math 901 fall 2011 course notess-kshulti1/fall2011/901.notes.pdf · 2014. 9. 18. · math 901 notes...

Math 901 Notes 3 January 2012 1 of 33

Math 901 ∼ Fall 2011 ∼ Course NotesInformation: These are class notes for the second year graduate level algebra course at UNL (Math 901) as takenin class and later typed by Kat Shultis. The notes are from the Fall of 2011, and that semester the course was taughtby Brian Harbourne. During the fall semester we covered the following topics:

B Category Theory, Functors, Natural TransformationsB Group Theory: Free (Abelian) Groups, Group Actions, Semi-Direct Products, Sylow Theorems, Nilpotent Groups,

Solvable GroupsB Field Theory: Algebraic Extensions, Separable Extensions, Separable Degree, Galois Theory, Galois Extensions,

Finite Fields, Transcendence Degree, Transcendence Bases

For each class day, I’ve indicated the topic at the top of that day’s notes.

Disclaimer: I created these notes in order to help me study, and so I’ve expanded proofs where I find it useful,shortened things I was comfortable with before this course, and changed at least one proof to one that I like betterthan the one presented in class. These notes are not meant to be a substitute for your own notes. They have beenproof-read, but are not guaranteed to be without errors. If you find errors, please email Kat at the following address:

s-kshulti1 “at” math.unl.edu

22 August 2011 Categories

B Comment: The purpose of categories and functors is to describe structure that is common in many parts ofmathematics.

B Comment: Intuitively, a category is a structure consisting of objects and arrows (morphisms), and the arrows aremaps between the objects.

B Examples:

Category Objects Arrows Typical Notation

Sets sets set maps SGroups groups group homomorphisms GAbelian Groups abelian groups group homomorphisms AbRings rings ring homomorphismsVector Spaces finite dimensional R-vector spaces linear transformations VManifolds topological manifolds continuous maps

differentiable manifolds differentiable mapsGroup, G ? elements of G

B Definition: A category, A, consists of a collection of objects, Ob(A), and for each pair of objects, A,B ∈ Ob(A),there is a set of morphisms, Mor(A,B), and a composition law defined for morphisms ϕ ∈ Mor(A,B) andψ ∈ Mor(B,C) when A,B,C ∈ Ob(A), such that ψ ◦ ϕ ∈ Mor(A,C), which satisfies the following:

1. Composition is associative when defined.2. For all A ∈ Ob(A), there exists a unique identity morphism, which is a left and right identity whenever composition

is defined. That is, there is a morphism 1A ∈ Mor(A,A) such that if ϕ ∈ Mor(A,B) and ψ ∈ Mor(B,A), thenϕ ◦ 1A = ϕ and 1A ◦ ψ = ψ.

3. Mor(A,B) ∩Mor(A′, B′) = ∅, unless A = A′ and B = B′,1 in which case Mor(A,B) = Mor(A′, B′).

24 August 2011 More Categories & Universal Properties

B Example: Let X be a topological space. We’ll use TX as the collection of open sets in X. We can turn thisinto a category, by setting Ob(X) = TX , and if U, V ∈ TX , we have either Mor(U, V ) = ∅ when U * V orMor(U, V ) = {U ⊆ V } when U ⊆ V .

B Definition: A morphism f : A → B in a category C is an isomorphism if and only if there is a morphismg : B → A such that g ◦ f = idA and f ◦ g = idB .

B Comment: We discussed the idea of direct products, and how a direct product is really just a function. This was abit confusing, but the general idea was to generalize n-tuples. I’ll only put it here in the most general context, not inthe more specific cases we discussed.

1These are actually equalities and not isomorphisms. In the category of groups, for example, my copy of Z is isomorphic, but notnecessarily equal to, Sarah’s copy of Z, and the morphisms from my copy to Caitlyn’s copy are hence fundamentally different from themorphisms from Sarah’s copy to Caitlyn’s copy.


B Definition: Let {Ai : i ∈ I} be a family of objects in a category C. We say P ∈ Ob(C) is a product of the objects

Ai, sometimes written∏i∈I

Ai, if P was given together with morphisms πi : P → Ai called projections such that they

satisfy the universal property. That is, given any B ∈ Ob(C) and morphisms fi : B → Ai there is a unique morphism∏i∈I

fi = ϕ ∈ Mor(B,P ) such that the diagram below commutes for each i ∈ I.

B P

Ai

ϕ

fiπi

B Aside: Suppose P and P ′ are both products, that is we have projections πi : P → Ai and π′i : P ′ → Ai. Then bythe universal property we have morphisms f : P → P ′ and g : P ′ → P , and these are inverse morphisms because thediagram below commutes.

P ′ P P ′

Ai

f g

πiπ′i π′i

In other words π′i ◦ g ◦ f = π′i and so we must have that g ◦ f = 1P ′ and similarly 1P = f ◦ g. Hence P and P ′

are isomorphic. However, note that we can actually only use this diagram for one direction, that is we can onlyshow that g ◦ f = 1P ′ . However, a similar argument will show the other direction. Once things are more formal,we’ll see that objects satisfying universal properties are unique.2

26 August 2011 Monomorphisms & Epimorphisms

B Review: From last time, we discussed the categorical product, and we’ve discussed the category of a topologicalspace, X. The product of U and V in this category is their intersection. For homework, we’ll ask more sophisticatedquestions about the product.

