be a commutative ring. The following are equivalent :

(i) A is artinian and Nil(A) = 0

(ii) A is artinian and radical-free

(iii) A is semisimple

(iv) There exist finitely many fields F1, · · · , Fn such that A '∏ni=1 Fi

(v) Every prime ideal in Spec (A) = m1, · · · ,mn is maximal and we have A '∏ni=1A/mi.

Proof. The equivalence of (i) and (ii) follows from Corollary 4.38 ; that of (ii) and (iii) follows fromCorollary 4.31. When A is semisimple, its simple components are fields, so since A is a direct product ofits simple components, (iii) implies (iv) ; the converse is obvious. The implication (v) ⇒ (iv) is clear by

taking Fi = A/mi. Conversely, it suffices to see that letting A =∏ni=1 Fi and mi

def=∏nj=1j 6=i

Fj , we have

Spec (∏ni=1 Fi) = m1, · · · ,mn and A '

∏ni=1A/mi.

Corollary 4.40. Let K be a field and A a commutative K-algebra which is radical-free (or equivalently,for which Nil(A) = 0). If B ⊆ A is a K-subalgebra which is finite-dimensional over K , then B is artinianand isomorphic to a finite direct product of finite field extensions of K . In particular, if K is algebraicallyclosed, then B ' Kn.

Proof. Since B is finite-dimensional over K , it has to be artinian (since a decreasing chain of ideals leadsto a decreasing chain of K-subspaces). It is also radical-free since Jac(B) = Nil(B) ⊆ Jac(A) = 0because any nilpotent element of B generates a nilpotent ideal of A, e.g. a nilideal of A ; such ideals arecontained in Jac(A) = 0 by Corollary 4.23 (iv). If Spec (B) = m1, · · · ,mn, the field extensions B/mi

are finite over K because B is. In particular, when K is algebraically closed, B/mi ' K .

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Chapter 4

Corollary 4.41. Let K be an algebraically closed field and A a finite-dimensional commutative K-algebra.Then every simple A-module is one-dimensional over K .

Proof. Every simple A-module M can be considered as a simple A-module over A/Jac(A). SinceA/Jac(A) is artinian and radical-free, we can assume without loss of generality that A is semisimple. Itfollows that M is a simple module over one of the simple components of A ; these are all isomorphic toK by Corollary 4.40, hence dimKM = 1.

We finish this section by considering the following application.

Theorem 4.42. LetK be a field and V be a finite-dimensionalK-vector space. For a subset S ⊆ EndK(V ),let A ⊆ EndK(V ) be the K-subalgebra generated by the set S (note that idV ∈ A by definition). SinceEndK(V ) is a finite K-algebra, A also is, hence A is an artinian K-algebra by Example 2.27. Furthermore,the following are equivalent :

(i) The K-algebra A is semisimple

(ii) The left A-module V is completely reducible

(iii) IfW ≤ V is a subspace such that for any ϕ ∈ S, ϕ(W ) ⊆W (or in other words, a subspaceW whichis stable under S), there exists a complement W ′ ≤ V such that V = W ⊕W ′ and W ′ is also stableunder S ; this is a direct sum of A-modules.

Proof. The equivalence of (i) and (ii) follows by Theorem 4.34 since idV ∈ A, hence V is a faithfulA-module (meaning that the inclusion map A ⊆ EndK(V ) induces the isomorphism A ' AV ). Theequivalence of (ii) and (iii) is clear since subspaces of V stable under S are the same as A-submodules.

Definition 4.43. Let K be a field and V a finite-dimensional K-vector space.

(i) A collection of endomorphisms S ⊆ EndK(V ) satisfying one of the equivalent conditions of Theo-rem 4.42 is called semisimple.

(ii) An endomorphism ϕ ∈ EndK(V ) is said to be semisimple if ϕ is semisimple.

(iii) If ϕii∈I is a family of endomorphisms of V , we say it is semisimple if the subset S ⊆ EndK(V ) ofall endomorphisms s ∈ EndK(V ) satisfying s = ϕi for some i ∈ I is semisimple.

Example 4.44. Let K be a field and V be a finite-dimensional K-vector space. An endomorphism ϕ ∈EndK(V ) is semisimple if and only if its minimal polynomial mϕ,K(t) ∈ K[t] is separable, i.e. factors as aproduct of distinct irreducible factors. To see this, write mϕ,K(t) =

∏mi=1 fi(t)

ni where f1(t), · · · , fm(t) ∈K[t] are distinct irreducible polynomials and ni ≥ 1. By the Chinese Remainder Theorem,

K[t]/(f(t)) 'n∏i=1

K[t]/(fi(t)ni),

and note that the latter is semisimple if and only if it has no nilpotent elements by Proposition 4.39, i.e. ifand only if n1 = · · · = nm = 1.

Theorem 4.45. Let K be a field and V be a finite-dimensional K-vector space. Suppose S ⊆ EndK(V )consists of commuting endomorphisms, i.e. ϕ ψ = ψ ϕ for all ϕ,ψ ∈ S. If S is semisimple and T ⊆ S,then T is semisimple. In particular, every ϕ ∈ S is semisimple in this case.

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Proof. The K-algebra A ⊆ EndK(V ) generated by S is commutative, semisimple and finite-dimensionalover K . It follows by Proposition 4.39 that it is artinian and has no nilpotents, so any K-subalgebra ofA is also artinian and has nilpotents, i.e. is semisimple by the same result. By Theorem 4.42, this meansthat T is semisimple (by considering the K-subalgebra of A generated by T ). The case of ϕ ∈ S is dealtwith by taking T = ϕ.

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Chapter 5

Tensor products of K-algebras

In this chapter, K is a field, A and B are K-algebras, M is a left A-module and N is a left B-module.

5.1 Modules over tensor products

Definition 5.1. In what follows, we will often consider a given abelian group as a module over differentrings and look at its radical. To distinguish between the radicals over its different module structures, givena left A-module M , we write radA(M) for the radical of the left A-module M .

We begin with a lemma which might seem irrelevant now, but to prevent confusion with future notation,we prove it now and use it later.

Lemma 5.2. Let C be a ring and B ⊆ C be a subring which turns C into a free right B-module. Let N bea left B-module and note that the left C ⊗B B-module C ⊗B N can be seen as a left C-module (resp. leftB-module) by multiplication on the first factor (resp. on the second factor, or using the canonical C ⊗B B-module structure of C⊗BN and using the isomorphism C⊗B B ' C). Define the map ιN : N → C⊗BNby ιN (n)

def= 1⊗ n, which is B-linear. We then have

ι−1N (radC(C ⊗B N)) ⊆ radB(N).

Proof. We first deal with the case where N is a simple left B-module, in which case radB(N) = 0,thus we have to show that ι−1

N (radC(C ⊗B N)) = 0. Let n ∈ N \ 0 ; we need to show thatιN (n) = 1 ⊗ n /∈ radC(C ⊗B N). By simplicity of N , we have B〈n〉 = N , hence C〈1⊗ n〉 = C ⊗B N .We know that C ⊗B N 6= 0 by the assumption that C is a free right B-module. The equalityC〈1⊗ n〉 = C ⊗B N also implies that C ⊗B N is a finitely generated left C-module, hence

radC(C ⊗B N) 6= C ⊗B N by Proposition 4.7. We deduce that 1 ⊗ n /∈ radC(C ⊗B N), proving thatι−1N (radC(C ⊗B N)) = 0.

Now suppose N arbitrary and let g : N → P be a morphism of left B-modules where P is simple. Wehave the following commutative diagram :

N P

C ⊗B N C ⊗B P

g

ιN ιP

id⊗g

By the first case, since radB(P ) = 0, the first paragraph shows that ι−1P (radC(C ⊗B P )) = 0. Let n ∈ N

74

be such that ιN (n) ∈ radC(C ⊗B N). By using the commutativity of the square and Proposition 4.8,

ιP (g(n)) = (1⊗ g)(ιN (n)) ∈ (id ⊗ g)(radC(C ⊗B N)) ⊆ radC(C ⊗B P ) =⇒ g(n) = 0.