B Example: Let C be the category whose objects are all the elements in the set R ∩ [0, 1], and Mor(x, y) ={∅, if x > y

{x ≤ y}, if x ≤ y. . Let {xi : i ∈ I} be a collection of objects in this category, then the categorical product is

p = inf{xi : i ∈ I}. This is easily seen to be what you want if you do the case of two objects and their product, sothat the product of x and y is min(x, y).

B Definition: Let C be a category and let A,B ∈ Ob(C), with ϕ ∈ Mor(A,B). We say ϕ is a monomorphism ifgiven any γ, δ ∈ Mor(C,A) with ϕ ◦ γ = ϕ ◦ δ then γ = δ.

B Proposition: In the category of sets, ϕ is a monomorphism if and only if it is injective.

B Definition: (Again, using pictures) A map ϕ is a monomorphism if whenever we have

C A Bγ

δ

ϕ

and it commutes, then γ = δ.

B Comment: An epimorphism is the categorical dual of a monomorphism, and also a generalization of a surjective map.

B Definition: A map ϕ is an epimorphism if whenever we have

C A Bγ

δϕ

and it commutes, then γ = δ.

2Also, the argument is frequently almost identical to this one.


B Example: Back to the order relations. Here, all morphisms are both monomorphisms and epimorphisms, but onlythe identity morphisms are isomorphisms. This indicates that the mono- and epi-morphisms are related to injectiveand surjective maps, but do not always act the same.

B Comment: The categorical dual of a product is a coproduct. Also, in most cases, categorical duals can be definedby simply reversing the directions of the arrows.

29 August 2011 Coproducts; Initial & Terminal Objects

B Comment: We’ll repeat the definition of a product here, so that we can compare it to the dual, the coproduct.

B Definition: Let {Ai : i ∈ I} be a family of objects in a category C.The product (if it exists) is an object

∏i∈I

Ai with a family of morphisms πj :∏i∈I

Ai → Aj such that given any object

B and family of morphisms pj : B → Aj , there is a unique morphism ϕ : B →∏i∈I

Ai such that the diagram below

commutes for each j ∈ I.

B

∏i∈I

Ai

Aj

ϕ

pj πj

The coproduct (if it exists) is an object∐i∈I

Ai with a family of morphisms πj : Aj →∐i∈I

Ai such that given any

object B and family of morphisms pj : Aj → B, there is a unique morphism ϕ :∐i∈I

Ai → B such that the diagram

below commutes for each j ∈ I.

B

∐i∈I

Ai

Aj

ϕ

pjπj

B Example: The category of abelian groups. In this category, the coproduct is the direct sum.

B Example: The category of sets. In this example, the coproduct is the disjoint union. Let’s make that concept a bit

more rigorous: Let {Ai : i ∈ I} be our collection of sets. Then set Ai = {i} ×Ai ⊆ I ×⋃i∈I

Ai, so that Ai ∩ Aj = ∅

if i 6= j. Then P =⋃i∈I

Ai works and πj : Aj → P is given by x 7→ (j, x).

B Definition: An object, I in a category C is said to be initial (aka universally repelling) if there is a uniquemorphism I → A for every object A ∈ Ob(C).

B Definition: An object T in a category C is said to be terminal (aka universally attracting) if there is a uniquemorphism A→ T for every object A ∈ Ob(C).

B Example: The initial object in the category of abelian groups is the trivial group, I = (0). In the categorycorresponding to a topological space, the empty set is an initial object, and the whole space is a terminal object.

B Theorem: If A and B are initial (respectively terminal) objects in a category C, then there is a unique isomorphismϕ : A→ B.

31 August 2011 Topology Review; (co)Products as (Initial)Terminal Objects

B Comment: We discussed a variety of possible definitions of continuity, and how these arose in history. We alsodiscussed the idea that topology was in some ways a natural abstraction of other areas of mathematics. We thenproceeded to define a topology (defined using open sets), closed sets, the closure and interior, the topological definitionof continuous, and a basis. We then proceeded to continue our discussion of initial and terminal objects. The purpose


of this discussion was to introduce ideas that show up in the first homework assignment due Friday. I omit the detailshere as I was already comfortable with these definitions.

B Definition: Let C be a category, and let F = {Ai : i ∈ I} be a family of objects in C. We define a new categoryCF . In this category, the objects are the objects of C together with a family of morphisms fi ∈ Mor(C,Ai) foreach i ∈ I. We’ll denote this object as C{fi}. Given two objects in this category C{fi} and D{gi}, a morphismϕ ∈ Mor(C{fi}, D{gi}) is a morphism ϕ ∈ Mor(C,D) in the original category such that the diagram (existing in thecategory C) below commutes for all i ∈ I.

C D

Ai

ϕ

figi

B Comment: The product in C is precisely a terminal object in CF .3

B Definition: Let C be a category, and let F = {Ai : i ∈ I} be a family of objects in C. We define a new categoryCF . In this category, the objects are the objects of C together with a family of morphisms fi ∈ Mor(Ai, C) foreach i ∈ I. We’ll denote this object as C{fi}. Given two objects in this category C{fi} and D{gi}, a morphismϕ ∈ Mor(D{gi}, C{fi}) is a morphism ϕ ∈ Mor(D,C) in the original category such that the diagram (existing in thecategory C) below commutes for all i ∈ I.

C D

Ai

ϕ

figi

B Comment: The coproduct in C is precisely an initial object in CF .4

2 September 2011 Fiber Products

B Comment: For intuition, we discuss the fiber product in the category of sets. Given X,Y, Z sets with f : X → Zand g : Y → Z the fiber product is the product of the fibers, that is, X ×Z Y = {f−1(z)× g−1(z) : z ∈ Z}, or moretraditionally though of as X ×Z Y = {(x, y) ∈ X × Y : f(x) = g(y)}.

B Definition: Let A,B ∈ Ob(C) for C some category, and let f ∈ Mor(A,C) and g ∈ Mor(B,C) be morphisms. Wesay an object F together with morphisms α ∈ Mor(F,A) and β ∈ Mor(F,B) is a fiber product of f and g if

– the diagram below commutes:

F

A B

C

α β

f g

– and whenever we have morphisms ϕ ∈ Mor(G,A) and ψ ∈ Mor(G,B) such that the diagram below commutes

G

A B

C

ϕ ψ

f g

3This is “obvious” if you go through the definitions of a product and a terminal object.4This is “obvious” if you go through the definitions of a product and a terminal object.


then there is a unique morphism θ ∈ Mor(G,F ) such that the third diagram below commutes.

G

F

A B

C

θϕ ψ

α β

f g

B Comment: The categorical fiber product of f ∈ Mor(A,C) and g ∈ Mor(B,C) in a category C is just a categoricalproduct of objects A{f} and B{g} in the category C{C}.5

B Comment: We next started to think about functors, and discussed the idea of a subcategory, but did not completethe definition of a functor.

B Definition: We say a category C is a subcategory of a category D if Ob(C) ⊆ Ob(D) and if given A,B ∈ Ob(C),then MorC(A,B) ⊆ MorD(A,B) and if moreover an identity in C is an identity in D, and the composition law in Cagrees with the composition law in D.

B Example: The category of abelian groups is a subcategory of the category of groups. The inclusion of the categoryof abelian groups in the category of groups is an example of a covariant functor.

7 September 2011 Functors

B Definition: Let C, D be categories.A covariant functor F : C → D is an assignment of an object F (A) in D for each object A in C, and for eachmorphism ϕ : A→ B of objects in C, a morphism F (ϕ) : F (A)→ F (B) such that

– F (1A) = 1F (A), and– if we have morphisms f : A→ B and g : B → C in C, then F (g ◦ f) = F (g) ◦ F (f).

A contravariant functor F : C → D is an assignment of an object F (A) in D for each object A in C, and for eachmorphism ϕ : A→ B of objects in C, a morphism F (ϕ) : F (B)→ F (A) such that

– F (1A) = 1F (A), and– if we have morphisms f : A→ B and g : B → C in C, then F (g ◦ f) = F (f) ◦ F (g).

B Examples:

– The inclusion of abelian groups into groups is a covariant functor.– The forgetful functor is a covariant functor. For example, we could have this functor from the category of groups

to the category of sets, where the functor sends a group to its underlying set and a group homomorphism issent to itself (as a set map).

B Comment: The next few items here are intended as a review of things necessary to understand the double dualas a covariant functor, and the dual as a contravariant functor.

B Definition: Given a vector space V , the dual of V is V ∗ = Mor(V,R).

B Definition: Given vector spaces V,W and a linear map f : V →W , we define the dual of the map as f∗ : W ∗ → V ∗

via f∗(ϕ) = ϕ ◦ f where ϕ : W → R is any linear transformation.

B Recall: If V is a real finite dimensional vector space, then Mor(V,R) is as well; moreover, they have the same dimen-sion, so there exists an isomorphism between V and V ∗∗ (and with V ∗ but we care significantly less about that here).

B Definition: Let V be the category whose objects are real finite dimensional vector spaces, with morphisms as lineartransformations.

B Example: We now define the functor F : V → V as follows. Given an object V ∈ Ob(V), then F (V ) = V ∗∗. Also,if V,W ∈ Ob(V), with f ∈ Mor(V,W ), then F (f) is a map from F (V ) to F (W ), that is from V ∗∗ to W ∗∗ givenby F (f) = (f∗)∗ = f∗∗.

B Lemma: Given f : V → W and g : W → Z, where V,W,Z are finite dimensional real vector spaces, and f, g arelinear transformations, then (g ◦ f)∗ = f∗ ◦ g∗.Proof. Let ϕ ∈ Z∗, that is ϕ : Z → R. Then it is clear that (g ◦ f)∗(ϕ) = ϕ ◦ (g ◦ f). Also, we have that(f∗ ◦ g∗)(ϕ) = f∗(g∗(ϕ)) = f∗(ϕ ◦ g) = (ϕ ◦ g) ◦ f . Now, as composition of morphisms is associative, we have thatthe action of (g ◦ f)∗ and of f∗ ◦ g∗ is the same, and hence, they are the same function.

5See 31 August 2011 for a reminder of what this category is.


B Proposition: As defined F : V → V is a covariant functor.

Proof. First, we’ll check that F (1V ) = 1F (V ) or equivalently, that (1V )∗∗ = 1V ∗∗ . So let ϕ ∈ V ∗∗ so thatϕ : Mor(V,R) → R. Then clearly, 1V ∗∗(ϕ) = ϕ. Also, 1V

∗∗(ϕ) = ϕ ◦ 1V∗. Both of these are maps from

V ∗ to V ∗, and we want to check they are the same map. So let λ ∈ V ∗. Then 1V ∗∗(ϕ)(λ) = ϕ ◦ λ. Also,1V∗∗(ϕ)(λ) = (ϕ ◦ 1V

∗)(λ) = ϕ(1V∗(λ)) = ϕ ◦ λ ◦ 1V = ϕ ◦ λ. Hence, we have F (1V ) = 1F (V ).