Since g was arbitrary, this means n ∈ radB(N), showing that ι−1N (radC(C ⊗B N)) ⊆ radB(N).

Definition 5.3. Let A, B be K-algebras, M be a left A-module and N a left B-module. It is understoodthat all four of A,B,M and N are K-vector spaces via their module/algebra structure. We understand thatthe algebra structures turn each of A,B,M and N into left K-modules, hence (K,K)-bimodules sinceK = Kop, which allows us to take tensor products without any left/right-related issues.

The tensor product of the K-algebras A and B is the K-vector space A⊗K B whose multiplicationis given by the K-multilinear map

A×B ×A×B → A⊗K B, (a, b, a′, b′) 7→ aa′ ⊗ bb′.

This extends to a K-bilinear map (A⊗K B)× (A⊗K B)→ A⊗K B and turns A⊗K B into a K-algebra.More generally, the K-multilinear map

A×B ×M ×N →M ⊗K N, (a, b,m, n) 7→ (am)⊗ (bn)

turns M ⊗K N into a left (A⊗K B)-module. Furthermore, we have canonical inclusion maps

ιAA,B : A→ A⊗K B, a 7→ a⊗ 1

and similarly, we haveιBA,B : B → A⊗K B, b 7→ 1⊗ b.

Because A and B are tensored over K which is a field, these maps are injective (the tensor products ofbases for A and B is a basis for A ⊗K B ; this is pure linear algebra, no theory of non-commutative ringsneeded). In particular, if M = A`, we see that A⊗K N is canonically endowed with an (A⊗K B)-modulestructure. We note that by Proposition 1.53, we have a natural isomorphism of (A⊗K B)-modules

A⊗K N ' A⊗K (B ⊗B N) ' (A⊗K B)⊗B N, a⊗ n 7→ (a⊗ 1)⊗ n

where A⊗K B is seen as a right B-module via the morphism of rings ιBA,B , so we can think of A⊗K N asbeing obtained by N via extension of scalars along the map ιBA,B : B → A⊗K B. We shall often need themap

ιA,N : N → A⊗K N, n 7→ 1⊗ n.

For the same reasons as with ιAA,B and ιBA,B , this map is injective. Finally, if N = B`, we note that the left(A⊗K B)-module structures on A⊗K (B`) and (A⊗K B)` coincide.

Remark 5.4. One can deduce that if N ′ ≤ N is a B-submodule of the left B-module N , then

(A⊗K N ′) ∩ ιA,N (N) = ιA,N (N ′) = ιA,N ′(N′)

using the following commutative diagram of injective maps :

N ′ N

A⊗K N ′ A⊗K N

⊆

ιA,N′ ιA,N

⊆

(The exactness of the tensor product A⊗K (−) (which gives the injectivity of the bottom map) follows fromthe fact that K is a field, so that any K-module is free, hence flat.) Another way to obtain the result isto extend the linearly independent set 1 ⊆ A to a K-basis of A, so that ιA,N (N) corresponds to thoseelements of A⊗K N which can be written in the form 1⊗ n = ιA,N (n) for some n ∈ N .

75

Chapter 5

Proposition 5.5. Let A and B be two K-algebras, M a left A-module and N a left B-module. Recall thatA⊗K N is a left (A⊗K B)-module (c.f. Definition 5.3). The equality

ι−1A,N (radA⊗KB(A⊗K N)) = radB(N)

holds in the following three cases :

(i) the K-algebra A is a finite-dimensional K-vector space

(ii) the left B-module N is finitely generated and there exists an increasing sequence Aii∈N of K-subalgebras of A which are finite-dimensional over K , e.g. Ai ⊆ Ai+1 and

⋃i≥1Ai = A (for instance,

when A is a field and the extension A/K is algebraic)

(iii) N = B` and Jac(B) = radB(N) is a nilpotent two-sided ideal of B (we already know it is a two-sidedideal by Proposition 4.15, and it is nilpotent when B is left-artinian for instance, c.f. Theorem 4.28).

Proof. Fix Cdef= A ⊗K B, so that B can be seen as a subring of C by using the injective morphism of

rings ιBA,B : B → A ⊗K B = C . By Lemma 5.2, the inclusion (⊆) holds, so it suffices to establish thereverse inclusion (⊇).

(i) Let g : A⊗KN → P be a morphism of (A⊗KB)-modules where P is a simple (A⊗KB)-module.It suffices to show that g|radA⊗KB(A⊗KN) = 0 to deduce that radB(N) ⊆ ι−1

A,N (radA⊗KB(A⊗KN)).

By restriction of scalars, we can consider P as a left B-module, which we denote by P[B].The composition g ιA,N : N → A ⊗K N → P[B] is a morphism of B-modules, hence(g ιA,N )(radB(N)) ⊆ radB(P[B]) by Proposition 4.8. Since ιA,N is injective, it remains to showthat radB(P[B]) = 0.

We note that since the subsets ιAA,B(A) and ιBA,B(B) of A⊗K B commute with one another :

(a⊗ 1)(1⊗ b) = a⊗ b = (1⊗ b)(a⊗ 1),

we have the inclusion AP ⊆ EndB(P[B]), i.e. multiplication by an element a ∈ A in P is an endo-morphism of B-modules of P[B]. It follows by Proposition 4.8 that rad(B)P[B] is a B-submoduleand an A-submodule of P[B], thus an (A⊗KB)-submodule of P . Since P is simple and A is finite-dimensional over K , the B-module P[B] is finitely generated (if p ∈ P satisfies A⊗KB〈p〉 = P ,fix a basis a1, · · · , an of A over K so that B〈(a1 ⊗ 1)p, · · · , (an ⊗ 1)p〉 = P[B]). We deduceradB(P[B]) 6= P[B] by Proposition 4.7, so the simplicity of the (A ⊗K B)-module P implies thatthe proper (A⊗K B)-submodule radB(P[B]) is zero.

(ii) Suppose B〈n1, · · · , nk〉 = N and take an increasing sequence Ann∈N as stated in the assump-tions. Let n ∈ radB(N), so that we have to show that 1 ⊗ n ∈ radA⊗KB(A ⊗K N). By right-exactness of the tensor product, we see that A⊗KB〈1⊗ n1, · · · , 1⊗ nk〉 = A ⊗K N is finitelygenerated, so by Proposition 4.12, it suffices to show that for any c1, · · · , ck ∈ A⊗K B, we have

A⊗KB〈1⊗ n1 + c1(1⊗ n), · · · , 1⊗ nk + ck(1⊗ n)〉 = A⊗K N.

However, the elements c1, · · · , ck are all contained within some Ai for a given i large enough, sothat attempting to prove the same result for Ai instead of A can be done via part (i) since Ai isfinite-dimensional over K because

n ∈ radB(N) = ι−1A,N (radAi⊗KB(Ai ⊗K N))

implies that

Ai⊗KB〈1⊗ n1 + c1(1⊗ n), · · · , 1⊗ nk + ck(1⊗ n)〉 = Ai ⊗K N.

76

by Proposition 4.12 applied again. Writing A⊗K N = A⊗Ai (Ai⊗K N), the right-exactness of thetensor product gives us the desired equality by applying the functor A⊗Ai (−) to the previous one.

(iii) We now want to show that

radB(B`) = Jac(B) ⊆ (ιBA,B)−1(Jac(A⊗K B)) = ι−1B,B`

(radA⊗KB(A⊗K B`)).

by assuming that Jac(B)k = 0 for some k ≥ 1. Let Jdef= A⊗KB〈Jac(B)〉A⊗KB

be the two-sidedideal of A⊗K B generated by Jac(B). A generic element of J will have the form

∑∗i ai⊗ bi where

the ∗ indicates that the sum has finitely many terms, ai ∈ A and bi ∈ Jac(B) since elements ofιAA,B(A) and ιBA,B(B) commute. It follows that Jk is generated as a two-sided ideal of A⊗K B byproducts of k elements of ιBA,B(B), hence Jk = 0. By Corollary 4.23, it follows that

ιBA,B(Jac(B)) ⊆ J ⊆ Jac(A⊗K B) =⇒ Jac(B) ⊆ (ιBA,B)−1(Jac(A⊗K B)).