Now, let f : V → W , and g : W → Z where V,W,Z ∈ Ob(V) and f, g are morphisms. Then we’ll check thatF (g◦f) = F (g)◦F (f). From the Lemma, we know that (g◦f)∗ = f∗ ◦g∗. So applying the lemma to f∗ ◦g∗ = (g◦f)∗

then gives that F (g ◦ f) = ((g ◦ f)∗)∗ = (f∗ ◦ g∗)∗ = g∗∗ ◦ f∗∗ = F (g) ◦ F (f) as required.

9 September 2011 More on Functors

B Last Time: The double dual gave us a covariant functor. Note here that we didn’t need finite dimensional or thefield we were working over to be R to get a covariant functor, but we’ll discuss this functor again later, and then we’llmake use of those properties.

B Example: We can get a contravariant functor D : V → V by taking duals. On objects, the functor is defined byD(W ) = W ∗ = {λ : W → R|λ is a linear transformation}. On morphisms, f : V → W , D(f) = f∗ known as the“pull back” morphism and is obtained by composing with f .6 We checked in class that this was a contravariantfunctor, but Sarah and I did that work already, and the explanation is already incorporated into the notes for theprevious class (7 Sept 2011).

B Comment: We get the covariant functor from last time by composing D with itself, or equivalently, using “pushforwards.” That is, we can define F : V → V by F (W ) = W ∗∗ and F (f) = f∗. We now describe this push forward.

B Definition: Given f : V →W we want to define f∗ : V ∗∗ →W ∗∗. Let ϕ ∈ V ∗∗. We have a morphism, ϕ : V ∗ → Rand we know what f∗ is, and so we define f∗(ϕ) = ϕ ◦ f∗.

B Alternate Viewpoint: Given f : V →W and ϕ : V ∗ → R, we want f∗(ϕ) to be an element of W ∗∗, so we shouldbe able to evaluate f∗(ϕ) at elements λ ∈W ∗. We then define (f∗(ϕ))(λ) = ϕ(λ ◦ f).7

B Definition: Let F be a covariant functor F : C → D. Then for any two objects C1, C2 in C, there is a mapMorC(C1, C2)→ MorD(F (C1), F (C2)) which is given by f 7→ F (f).

– If the map is injective for all C1, C2 in C, then we say F is faithful.– If the map is surjective for all C1, C2 in C, then we say F is full.– If the map is bijective for all C1, C2 in C, then we say F is fully faithful.

B Examples:

– The inclusion functor from the category of abelian groups to the category of groups is fully faithful.– The forgetful functor from the category of groups to the category of sets is faithful but not full.– The functor from the category of sets to the category with a single object, and single morphism is full, but not

faithful.

12 September 2011 Functor Categories & Representability of Functors

B Motivation: We discussed the motivation for functor categories. The basic idea was to stick in things that we wantto have in our categories, and then maybe they were already there, but we have them now. We then defined tworepresentation functors.

B Definition: Let A be a category, and let A be an object in A. Then we define a covariant functor, MA from Ato the category of sets, S. On objects, this functor is MA(B) = MorA(A,B), and if f ∈ MorA(C,D), then MA(f)should be a morphism from MorA(A,C) to MorA(A,D), so we set MA(f) = f∗ where f∗ is the push forward map,f∗(g) = f ◦ g.

B Proposition: The above defines a covariant functor.8

B Definition: Similarly, we define a contravariant functor, MA from A to S. On objects, the functor is MA(B) =MorA(B,A), and if f ∈ MorA(C,D), then MA(f) should be a map from MorA(D,A) to MorA(C,A), so we setMA(f) = f∗ where f∗ is the pull back map, f∗(g) = g ◦ f .

B Proposition: The above defines a contravariant functor.9

6and was also defined on 7 Sept 20117I’m not sure how this helps, it may make more sense if we had other examples of a push forward, right now, it seems like a pull back

still and to be honest, I’m a bit confused.8The proof is not difficult, but it can be confusing to think about.9The proof is not difficult, but it can be confusing to think about.


B Definition: Let C and D be categories. Also, let F and G be covariant functors from C to D. A naturaltransformation from F to G is a rule, Φ which assigns to every object C of C a morphism ΦC ∈ MorD(F (C), G(C))such that given any morphism f ∈ Mor(C1, C2) in C, the diagram below commutes.10

F (C1) F (C2)

G(C1) G(C2)

F (f)

G(f)

ΦC1ΦC2

B Fact: Let C and D be categories. Let CD be the collection of covariant functors from C to D. This becomes a categoryif given two covariant functors F and G from C to D we define MorCD (F,G) to be the natural transformations fromF to G.11

B Definition: A natural equivalence is a natural transformation which is an isomorphism in the functor category.

B Definition: A covariant functor, F , from a category C to the category S is said to be representable if F isnaturally equivalent to MA for some object A in C.A contravariant functor, G from C to the category S is representable if G is naturally equivalent to MA for someobject A in C.

B Comment: We’re now ready to discuss the vector space example again. Next time, we’re going to show that thedouble dual functor is naturally equivalent to the identity functor on the category V as defined on 7 September 2011.