Corollary 5.6. Let A,B be two K-algebras such that B is left-artinian. If A⊗K B is radical-free, then Bis semisimple.

Proof. We see that Jac(B) = 0 by Proposition 5.5, so B is left-artinian and radical-free. By Corollary 4.31,it is semisimple.

Remark 5.7. Given a ring A, we can consider the group of ring automorphisms of A under composition ;we denote it by Aut(A). We always have a morphism of groups

ϕ : A× → Aut(A), a 7→ (ϕa : x 7→ axa−1).

(The check that ϕa respects addition, multiplication and unit element is trivial.) This morphism is trivialif and only if A× ⊆ Z(A). In particular, if A is a division ring, then the set of elements x ∈ A such that

ϕa(x) = x for all a ∈ A is precisely Z(A). Another way to see this last statement is that Gdef= A× acts on

A via ϕ, and the set of fixed points is AG = Z(A). In the case where A is a central K-algebra which is adivision ring, this means AG = K . We also call AG the ring of invariants of G acting on A.

Proposition 5.8. Let K be a field, A be a K-algebra which is a division ring and B a K-algebra. For anyleft B-module N , the equality

radA⊗KB(A⊗K N) = A⊗K ι−1A,N (radA⊗KB(A⊗K N))

holds in the two following cases :

(i) There exists a group action G A acting by ring automorphisms (i.e. is given by a morphism ofgroups ϕ : G → Aut(A)) such that K = AG is the ring of elements left invariant by G. (ByRemark 5.7, this is the case if A is a central K-algebra.)

(ii) The B-module N is finitely generated and the division ring A is a field such that the extension A/Kis separable (algebraic or transcendental).

Proof. Without loss of generality, assume K is a subring of A and that the K-algebra structure mapK → A is given by inclusion, so that in case (i), we have AG = K .

(i) Note that if N ′ ≤ N is a B-submodule such that radA⊗KB(A ⊗K N) = A ⊗K N ′ ⊆ A ⊗K N ,then ι−1

A,N (radA⊗KB(A ⊗K N)) = N ′. So it suffices to prove the existence of a B-submodule N ′

77

Chapter 5

such that radA⊗KB(A ⊗K N) = A ⊗K N ′. By Corollary 3.16, since AG = K , it suffices to showthat radA⊗KB(A⊗K N) is stable under every g⊗ idN ∈ EndB(A⊗K N) (flip the factors tensoredin Corollary 3.16 to obtain the required statement!).

Consider the group action G

A acting by K-linear automorphisms of A and pick g ∈ G. Theelement g⊗ idN acts by B-linear automorphisms on the (A⊗KB)-module A⊗KN by the formula

(g⊗ idN )(a⊗ n)def= g(a)⊗ n. It follows that if P ≤ A⊗K N is an (A⊗K B)-submodule, so does

(g ⊗ idN )(P ) :

a(g ⊗ idN )

( ∗∑i

ai ⊗ ni

)=

∗∑i

ag(ai)⊗ ni

=

∗∑i

g(g−1(a)ai)⊗ ni

= (g ⊗ idN )

(g−1(a)

( ∗∑i

ai ⊗ ni

)).

(The latter shows that the B-submodule (g ⊗ idN )(P ) is also an A-submodule of A⊗K N , hencean (A ⊗K B)-submodule.) The action of G on A ⊗K N thus permutes the set of (A ⊗K B)-submodules of A⊗K N in an inclusion-preserving way, so that it also permutes the set of maximal(A ⊗K B)-submodules of A ⊗K N . This implies that radA⊗KB(A ⊗K N) is stable under theaction of G since it is the intersection of all those maximal submodules.

(ii) Let N ′def= ι−1

A,N (radA⊗KB(A⊗K N)). It is clear that radA⊗KB(A⊗K N) ⊇ A⊗K N ′, so it suffices

to show that radA⊗KB(A⊗K N) ⊆ A⊗K N ′. Pick zdef=∑n

i=1 ai ⊗ ni ∈ radA⊗KB(A⊗K N), sothat we wish to show that ni ∈ N ′.

Let Ω be an algebraic closure for A and σ ∈ Aut(Ω/K). The automorphism σ⊗ idN ∈ Aut(Ω⊗KN) can be restricted to an isomorphism of B-modules σ ⊗ idN : A ⊗K N → σ(A) ⊗K N : itbecomes an isomorphism of A⊗K B-modules if we see σ(A)⊗K N as a σ(A)⊗K B-module andtwist this module structure via the isomorphism σ ⊗ idB : A⊗K B → σ(A)⊗K B. It follows that

(σ ⊗ idN )(z) ∈ radA⊗KB(σ(A)⊗K N) = radσ(A)⊗KB(σ(A)⊗K N).

(The latter equality holds because the σ(A) ⊗K B-submodules of σ(A) ⊗K N and its A ⊗K B-submodules agree.) We have an isomorphism of K-algebras

Ω⊗K B ' Ω⊗σ(A) (σ(A)⊗K B),

so since the extension Ω/σ(A) is algebraic, we see that (σ ⊗ idN )(z) ∈ radΩ⊗KB(Ω ⊗K N) byProposition 5.5 (ii) ; to see this, let the extension Ω/σ(A) play the role of A/K in Proposition 5.5,so that

radΩ⊗KB(Ω⊗K N) = Ω⊗σ(A)

(radσ(A)⊗KB(σ(A)⊗K N)

).

(The K-algebra σ(A) ⊗K B plays the role of B in Proposition 5.5 and the role of N is playedby σ(A) ⊗K N which is a finitely generated σ(A) ⊗K B-module since N is a finitely generatedB-module.)

By [Commutative Algebra,Theorem 13.52], there exists σ1, · · · , σn ∈ Gal(Ω/K) such that thematrix (σi(aj))ij ∈ Matn×n(Ω) is invertible : let (µij)ij ∈ Matn×n(Ω) be its inverse. We see that

78

for each 1 ≤ i ≤ n,

n∑j=1

µij(σj ⊗ idN )(z) =

n∑j=1

µij(σj ⊗ idN )

(n∑k=1

ak ⊗ nk

)=

n∑j,k=1

µijσj(ak)

⊗ nk = 1⊗ ni

is an element of radΩ⊗KB(Ω⊗KN). By Lemma 5.2, since A⊗KN is a finitely generated (A⊗KB)-module and the extension Ω/A is algebraic, considering the map ιΩ,N extending the (A ⊗K B)-module A⊗K N to an (Ω⊗K B)-module Ω⊗K N , we have

1⊗ ni ∈ ι−1Ω,N (radΩ⊗KB(Ω⊗K N)) = radA⊗KB(A⊗K N),

or in other words, ni ∈ N ′, which completes the proof.

Corollary 5.9. Let K be a field, A be a K-algebra which is a division ring containing K in its center andB a K-algebra. For any left B-module N , the inclusion

radA⊗KB(A⊗K N) ⊆ A⊗K radB(N)

holds in the two following cases :

(i) There exists a group action G

A acting by ring automorphisms (i.e. is given by a morphism ofgroups ϕ : G → Aut(A)) such that K = AG is the ring of elements left invariant by G. (ByRemark 5.7, this is the case if A is a central K-algebra.)

(ii) The B-module N is finitely generated and the division ring A is a field such that the extension A/Kis algebraic and separable.

Proof. This is clear by the proposition and Lemma 5.2 (which applies because K is a field, so A is a freeK-module).

Corollary 5.10. Let A/K be a separable field extension and B a K-algebra. For any left B-module N , theequality

radA⊗KB(A⊗K N) = A⊗K radB(N)

holds in the three following cases :

(i) The field extension A/K is finite

(ii) The field extension A/K is algebraic and N is a finitely generated B-module

(iii) The ideal rad(B) is a nilpotent ideal of B and N = B`.

Proof. The cases (ii) and (iii) are straightforward from Proposition 5.5 and Proposition 5.8. As for thecase (i), the result is certainly true if the extension A/K is Galois (by Proposition 5.5 and Proposition 5.8again). If A/K is not necessarily Galois, let E be a normal closure for the finite separable extensionA/K in some algebraic closure for A, so that E/K is finite and Galois. By the Galois case of part (i), wehave

radE⊗KB(E ⊗K N) = E ⊗K radB(N).

Since E/A is finite, by Proposition 5.5, we also have

ι−1E,N (radE⊗A(A⊗KB)(E ⊗A (A⊗K N))) = radA⊗KB(A⊗K N)

79

Chapter 5

Identifying the functors E ⊗K (−) and E ⊗A (A⊗K (−)), we see that

radA⊗KB(A⊗K N) = ι−1E,N (radE⊗KB(E ⊗K N)) = ι−1

E,N (E ⊗K radB(N)) = A⊗K radB(N).

5.2 Tensor product of K-fields

In this section, K is a field. A K-field is a field E which is a K-algebra, so the data is equivalent to that ofa field extension E/K .

Proposition 5.11. Let E,F be two K-fields. Then Jac(E ⊗K F ) = Nil(E ⊗K F ).

Proof. Note that if A is any commutative ring, we have

Nil(A) =⋂

p∈Spec(A)

p ⊆⋂

m∈MaxSpec(A)

m = rad(A).

For the reverse inclusion, let Ω be an algebraic closure for E and recall that Kp−∞ is the perfect closureof K in Ω. Suppose ch(K) = p > 0 for the moment. Note that by applying Proposition 5.5 (iii) byreplacing (A,K,B) by (Ω, E,E ⊗K F ), we obtain

ι−1Ω,E⊗KF

(Jac(Ω⊗K F )) = Jac(E ⊗K F )

henceΩ⊗E Jac(E ⊗K F ) ⊆ Jac(Ω⊗E (E ⊗K F )) = Jac(Ω⊗K F ).

By Proposition 5.8 applied by replacing (A,K,B) by (Ω,Kp−∞ ,Kp−∞⊗KF ), since ΩAut(Ω/K) = Kp−∞

(c.f. [Commutative Algebra, Theorem 13.16 (iv)]), we obtain

Ω⊗E Jac(E ⊗K F ) ⊆ Jac(Ω⊗K F ) ⊆ (Jac(Kp−∞ ⊗K F ))Ω⊗KF .

If Jac(Kp−∞⊗KF ) ⊆ Nil(Kp−∞⊗KF ), this will imply the nilpotency of every element of Jac(E⊗KF ).Therefore, without loss of generality, it suffices to prove the result when E = Kp−∞ . By repeatingthe argument with F instead of E, it suffices to show that Jac(Kp−∞⊗KKp−∞) ⊆ Nil(Kp−∞⊗KKp−∞).

Let x =∑n

i=1 ai⊗ bi ∈ Jac(Kp−∞⊗KKp−∞) where ai, bi ∈ Kp−∞ =⋃m≥1K

p−m. Fix m large enough

so that ai, bi ∈ Kp−mfor all 1 ≤ i ≤ n. It follows that

xpm

=n∑i=1

apm

i ⊗ bpm

i ∈ K.

It follows that xpm ∈ K \K× = 0 because no element of the Jacobson radical can be a unit, hence x

is nilpotent.

(When ch(F ) = 0, the same proof works but one replaces Kp−∞ by K and the ring Kp−∞ ⊗K Kp−∞

just becomes K , so the result is trivial.)

Theorem 5.12. Let K be a field and E,F be two K-fields. The following are equivalent :

(i) The field extension E/K is separable, i.e. if L is a K-field containing E and Kp−1, the map

E ⊗K Kp−1 → L,

80

given by multiplication, is injective

(ii) For every K-field F , Jac(E ⊗K F ) = 0

(iii) For every K-field F , Nil(E ⊗K F ) = 0

(iv)∗ Nil(E ⊗K Kp−1) = 0.

(∗) This statement only makes sense in characteristic p. For characteristic zero, the statements (i) to (iii) arealways true.

Proof. ( (i)⇒ (ii) ) This follows by Corollary 5.9 since it implies

Jac(E ⊗K F ) ⊆ E ⊗K Jac(F ) = E ⊗K 0 = 0.

( (ii) ⇐⇒ (iii) ) This is clear by Proposition 5.11.

( (iii)⇒ (iv) ) This is obvious.

( (iv) ⇒ (i) ) Suppose that there exists x1, · · · , xn ∈ E and a1, · · · , an ∈ Kp−1such that

∑ni=1 xiai = 0

in L. Raising to the pth power gives∑n

i=1 xpi api = 0, an equation in E since api ∈ K . This implies(

n∑i=1

xi ⊗ ai

)p=

n∑i=1

xpi ⊗ api =

(n∑i=1

xpi api

)⊗ 1 = 0,

hence∑n

i=1 xi ⊗ ai ∈ Nil(E ⊗K Kp−1) = 0, i.e. E and Kp−1

are linearly disjoint over K , so thatE/K is separable.

Corollary 5.13. Let E1, · · · , En be K-fields. Assume that n − 1 of those fields are separable over K and

that n− 1 of them are finite over K (possibly different ones). Then⊗n

i=1Eidef= E1 ⊗K E2 ⊗K · · · ⊗K En

is isomorphic to a direct product of a finite number of K-fields. Furthermore, if E1, · · · , En are finiteextensions of K , the K-fields appearing in the product are also finite extensions of K .

Proof. We argue by induction on n ≥ 2. For n = 2, if E1 is separable, we know that Jac(E1⊗KE2) = 0.One of those fields is a finite extension, hence E1⊗KE2 is a finite algebra over a field. By Corollary 4.40,since E1 ⊗K E2 is radical-free, it is a direct product of fields which are finite over either E1 or E2

(depending on of E2 or E1 is a finite extension of K). If both are finite extensions of K , so do the directfactors.

If n > 2, we can assume that E3, · · · , En are finite separable extensions of K , so by induction on n,we see that

⊗ni=2Ei is a direct product of a finite number of K-fields. Since tensor products and direct

products commute, we conclude by the case where n = 2 on each direct factor.

Example 5.14. (Some separable field theory) Let E/K be a field extension, f(t) ∈ K[t] be an irreducible

polynomial and Fdef= K[t]/(f(t)). When E or F is separable overK , theK-algebra E⊗KF ' E[t]/(f(t))

is isomorphic to a direct product of fields by Corollary 5.13. If f(t) =∏ni=1 fi(t) is the factorization of f(t)

over E[t] into distinct irreducible factors, then the isomorphism of Corollary 5.13 gives

E[t]/(f(t)) 'n∏i=1

E[t]/(fi(t)).

To see this, note that if E[t]/(f(t)) '∏ni=1 Fi are the fields appearing in Corollary 5.13, the projection to Fi

gives a surjective morphism E[t] → Fi, so Fi ' E[t]/(fi(t)) for some irreducible polynomial fi(t) ∈ E[t].

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Chapter 5

This gives the equality (f(t))E[t] =⋂ni=1(fi(t))E[t] between ideals of E[t]. Note that f(t) divides

∏ni=1 fi(t)

since the latter annihilates E[t]/(f(t)), hence∏ni=1 fi(t) ∈ (f(t))E[t]. Since deg f =

∑ni=1 deg fi by the

isomorphism E[t]/(f(t)) '∏ni=1E[t]/(fi(t)), this means f(t) =

∏ni=1 fi(t) (by assuming that f(t) and

each fi(t) are monic polynomials).

Note that we already knew this when F/K was separable (i.e. when f(t) was an irreducible separablepolynomial). The new statement here is that this is also true when E is separable, regardless of whetherf(t) is separable or not ; in other words, to factor an irreducible inseparable polynomial into inseparablefactors, you need to use an inseparable field extension, otherwise the irreducible factors you will obtain willappear with multiplicity 1.

5.3 Tensor product of completely reducible modules

In this section, K is a field, A,B are two K-algebras, M is a left A-module and N is a right B-module.