14 September 2011 Natural Equivalence Example

B Example: We spent all of this day explaining why the identity functor on V, which we’ll denote 1V , is naturallyequivalent to the double dual functor, F (as defined on 7 September 2011).

– Step 1: First we defined an evaluation map eWw : W ∗ → R for any fixed object W in V and any fixed elementw ∈W , by λ 7→ λ(w). We showed this is a linear map, and hence is an element of W ∗∗.

– Step 2: We next defined a map eW : W →W ∗∗ for any finite dimensional real vector space, W via w 7→ eWw .We again showed12 this is a linear map, and hence an object of MorV(W,W ∗∗).

– Step 3: We next defined a natural transformation Φ from 1V to F . For each object W of V we need a morphismΦW : 1V(W )→ F (W ). Note that 1V(W ) = W and F (W ) = W ∗∗. We’ll take ΦW = eW . We must show thatthis is natural. That is, given any morphism L : W1 →W2 of real finite dimensional vector spaces, we must showthat the following diagram commutes:

1V(W1) 1V(W2)

F (W1) F (W2)

L

eW1 eW2

L∗∗ = L∗

We’ll rewrite the diagram with “easier” notation:

W1 W2

W ∗∗1 W ∗∗2

L

eW1 eW2

L∗∗ = L∗

We checked that the diagram commutes in class, and it was not difficult.

B Aside: For a natural transformation Φ to be a natural equivalence, we just need for each object W , that ΦW is anisomorphism. We can then define an inverse natural transformation Ψ as ΨW = (ΦW )−1 and then we just need tocheck Ψ is a a natural transformation and that it is an inverse to Φ. However, I believe Brian Harbourne said this willALWAYS happen if each ΦW is an isomorphism.

10You can similarly define a natural transformation of contravariant functors by changing the direction of the horizontal arrows.11I believe that we call the category CD the functor category.12We actually just claimed this, and didn’t show it; but it took me all of 60 seconds to write it down including finding a working marker.


B Example: Back to our example, it is enough then to show that eW is an isomorphism for each W . We’ll assumewithout proof that dim(W ) = dim(W ∗) = dim(W ∗∗). However, this fails unless dim(W ) <∞. We’ll also use thefact that if dim(W ) = dim(W ∗∗) and eW is injective, then eW is an isomorphism.13 We then made a clever choice inorder to show that ker(eW ) = 0.

B Comment: “People” often say that W and W ∗∗ are canonically isomorphic. They’re referring to the isomorphismeW because it defines a natural equivalence from 1V to F .

16 September 2011 Examples of Representability

B Example: Let F be the forgetful functor from V to S. Note that F is a covariant functor. We showed that Fis representable. That is, there is a vector space U such that F is naturally equivalent to MU = MorV(U,−). Orrestated again, we showed there is a vector space U such that for all objects W in V, there is a natural bijectionbetween W = F (W ) and MU (W ) = MorV(U,W ). In fact, we can say that R represents F , i.e., that F is naturallyequivalent to MR. We define the natural transformation Φ : MR → F via ΦW : MorV(R,W ) → W is given byL 7→ L(1). It’ll be a homework problem to actually check that ΦW is a bijection for all objects W in V and also thatΦ is natural.

B Example: Now, let W1 and W2 be fixed objects in V. We’ll define F to be the contravariant functor from V toS which is given by F (W ) = MorV(W,W1)×MorV(W,W2), and F (f) will be given by (λ′1, λ

′2) 7→ (λ′1 ◦ f, λ′2 ◦ f).

This functor is also representable. In this case, it is represented by U = W1 ×W2. Again, we didn’t actually show thenatural transformation, and I don’t feel like typing the details of the setup.

B Example: Let F = {Wi : i ∈ I} be a family of objects in V. We define a contravariant functor F from V toS similarly to the previous example. That is, on objects, F (W ) =

∏i∈I MorV(W,Wi) and given a morphism

L : W →W ′, we define F (f) to be the map from F (W ′) to F (W ) which is given by∏i∈I

λ′i : i ∈ I 7−→∏i∈I

λ′i ◦ L : i ∈ I.

However, this functor is not representable if∑i∈I dimWi = ∞, as the vector space we would want it to be

representable by (namely U =∏i∈IWi is not an object in V).

B Comment: Grothendieck had a solution to this issue of non-representability. The solution basically boils down toembedding the category into a larger category where the object you want exists. We’ll start explaining this with theYoneda Lemma next time.

19 September 2011 Yoneda Lemma

B Yoneda Lemma Setup: Let C be a category. We can define a covariant functor F from C to CS where CS is thecategory of contravariant functors C to S.14 On objects, we define F (C) = MC .15 Given a morphism f ∈ MorC(C,D)we define F (f) to be the natural transformation F (f) : MC →MD. In particular, if B is an arbitrary object in C,then we have (F (f))(B) : MC(B)→ MD(B) or equivalently (F (f))(B) : MorC(B,C)→ MorC(B,D) is given byg 7→ f ◦ g.

B Yoneda Lemma: F as defined above is a fully faithful functor.

Proof. We need to check four things here. They are:

1. F (f) is a natural transformation:Let A,B be objects in C with a morphism g : B → A. We must show the following diagram commutes:

MC(A) MC(B)

MD(A) MD(B)

MC(g)

F (f)(A) F (f)(B)

MD(g)

13This is because the dimensions are equal, so a basis must map to a basis to get injectivity.14See Fact on 12 September.15Recall that MC is a contravariant functor.