Definition 5.15. For a left (resp. right) A-module M , let LsubmodA(M) (resp. RsubmodA(M)) denotethe set of left (resp. right) A-submodules of M . Let B be a ring and N a left B-module.

(i) A map Φ : LsubmodA(M) → LsubmodB(N) is said to preserve the radical if Φ(radA(M)) =radB(N).

(ii) A map Φ : LsubmodA(M)→ LsubmodB(N) is said to preserve direct sums if for two submodulesM1,M2 ∈ LsubmodA(M), we have Φ(M1 ⊕M2) = Φ(M1)⊕ Φ(M2).

(iii) A map Φ : LsubmodA(M)→ LsubmodB(N) is said to be inclusion-preserving if for two submod-ules M1,M2 ∈ LsubmodA(M) with M1 ⊆M2, we have Φ(M1) ⊆ Φ(M2).

Analogous definitions apply if we replace LsubmodA(M) or LsubmodB(N) by their right counterpartsRsubmodA(M) and RsubmodB(N).

Lemma 5.16. Let A and B beK-algebras andN a simple left B-module. Interpret the right CB(N)-moduleCB(N)r as a (CB(N)opp, CB(N))-bimodule. For every left A-module M , consider the isomorphism ofA⊗K B-modules

ϕN : M ⊗K CB(N)r ⊗CB(N) N →M ⊗K N, m⊗ ϕ⊗ n 7→ m⊗ ϕ(n).

The following map is a bijection between the set of left (A⊗K CB(N)opp)-submodules of M ⊗K CB(N)rand the set of left (A⊗K B)-submodules of M ⊗K N :

Φ : LsubmodA⊗KCB(N)opp(M ⊗K CB(N)r)→ LsubmodA⊗KB(M ⊗K N),

P 7→ ϕN (P ⊗CB(N) N).

Furthermore, this map preserves direct sums, inclusions and the radical.

Proof. We wish to apply Corollary 3.15. In this result, make the following substitutions :

(P,A,D, V,N) 7→ (N,B,CB(N),M ⊗CB(N) CB(N)r, P ).

Furthermore, replace the subset S ⊆ EndD(V ) = EndCB(N)(M ⊗CB(N) CB(N)r) in the result by theset of homotheties of the left A-module M ⊗CB(N) CB(N)r, which are right CB(N)-linear since this isa left A ⊗K CB(N)opp-module. It follows that there is a bijection between the set of right CB(N)opp-submodules of M ⊗CB(N) CB(N)r stable under left multiplication by elements of A and the set ofB-submodules of M ⊗CB(N) CB(N)opp ⊗K N stable under those same multiplications. In other words,

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the applicationP 7→ P ⊗CB(N) N

puts in bijection the (A⊗KCB(N)opp)-submodules ofM⊗CB(N)CB(N)r and the (A⊗KB)-submodulesof M ⊗CB(N) CB(N)opp ⊗K N . Since ϕN is an isomorphism, the first part of the proof follows. Thesecond part is obvious by properties of the functor (−)⊗CB(N) N and Remark 5.17.

Remark 5.17. • In Definition 5.15, if Φ is an inclusion-preserving bijection where Φ−1 is also inclusion-preserving, then it preserves the radical ; this is because it puts in correspondence the set of maximalsubmodules too, so Φ(radA(M)) ⊆ radB(N) ; the reverse inclusion comes from applying the samereasoning to Φ−1. It also preserves direct summands ; we have M = M ′ ⊕M ′′ if and only if theonly submodule M0 ⊆ M ′,M ′′ is zero and the only submodule M ′,M ′′ ⊆ M1 is M1 = M , andthis property is preserved under inclusion-preserving bijections. It follows that it also preserves directsums by applying the result toM ′⊕M ′′ ⊆M , i.e. by restricting Φ to those submodules ofM ′⊕M ′′.

• We have the equality LsubmodA(M) = RsubmodAopp(Mopp) by definition of the opposite ring/module.This implies

LsubmodA⊗KB(M ⊗K N) ' RsubmodBopp⊗KAopp(Nopp ⊗K Mopp).

defined by using the isomorphism (M ⊗K N)opp ' Nopp ⊗K Mopp.

• Using the fact that left (A ⊗K B)-submodules of M ⊗K N are the same as right Aopp ⊗K Bopp-submodules of Mopp ⊗K Nopp, we also have a bijection

LsubmodA⊗KB(M ⊗K N) ' RsubmodAopp⊗KBopp(Mopp ⊗K Nopp).

• If M is a (A,B)-bimodule, then Mopp is a (Bopp, Aopp)-bimodule. Seeing this when A = B andthe left A-module M is given its canonical (A,Aopp)-bimodule structure, we see that Mopp is an((Aopp)opp, Aopp)-bimodule, i.e. Mopp = M as (A,Aopp)-bimodules. Similar remarks apply tothe case when M is a right A-module, and we can apply this to the (A,Aopp)-bimodule A` or the(Aopp, A)-bimodule Ar .

• If M is a left A-module, then CAopp(Mopp) = CA(M)opp (this can be checked trivially ; the reasonwhy it is equal to CA(M)opp and not CA(M) is because in the computation of CAopp(Mopp), we mustsee Mopp as a right Aopp-module, hence a right CAopp(Mopp)-module, which means composition isin the reverse order). It follows that the (A,Aopp)-bimodule A` is such that (A`)

opp = A` = (Aopp)`as (A,Aopp)-bimodules. Similar remarks apply to Ar .

Corollary 5.18. Let A and B be rings and M a simple left A-module. Interpret the right CA(M)-moduleCA(M)r as a (CA(M)opp, CA(M))-bimodule and the left CA(M)-module M as a (CA(M), CA(M)opp)-bimodule. For every left B-module N , consider the isomorphism of A⊗K B-modules

ψM : M ⊗CA(M)opp CA(M)r ⊗K N →M ⊗K N, m⊗ ψ ⊗ n 7→ ψ(m)⊗ n.

The following map is a bijection between the set of left (CA(M)opp ⊗K B)-submodules of CA(M)r ⊗K Nand the set of left (A⊗K B)-submodules of M ⊗K N :

Φ : LsubmodCA(M)opp⊗KB(CA(M)r ⊗K N)→ LsubmodA⊗KB(M ⊗K N),

Q 7→ ψN (M ⊗CA(M)opp Q).

Furthermore, this map preserves direct sums, inclusions and the radical.

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Proof. Apply Lemma 5.16 by substituting the following data

(A,M,B,N) 7→ (B,N,A,M).

By Remark 5.17, this gives us a bijection

Φ : LsubmodB⊗KCA(M)opp(N ⊗K CA(M)r)→ LsubmodB⊗KA(N ⊗K M),

P 7→ ϕM (P ⊗CA(M) M)

where ϕM is obtained by applying ψ ∈ CA(M) to m ∈ M as in Lemma 5.16. Using Remark 5.17 again,we get

LsubmodCA(M)opp⊗KB(CA(M)r ⊗K N) = RsubmodCA(M)⊗KBopp(CA(M)r ⊗K Nopp)

' LsubmodB⊗KCA(M)opp(N ⊗K CA(M)r)

' LsubmodB⊗KA(N ⊗K M)

' RsubmodAopp⊗KBopp(Mopp ⊗K Nopp)

' LsubmodA⊗KB(M ⊗K N).

We merely flipped the tensor factors in each of those bijections by using the (−)opp operation, so thecomposition of all of them still amounts to applying ψ ∈ CA(M) to m ∈M .

Corollary 5.19. Let A,B be rings, M a simple left A-module and N a simple left B-module. Considerthe K-algebra CA(M) ⊗K CB(N), a tensor product of two division K-algebras. The (A ⊗K B)-module

M ⊗K N is canonically a CA(M)⊗K CB(N)-module by the definition (ϕ⊗ ψ)(m⊗ n)def= ϕ(m)⊗ ψ(n).

This gives an isomorphism

(CA(M)⊗K CB(N))r ⊗CA(M)⊗CB(N) (M ⊗K N) 'M ⊗K N

given by multiplication. For a right ideal a E CA(M) ⊗K CB(N), we let a(M ⊗K N) be the image ofa⊗ (M ⊗K N) under this map. The application

a 7→ a(M ⊗K N)

is a bijection between the set of right ideals of CA(M)⊗K CB(N) and the set of left (A⊗K B)-submodulesof M ⊗K N , i.e.

RsubmodCA(M)⊗KCB(M)(CA(M)⊗K CB(N)) ' LsubmodA⊗KB(M ⊗K N).

Proof. It suffices to apply Lemma 5.16 and Corollary 5.18 successively on the right and left tensor factorsof M ⊗K N :

LsubmodA⊗KB(M ⊗K N) ' LsubmodA⊗KCB(N)opp(M ⊗K CB(N)r)

' LsubmodCA(M)opp⊗KCB(N)opp(CA(M)r ⊗K CB(N)r)

since the latter is easily seen to be RsubmodCA(M)⊗KCB(N)(CA(M)r ⊗K CB(N)r), i.e. the set of rightideals of CA(M) ⊗K CB(M). Up to permuting the tensor factors, this map was obtained by tensoringthe two operations

CA(M)⊗K M →M, ψ ⊗m 7→ ψ(m)

CB(N)⊗K N → N, ϕ⊗ n 7→ ϕ(n),

so the result is clear.

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Theorem 5.20. Let K be a field, A,B two K-algebras, M a left A-module and N a left B-module.

(i) If M,N are non-zero and M ⊗K N is a completely reducible (resp. simple) left (A ⊗K B)-module,then M and N are completely reducible (resp. simple).

(ii) Assume M and N are simple modules. The following are equivalent :

• The (A⊗K B)-module M ⊗K N is radical-free

• The K-algebra CA(M)⊗K CB(N) is radical-free

• The K-algebra Z(CA(M))⊗K Z(CB(N)) is radical-free.

(iii) Assume M and N are simple modules. The following are equivalent :

• The (A⊗K B)-module M ⊗K N is simple (resp. completely reducible)

• The K-algebra CA(M)⊗K CB(N) is simple (resp. semisimple).

Proof. (i) Let N ′ ≤ N be a B-submodule. It follows that M ⊗K N ′ is a submodule of M ⊗K Nby flatness (because M is a K-vector space), so there exists an (A ⊗K B)-linear projection π :M ⊗K N → M ⊗K N ′. Since M 6= 0, there exists x ∈ M and α ∈ HomK(M,K) such thatα(x) = 1. The map α : M → K can be extended to an endomorphism α : M ⊗KN →M ⊗KN ;

explicitly, we have α(m⊗n)def= α(m)x⊗n. Note that the map n 7→ x⊗n gives an isomorphism of

N with 〈x〉K ⊗N ' K⊗KN ' N which also maps N ′ isomorphically onto 〈x〉K ⊗KN ′. BecauseK is a field, we can fix K-bases for M and N (where the basis of M contains x) and deduce

π(〈x〉K ⊗K N) ⊆ (M ⊗K N ′) ∩ (〈x〉 ⊗K N) = 〈x〉k ⊗K N ′,

so π|〈x〉K⊗KN is a projection of 〈x〉K ⊗K N . Since π|M⊗KN ′ = idM⊗KN ′ , it follows thatimπ|〈x〉K⊗KN = 〈x〉K ⊗K N ′, hence 〈x〉K ⊗K N ′ is a direct summand of 〈x〉K ⊗K N , meaningthat N ′ is a direct summand of N .

If M ⊗K N is simple but N is not, it is still semisimple, so we can find two direct summandsN = N ′ ⊕N ′, which implies M ⊗K N = (M ⊗K N ′)⊕ (M ⊗K N ′′), a contradiction. Therefore,N is simple ; similarly, M is simple.

(ii),(iii) The entire result is clear except the equivalence between the radical-freeness of CA(M)⊗KCB(N)and that of Z(CA(M)) ⊗K Z(CB(N)) (because the first two points are obviously equivalent). Infact, we can show that

Jac(CA(M)⊗K CB(N)) = Jac(Z(CA(M))⊗K Z(CB(M)))(CA(M)⊗K CB(N)

).

(⊆) Consider the isomorphism

CA(M)⊗K CB(N) ' CA(M)⊗Z(CA(M))

(CB(N)⊗Z(CB(N)) (Z(CA(M))⊗K Z(CB(N)))

),

ϕ⊗ ψ 7→ ϕ⊗ (ψ ⊗ (1⊗ 1)).

It then suffices to apply Corollary 5.9 twice (c.f. Remark 5.7) and the multiplication map twice todeduce the inclusion since the right-hand side is a two-sided ideal in CA(M)⊗K CB(N) because

Jac(Z(CA(M))⊗K Z(CB(M)))(CA(M)⊗K CB(N)

)=(

CA(M)⊗K CB(N))

Jac(Z(CA(M))⊗K Z(CB(M)))

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follows from the fact that Jac(Z(CA(M))⊗K Z(CB(M))) ⊆ Z(CA(M)⊗K CB(N)).

(⊇) The elements of Jac(Z(CA(M)) ⊗K Z(CB(M))) are nilpotent and contained in the centerof CA(M) ⊗K CB(N), which means that the right-hand side of the statement is a nilideal ofCA(M) ⊗K CB(N) ; it is therefore contained in the Jacobson radical of CA(M) ⊗K CB(N) (c.f.Corollary 4.23 (iv)), i.e. the left-hand side.

Theorem 5.21. Let K be a field, A,B be K-algebras,M a left A-module and N a left B-module. SupposeM and N are isotypically simple and that one of the two is finite-dimensional over K . If CB(N) is a centralK-algebra (meaning that Z(CB(N)) = K), then M ⊗K N is an isotypically simple (A⊗K B)-module.

Proof. Suppose M is isotypically simple of type P and N is isotypically simple of type Q. We reduceto the case where M = P and N = Q. If M ' P⊕I and N ' Q⊕I , we have an isomorphism of(A⊗K B)-modules

M ⊗K N ' P⊕I ⊗K Q⊕J ' (P ⊗K Q)⊕(I×J).

It suffices to show that the hypotheses on N carry to Q. The ring CB(Q) is a division ring, soHomB(Q,N) is a right CB(Q)-vector space ; this means it is free. By Corollary 1.38 and Proposi-tion 3.14 (ii), this implies

Z(CB(Q)) = Z(EndCB(Q)(HomB(Q,N))) = Z(CB(N)) = K

by the assumption on CB(N).

Assume that M and N are simple. The K-algebra

Z(CA(M))⊗K Z(CB(N)) = Z(CA(M))⊗K K ' Z(CA(M))

is a field (and in particular radical-free), so M ⊗K N is a radical-free (A ⊗K B)-module by Theo-rem 5.20 (ii). IfM (resp. N ) is finite-dimensional over K , thenM ⊗K N is a B-module (resp. A-module)of finite length, and thus an (A ⊗K B)-module of finite length. This implies that it is a radical-freeartinian (A ⊗K B)-module, hence is completely reducible by Theorem 4.29. By Theorem 5.20 appliedagain, we deduce that CA(M) ⊗K CB(N) is semisimple ; by this result applied a third time, it nowsuffices to show that CA(M)⊗K CB(N) is iso simple to finish the proof.

Let a E CA(M) ⊗K CB(N) be a non-zero two-sided ideal. It is a right CB(N)-vector subspace stableunder the subgroup G of automorphisms of the form 1⊗ϕa where ϕa ∈ Aut(CB(N)) is given by ϕa(x) =axa−1, so since CB(N)G = Z(CB(N)) = K , we deduce by Corollary 3.16 that a = V ⊗K CB(N) whereV ≤ CA(M) is a K-vector subspace of CA(M). Since a is a two-sided ideal, V is a left ideal of CA(M)which is non-zero (because a 6= 0), hence V = CA(M), i.e. a = CA(M)⊗K CB(N).

Corollary 5.22. Let A,B be two simple K-algebras, one of the two being finite-dimensional over K andone of the two being a central K-algebra. Then A⊗K B is a simple K-algebra.

Proof. Theorem 5.21 proves that (A⊗K B)` is isotypically simple, hence A⊗K B is a simple K-algebraby definition.