So let ϕ ∈MC(A) = MorC(A,C). Then we have (F (f)(B) ◦MC(g))(ϕ) = F (f)(B)(ϕ ◦ g) = f ◦ ϕ ◦ g. Similarly,we get (MD(g) ◦ F (f)(A))(ϕ) = MD(g)(f ◦ ϕ) = f ◦ ϕ ◦ g, so that the diagram commutes. However, this is“obvious” as MC(g) and MD(g) act by composing with g on the right, and F (f)(A) and F (f)(B) act by composingwith f on the left, and composition is associative.

2. F is a covariant functor: Let C be an object in C, and let 1C be the identity morphism on C in C. Then,F (1C) is a natural transformation from MC to MC . So let B ∈ Ob(C), then F (1C)(B) : MC(B)→MC(B) isgiven by g 7→ g ◦ 1C = g. Thus, F (1C) = 1F (C). We now need to look at compositions. So let f ∈ MorC(A,B) andg ∈ MorC(B,C). Then we want to show that F (g ◦ f) = F (g) ◦ F (f). For any object D in the category C, thenF (g◦f)(D) : MA(D)→MC(D) is given by composition with g◦f on the left, so we get that h 7→ (g◦f)◦h. Similarly,F (f)(D) : MB(D)→MC(D) is given by composition with f on the left, so h 7→ f ◦ h, and F (g)(D) : MA(D)→MB(D) is given by composition with g on the left, so h 7→ g ◦ h. Now, if we have h ∈ MA(D), then (F (g) ◦F (f))(h) = F (g)(f ◦h) = g◦(f ◦h). However, composition is associative, and so we have that F (g◦f) = F (g)◦F (f).

3. F is faithful: Let C,D be objects in C and consider the map MorC(C,D)→ MorCS (MC ,MD) which is givenby f 7→ F (f). We must show that this map is injective. So let g, h ∈ MorC(C,D) such that g 6= h. We mustshow that F (g) 6= F (h). We’ll show this on C. So consider F (g)(C), F (h)(C) : MC(C) → MD(C). Then,F (g)(C)(1C) = g ◦ 1C = g 6= h = h ◦ 1C = F (h)(C)(1C), so that F (g)(C) 6= F (h)(C) and hence F (g) 6= F (h).

4. F is full: Again, Let C,D be objects in C and consider the map MorC(C,D)→ MorCS (MC ,MD) which is given byf 7→ F (f). We must show that this map is surjective. So let η ∈ MorCS (MC ,MD). We’ll set α = η(C)(1C). Notefirst that η(C) : MC(C)→MD(C) so that α ∈ MorC(C,D). We now claim that F (α) = η. To show this, let A bean object in C and ϕ ∈ MorC(A,C). Then as η is a natural transformation, we know the following diagram commutes:

MC(C) MD(C)

MC(A) MD(A)

η(C)

ϕ∗ = MC(ϕ)

η(A)

MD(ϕ) = ϕ∗

Now, tracing the identity morphism on C will give us what we want. Note here that the red equality is due to the defi-nition of α, the blue equality is due to the diagram commuting, and the green equality is due to the definition of F (α).

idC η(C)(idC)= α

idC ◦ ϕ = ϕ η(A)(ϕ)=α ◦ ϕ=F (α)(A)(ϕ)

Thus the diagram tells us that η(A)(ϕ) = F (α)(A)(ϕ) and so we have that η = F (α) as desired so that the mapis surjective.

21 September 2011 Representability and Free Groups

B Example: Let X be a set. We’ll define a covariant functor FX from Ab to S. On objects, FX(A) = MorS(X,A)and on morphisms, if g : A → B, then FX(g) : F (A) → F (B) is given by ϕ 7→ g ◦ ϕ. We wonder now if FX isrepresentable. It is, and I’ll show how we got there in class.

Proof. We started by assuming that FX is representable and seeing what that meant about the functor and aboutthe object that represents it.

– So we’ll assume that B represents FX , i.e. that FX is naturally equivalent to MB . Let Φ : MB → FX be thenatural transformation.

– Then in particular we’d have ΦB : MB(B)→ FX(B) which is given by 1B 7→ ΦB(1B) = ϕ, where ϕ is simplydefined to be the image of ΦB applied to the identity on B.

– However, we really want to know what Φ is in general. So let D be an abelian group, and g : B → D a grouphomomorphism. We create the following commutative diagram to help us figure out what ΦD(B) is.


MB(B) MB(D)

FX(B) FX(D)

MB(g) = g∗

ΦB ΦD

FX(g) = g∗

– Tracing the element 1B tells us what we need, namely, that ΦD(g) = g ◦ ϕ = g∗(ϕ). See the below diagram wherethe red equality is what will make the diagram commute.

1B g∗(1B) = g

ϕ g∗(ϕ) = g ◦ ϕ = ΦD(g)

– Claim: The pair (B,ϕ : X → B) satisfies the following universal property: for any abelian group D and anyset map ψ : X → D there is a unique homomorphism hψ : B → D such that the following diagram commutes:

B D

X

hψ

ϕ ψ

Sub-Proof. This proof was fairly simple and was based entirely on the fact that we’re assuming that Φ is a naturalequivalence, i.e., that ΦD is an isomorphism of sets (which is just a bijection).X

B Comment: The pair (B,ϕ) exists, and is referred to as the free abelian group on the set X.