Corollary 5.23. Let A be a CSA over K of dimension m ≥ 1 and Ω an algebraic closure for K . Thenm = n2 is a perfect square (i.e. n ∈ N) and we have an isomorphism of Ω-algebras Ω⊗K A ' Matn×n(Ω).

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Proof. By the previous corollary, Ω⊗KA is simple and has center isomorphic to Z(Ω)⊗K Z(A) = Ω⊗KK ' Ω, i.e. it is a CSA over the algebraically closed field Ω. The result follows by Corollary 3.40 (ii).

5.4 Separable algebras and modules over a field

In this section, K is a field. Note that in characteristic zero, every K-algebra A and every A-module M isseparable over K , so the results of this section give strong statements about algebras and modules.

Definition 5.24. Let K be a field, A a K-algebra and M a left A-module.

(i) The A-moduleM is said to be separable over K if for every field extension E/K , the left (E⊗KA)-module E ⊗K M is radical-free.

(ii) The K-algebra A is said to be separable over K if A` is a separable module over K , i.e. ifJac(E ⊗K A) = 0 for all field extensions E/K .

Proof. Let K be a field, A a K-algebra and M a left A-module.

(i) If M is separable over K , then every A-submodule of a M is separable over K .

(ii) If Mii∈I is a family of separable A-modules over K , then⊕

i∈IMi is separable over K .

(iii) If M is separable over K , then for every field extension E/K , the (E ⊗K A)-module E ⊗K M isseparable over E.

Proof. Consider a field extension E/K .

(i) LetM ′ ≤M be such an A-submodule. By Proposition 4.8 applied to the inclusion map E⊗KM ′ →E ⊗K M ,

radE⊗KA(E ⊗K M ′) ⊆ radE⊗KA(E ⊗K M) = 0.

(ii) Recalling that E ⊗K⊕

i∈IMi '⊕

i∈I E ⊗K Mi, by Proposition 4.9, we obtain

radE⊗KA(E ⊗K⊕i∈I

Mi) ' radE⊗KA(⊕i∈I

E ⊗K Mi) =⊕i∈I

radE⊗KA(E ⊗K Mi) = 0.

(iii) This follows since for every field extension E′/E, we have an isomorphism of rings/modules

E′ ⊗E (E ⊗K A) ' E′ ⊗K A, E′ ⊗E (E ⊗K M) ' E′ ⊗K M

and the module structures correspond, so the separability of E ⊗K M over E follows from that ofM over K .

Proposition 5.25. Let K be a field, A a K-algebra and M a finitely generated A-module. Suppose thereexists an algebraic field extension F/K with F perfect such that P⊗KM is a radical-free (P⊗KA)-module.Then M is separable over K .

Proof. Let E/K be a field extension and Ω be an algebraic closure of E. Since P/K is algebraic, wecan assume without loss of generality that P ⊆ Ω, which implies that Ω/P is a separable extension. Weare in the conditions of Corollary 5.9 (M is a finitely generated A-module and Ω/P is separable), which

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Chapter 5

means thatradΩ⊗KA(Ω⊗K M) ⊆ Ω⊗P radP⊗KA(P ⊗K M) = Ω⊗K 0 = 0.

Considering the isomorphism Ω⊗K M ' Ω⊗E (E ⊗K M), letting ι : E ⊗K M → Ω⊗E (E ⊗K M) 'Ω ⊗K M denote the canonical map (which is injective because Ω/E is a field extension, e.g. Ω is anE-vector space), Proposition 5.5 (ii) gives

ι(radE⊗KA(E ⊗K M)) ⊆ radΩ⊗KA(Ω⊗K M) = 0 =⇒ radE⊗KA(E ⊗K M) = 0.

Corollary 5.26. Let K be a field and A a K-algebra. Consider the two following statements :

(i) There exists a field extension P/K such that P is perfect and Jac(P ⊗K A) = 0

(ii) A is a separable K-algebra.

Then (i) implies (ii), and if A is finite-dimensional over K , (i) and (ii) are equivalent.

Proof. The first part of the statement follows directly from Proposition 5.25.

For the second part of the statement, apply the proof of Proposition 5.25 to the case where M = A`,but instead of using Proposition 5.5 (ii) (which is where we need the fact that Ω/E is algebraic), wepick an arbitrary field Ω containing both E and P as K-subfields (it could be a residue field of somep ∈ Spec (E ⊗K P ) for instance, c.f. Definition 10.28) ; this preserves the fact that Ω/P is separable andsince E ⊗K A is finite-dimensional over E, Proposition 5.5 (iii) applies (because E ⊗K A is left-artinianby Example 2.27) and gives us the same result.

Proposition 5.27. Let K be a field, A a K-algebra andM a completely reducible A-module. The followingare equivalent :

(i) M is separable over K

(ii) For each simple A-submodule P of M , Z(CA(P ))/K is a separable field extension.

Proof. Recall (c.f. Definition 2.56) that if P is a simple A-module, the isotypical component of M isdenoted by MP . By writing

M '⊕

P |MP 6=0

MP ,

Definition 5.4 (i) and (ii) implies that we can assume without loss of generality that M = P where I is aset and P is a simple A-module. Consider a field extension E/K and note that E is a simple E-modulesatisfying Z(E) = E. By Theorem 5.20 (ii), the following are equivalent :

• M is separable over K

• For every field extension E/K , E ⊗K M is radical-free

• For every field extension E/K , E ⊗K Z(CA(P )) is radical-free

• The K-algebra Z(CA(P )) is separable

• The field extension Z(CA(P ))/K is separable.

This completes the proof.

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Corollary 5.28. Let K be a field and A a semisimple K-algebra. The following are equivalent :

(i) A is separable over K

(ii) Z(A) is separable over K

(iii) Write A =∏ni=1Ai where the Ai are simple K-algebras (c.f. Theorem 3.38). Then Z(Ai)/K is a

separable field extension for i = 1, · · · , n.

Proof. This follows from Theorem 3.38 and Proposition 5.27.

Definition 5.29. Let K be a field, A a K-algebra and M an A-module.

(i) We say that M is absolutely completely reducible over K if for every field extension E/K , the(E ⊗K A)-module E ⊗K M is completely reducible.

(ii) We say that A is absolutely semisimple overK if A` is an absolutely completely reducible A-moduleover K .

Proposition 5.30. Let K be a field, A a K-algebra and M a left A-module.

(i) IfM is absolutely completely reducible overK , then every A-submodule ofM is absolutely completelyreducible over K .

(ii) If Mii∈I is a family of absolutely completely reducible A-modules over K , then⊕

i∈IMi is abso-lutely completely reducible over K .

(iii) If M is absolutely completely reducible over K , then for every field extension E/K , the (E ⊗K A)-module E ⊗K M is absolutely completely reducible over E.

Proof. Consider a field extension E/K .

(i) Let M ′ ≤ M be such an A-submodule. Since E ⊗K M ′ is an (E ⊗K A)-submodule of E ⊗K Mwhich is completely reducible, E ⊗K M ′ is also completely reducible.

(ii) Since E ⊗K⊕

i∈IMi '⊕

i∈I E ⊗K Mi can be written as a direct sum of completely reducible(E ⊗K A)-submodules, it is completely reducible by definition.

(iii) This follows since for every field extension E′/E, we have an isomorphism of rings/modules

E′ ⊗E (E ⊗K A) ' E′ ⊗K A, E′ ⊗E (E ⊗K M) ' E′ ⊗K M

and the module structures correspond, so the absolute complete reducibility of E ⊗K M over Efollows from that of M over K .

Theorem 5.31. Let K be a field, A a K-algebra and M an A-module. Consider the following statements :

(i) M is absolutely completely reducible over K .

(ii) M is separable over K

Then (i) implies (ii), and if M is finite-dimensional over K , both statements are equivalent.

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Proof. Completely reducible modules are radical-free by Proposition 4.9 (and the fact that the radical ofa simple module is zero), so absolute complete reducibility implies separability. Conversely, in the casewhere M is finite-dimensional over K , let E/K be a field extension. The (E ⊗K A)-module E ⊗KM isfinite-dimensional over E, hence is left-artinian by Example 2.27 ; since it is radical-free, it is completelyreducible by Theorem 4.29. This implies that M is absolutely completely reducible over K .

Corollary 5.32. Let K be a field and A a K-algebra. Consider the following statements :

(i) A is absolutely completely reducible over K .

(ii) A is separable over K

Then (i) implies (ii), and if A is finite-dimensional over K , both statements are equivalent.

Proof. This is straightforward from Theorem 5.31 applied to the case where M = A`.

Theorem 5.33. Let A,B be K-algebras, M be a separable left A-module over K and N a left B-module.Then

radA⊗KB(M ⊗K N) ⊆M ⊗K radB(N).

In particular, the tensor product of a radical-free module with a module which is separable over K gives aradical-free module.

Proof. First, we treat the case where N is simple, so that radB(N) = 0. By Lemma 5.16, it suffices toshow that the left (A ⊗K CB(N)opp)-module M ⊗K CB(N)r is radical-free. We have an isomorphismof (A⊗K CB(N)opp)-modules

M ⊗K CB(N)r ' (M ⊗K Z(CB(N)))⊗Z(CB(N)) CB(N)r.

SinceM is separable over K and N is simple, considering the field extension Z(CB(N))/K implies thatM ⊗K Z(CB(N)) is radical-free. By Corollary 5.9 (i), we deduce that M ⊗K CB(N)r is radical-free.

Onto the general case. Let z =∑n

i=1mi⊗ni ∈ radA⊗KB(M⊗KN) wheremi ∈M and ni ∈ N . Withoutloss of generality, we can suppose that the mi are linearly independent over K . Let Q be a simple B-module ; to show that ni ∈ radB(N), we have to show that whenever f : N → Q is a morphism ofB-modules, we have f(ni) = 0. Extend this morphism to M , so that idM ⊗ f : M ⊗K N → M ⊗K Qsatisfies

(idM ⊗ f)(radA⊗KB(M ⊗K N)) ⊆ radA⊗KB(M ⊗K Q) = 0

by Proposition 4.8 and the first part of this proof. This means

n∑i=1

mi ⊗ f(ni) = (idM ⊗ f)(z) = 0 =⇒ f(ni) = 0

by the linear independence of themi overK . Therefore ni ∈ radB(N), which completes the argument.

Corollary 5.34. Let A be a separable K-algebra and B a K-algebra. For every left B-module N , we haveradA⊗KB(A⊗K N) ⊆ A⊗K radB(N).

Proof. This is Theorem 5.33 in the case where M = A`.

Corollary 5.35. Let A, B be K-algebras, M a left A-module and N a left B-module, both modules beingseparable over K . Then then (A⊗K B)-module M ⊗K N is separable over K .

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Proof. Let E/K be a field extension. By separability of N , we see that the (B ⊗K E)-module N ⊗K Eis radical-free. By separability of M and Theorem 5.33,

radA⊗K(B⊗KE)(M ⊗K (N ⊗K E)) ⊆M ⊗K radB⊗KE(N ⊗K E) = 0.

Since the A⊗K (B⊗K E)-moduleM ⊗K (N ⊗K E) can be identified with the E⊗K (A⊗K B)-moduleE ⊗K (M ⊗K N) (via m⊗ (n⊗ e) 7→ e⊗ (m⊗ n)), we see that the latter is radical-free, which impliesthat M ⊗K N is separable over K .

Corollary 5.36. Let A and B be two K-algebras. If A is separable over K and Jac(B) = 0 (i.e. if B isradical-free), then Jac(A⊗K B) = 0 (i.e. A⊗K B is radical-free).

Proof. It suffices to set M = A` and N = B` in the previous corollary.

Corollary 5.37. Let K be a field and A,B two K-algebras.

(i) If A⊗K B is semisimple, then A and B are semisimple.

(ii) If A and B are semisimple, A is separable over K and one of A or B is finite-dimensional over K ,then A⊗K B is semisimple. In particular, if K is a perfect field (which is the case when ch(K) = 0),the tensor product of finite-dimensional semisimple K-algebras is semisimple.

Proof. Part (i) follows by applying Theorem 5.20 (i) to M = A` and N = B`. For part (ii), since A (resp.B) is finite-dimensional over K , A ⊗K B is a free left A-module (resp. right B-module) of finite rank.The semisimplicity of A and B implies that both are left-artinian, right-artinian and radical-free (c.f.Theorem 4.29), so by Proposition 2.28, A⊗K B is left-artinian (resp. right-artinian). Since one algebra isseparable and the other is radical-free, A⊗K B is radical-free by Corollary 5.36, so that it is semisimpleby Theorem 4.29.

Theorem 5.38. Let K be a field, A,B be two K-algebras and write C (A) for the set of isomorphismclasses of all simple left A-modules which are finite-dimensional over K ; given α ∈ C (A), write α = [P ]for a representative P of this class, i.e. P is a simple left A-module in the isomorphism class.

For each isomorphism class [R] ∈ C (A⊗K B) represented by, there exist [P ] ∈ C (A) and [Q] ∈ C (B)such that the (A ⊗K B)-module R is isomorphic to a quotient of P ⊗K Q. Furthermore, [P ] and [Q] areunique with this property, giving a well-defined map C (A ⊗K B) → C (A) × C (B) defined by [R] 7→([P ], [Q]).

Proof. Let R ∈ C(A ⊗K B) be a simple left (A ⊗K B)-module. Since R is finite-dimensional overK , it is a left-artinian B-module, so there exists a simple B-submodule ; call it Q. Instead of seeingR as an (A ⊗K B)-module, we interpret it as a left A-module and a left B-module such that the two

module structures are compatible. This gives Mdef= HomB(Q,R) the structure of a left A-module (c.f.

Remark 1.47). Consider the map ϕ : M ⊗K Q→ R defined by evaluation, i.e. µ⊗ q 7→ ϕ(µ⊗ q) def= µ(q).

This is a morphism of (A ⊗K B)-modules since for a ∈ A, b ∈ B,µ ∈ HomB(Q,R) = M and q ∈ Q,we have

ϕ((a⊗ b)(µ⊗ q)) = ϕ(aµ⊗ bq) = aµ(bq) = a(bµ(q)) = (a⊗ b)ϕ(µ⊗ q).

The left A-module HomB(Q,R) is finite-dimensional over K (because Q and R are), so it is left-artinian ; let P be a simple left A-submodule of HomB(Q,R). We have a canonical inclusion of(A ⊗K B)-submodules P ⊗K Q ⊆ HomB(Q,R) ⊗K Q (tensor up with Q over K), and the restrictionof ϕ to P ⊗K Q is not identically zero (for any ϕ ∈ HomB(Q,R) \ 0, there exists at least one

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q ∈ Q such that ϕ(q) 6= 0). This implies that ϕ|P⊗KQ is surjective by simplicity of R, which meansthat ϕ induces an isomorphism P⊗KQ/S ' R where S ≤ P⊗KQ is an (A⊗KB)-submodule of P⊗KQ.

Because Q is aK-vector space, the left A-module P⊗KQ is isotypically simple of type P ; symetrically, itis an isotypically simple left B-module of type Q. It follows by the (A⊗K B)-linearity of ϕ that the sameis true for R, i.e. it is an isotypically simple A-module of type P and an isotypically simple B-module oftype Q). It follows that [P ] and [Q] are the unique isomorphism classes with the property that a quotientmap P ⊗K Q→ R exists.

Corollary 5.39. Let K be an algebraically closed field, A,B be two K-algebras and write C (A) for the setof all simple left A-modules which are finite-dimensional over K . The map

Φ : C (A)× C (B)→ C (A⊗K B), (P,Q) 7→ P ⊗K Q

is a well-defined map (i.e. the tensor product of two simple modules becomes simple) which is a bijection.

Proof. By Corollary 5.19, it suffices to show that if [P ] ∈ C (A), then CA(P ) ' K as K-algebras. This isclear by Lemma 2.66.

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