B Comment: The above shows that the free abelian group on the set X represents the functor FX .

B Construction/Definition: A much more explicit construction of a free abelian group can also be given. Here,

we did this as the free abelian group on the set X is B =⊕x∈X

Z, with the map ϕ : X → B. More generally, if

{Bx : x ∈ X} is a family of abelian groups such that Bx = Z for all x ∈ X, then B =⊕x∈X

Bx and ϕ : X → B is

given by x 7→ ex where ex : X → Z is the map ex(y) = δxy =

{1, x = y

0, x 6= y.With all this, we denote the canonical inclusions by πx : Bx → B.

B Claim: We claim now that B is a free abelian group, i.e. that it satisfies the appropriate universal property. Theidea is to either:

– Show directly that (B,ϕ) satisfies the universal property OR– Use the universal property of B as a categorical coproduct (sum) to show that (B,ϕ) satisfies the universal

property of a free abelian group.

23 September 2011 Free Groups & Coproducts

B Claim: (B,ϕ) as defined in the last class is a free abelian group, that is, it satisfies the following universal property:given any set map Ψ : X → D, where D is an abelian group, there is a unique homomorphism hΨ : B → D such thatthe following diagram commutes:

B D

X

hΨ

ϕ Ψ

Proof. We did most of the details of this in class. However, we used the fact that B is a categorical coproduct,and that basically gave us what we wanted, so I’m going to not rewrite down the details since neither existencenor uniqueness seemed at all tricky.


B Definition: Let X be a set, F a group, and ϕ : X → F a set map. We say (F,ϕ) is a free group on the set Xif for every group G and every map Ψ : X → G, there exists a unique homomorphism hΨ : F → G such that thediagram below commutes.

F G

X

hΨ

ϕ Ψ

B Fact: Coproducts and free groups exist in the category of groups.

B Proposition: Let A and B be groups with A ∩B = ∅ (for simplicity). Define the free product A ◦B = A ∗B to bethe set of sequences (a1, . . . , am) for m ≥ 0 such that ai ∈ (A \ {1A}) ∪ (B \ {1B}) and such that if ai ∈ A, thenai+1 /∈ A and if ai ∈ B then ai+1 /∈ B. Then A ◦B is a group under the following:

– 1A◦B is the unique sequence of length 0, namely (),– (a1, . . . , am)−1 = (a−1

m , . . . , a−11 ), and

– ab = (a1, . . . , am) · (b1, . . . , bn) =

(a1, . . . , am, b1, . . . , bn) if am, b1 are in different groups, A,B

(a1, . . . , amb1, . . . , bn) if am, b1 ∈ A (or B), and amb1 6= 1A or 1B(a1, . . . , am−1) · (b2, . . . , bn) if amb1 = 1A or 1B

.

Proof. We discussed the ideas behind this proof in the next class, but I don’t think that any of them are really worthtyping up. Basically, these things make sense. The only difficulty arose in showing associativity,16 and a sort ofclever induction argument made it not so bad.

26 September 2011 More on Coproducts & Free Groups

B Comments:

– A ◦B = B ◦A– We have the obvious injective homomorphisms:17 πA : A → A ◦ B, and πB : B → A ◦ B where x 7→ (x) forx ∈ A (or x ∈ B as is appropriate), and obviously in each case the identity maps to the identity in A ◦B whichis the sequence of length 0.

– Given any group homomorphisms, ϕ∗ : ∗ → G for ∗ = A,B, then we claim there is a unique18 group homomorphismϕ : A ◦B → G such that the following diagrams commute for ∗ = A,B:

A ◦B G

∗

ϕ

π∗ ϕ∗

In fact, we define ϕ as follows: Clearly we set ϕ(()) = eG. Also, ϕ((x)) = ϕ∗(x) where x ∈ ∗ and ∗ = A,B, andin general, if x = (x1, . . . , xn), then ϕ(x) = ϕ((x1)) · . . . · ϕ((xn)). There are a bunch of cases to check it is ahomomorphism, but it seems clear enough without checking them. Also, the fact that the diagrams commuteseems clear, so I won’t write down any of the details of either.19

B Corollary: Let A and B be groups, which we regard as having disjoint sets of elements. Then (A ◦B, πA, πB) is acoproduct of A and B in the category of groups.

Proof. It only remains to show uniqueness of the map ϕ. However, this is easy to show since the diagrams mustcommute and ϕ must be a homomorphism, so I’ll omit the details from these typed notes.

B Examples: The coproduct of A = Z/2Z = {0, 1A} and B = Z/2Z = {1B ,−1} contains the elements: A ◦ B ={(), (1A), (−1), (1A,−1), (−1, 1A), (1A,−1, 1A), (−1, 1A,−1), . . .}. Note here that there are exactly two elements ofeach length. However, in the category of abelian groups, the coproduct is the direct sum, namely A⊕B = Z/2Z⊕Z/2Zwhich is the Klein Vier Group.20

B Lemma: Let X1, and X2 be disjoint sets, and assume (Fi, ϕi) is a free group on Xi for i = 1, 2. Then F1 ◦ F2 is thefree group on the set X = X1 ∪X2 with respect to the map ϕ : X → F1 ◦ F2 defined by x 7→ πi(ϕi(x)).

16as always!17that they are injective and homomorphisms is “clear.”18“I haven’t asserted uniqueness, but it will be unique. It’s the freeness that makes it unique.” –B.H.19I’m aware that this silly ∗ notation sucks, but I’m lazy and don’t want to do everything twice for both A and B.20Vier is 4 in German.


28 September 2011 Free (Abelian) Groups

B Lemma: If X1, X2 are disjoint sets, and (Fi, ϕi) is a free group on Xi for i = 1, 2, then F1 ◦ F2 is a free group onX = X1 ∪X2.

Proof. Let πi : Fi → F1 ◦ F2 be the coproduct canonical homomorphisms, and define ϕ : X → F1 ◦ F2 viax 7→ πi(ϕi(x)) for x ∈ Xi. We need to show that (F1 ◦ F2, ϕ) is a free group on X. So let ψ : X → G be any setmap to any group G. We must show there is a unique homomorphism hψ : F1 ◦ F2 → G such that hψ ◦ ϕ = ψ.In order to work the coproduct into the proof, we’ll need homomorphisms Γi : Fi → G. Let Γi be the uniquehomomorphism guaranteed by the universal property of Fi free from the set map γi : Xi → G, where γi = ψ|Xi .The maps Γi then induce the map Γ : F1 ◦ F2 → G since F1 ◦ F2 is a coproduct. Set hψ = Γ. We must show thathψ ◦ ϕ = ψ and that this map is unique. So, let x ∈ X = X1 ∪X2. Without loss of generality (WLOG), let x ∈ X1.Then,

(hψ ◦ ϕ)(x) = Γ ◦ πi ◦ ϕi(x) = Γ1 ◦ ϕ1(x) = γ1(x) = ψ|X1(x) = ψ(x).21

B Examples:

1. The free group on the empty set is the trivial group.2. The free group on X = {1} is isomorphic to Z.3. Inductively, using the lemma, we have the the free group on a set X = {1, . . . , n} is Z ◦ . . . ◦ Z︸︷︷︸

n times

.

4. It is possible to define a free group on an infinite set as well, and Lang describes this. It is similar in idea towhat we’ve done so far, but is a bit messier.22

B Lemma 1: Let (F,ϕ : X → F ) be a free group. Then ϕ is injective, and ϕ(X) generates F .

Proof. Notice that the injectivity is trivial if |X| < 2, so we may let x1, x2 be distinct elements of X. Define ψ : X →

Z/2Z as ψ(x) =

{1, x 6= x1

0, x = x1. Then by the universal property, there is a unique homomorphism hψ : F → Z/2Z

such that hψ ◦ϕ = ψ. However, ψ(x1) 6= ψ(x2), so we must have that ϕ(x1) 6= ϕ(x2) due to the fact that hψ ◦ϕ = ψ.Next, let G be the subgroup of F generated by ϕ(X). We’ll show G = F . Consider the diagram below:

F G F

X

hϕ i

ϕϕ

ϕ

Notice that by definition of F as a free group, there is a unique homomorphism hϕ : F → F such that hϕ ◦ϕ = ϕ. As1F works for hϕ, then we must have that 1F = i ◦ hϕ. Since 1F is a bijection, then we must have that i is surjective,meaning that F ⊆ G, and hence F = G.

B Lemma 2: Let (A,ϕ : X → A) be a free abelian group. Then ϕ is injective and ϕ(X) generates A.

Proof. The proof is the same as the proof of Lemma 1, mutatis mutandis.23

30 September 2011 Free (Abelian) Groups

B Example: If X is a set, then A =⊕x∈X

Z together with e : X → A given by x 7→ ex is a free abelian group, where

ex : X → Z is given by ex(y) = 0 for y 6= x, and ex(x) = 1.

B Lemma 3: Let (A,ϕ : X → A) be a free abelian group on X. Then every element a ∈ A has a unique expression

a =∑x∈X

mxϕ(x), where mx ∈ Z.

21This is “obvious” once you draw the pictures, but there are too many to be worth putting in here.22obviously!23See Hungerford page xv. This is latin for (roughly) “by changing the things which (obviously) must be changed (in order that the

argument will carry over and make sense in the present situation).”


Proof. Notice that existence here is due to Lemma 2 since ϕ(X) generates A as an abelian group. So we really only

need to show uniqueness. Since

(⊕x∈X

Z, e

)is a free abelian group, then we have homomorphisms he and hϕ such

that he ◦ ϕ = e and hϕ ◦ e = ϕ. Now, suppose∑x∈X

mxϕ(x) =∑x∈X

nxϕ(x). We’ll apply he to get the following:

∑mxex =

∑mxhe(ϕ(x)) = he

(∑mxϕ(x)

)= he

(∑nxϕ(x)

)=∑

nxhe(ϕ(x)) =∑

nxex.

We can apply these functions to any element y ∈ X to get that

my =∑

mxex(y) =∑

nxex(y) = ny

for all y ∈ X. Hence, the two expansions are equivalent, which gives us uniqueness.

B Definition: A subset, B, of an abelian group, A, is a basis for A if B generates A in a unique way.

B Definition: If B is a basis for A, then we say |B| is the rank of A.

B Lemma 4: If (F,ψ : X → F ) is a free abelian group and we have a commutative diagram:

F G

X

'

ψ ϕ

then (G,ϕ) is free abelian on X as well.

B Proposition: If B is a basis for an abelian group A, then (A,ϕ) is a free abelian group where ϕ : B ↪→ A is theinclusion of B into A.

Proof. Let (F,ψ) be a free abelian group on the set B. Then we have the induced commutative diagram:

F A

B

hϕ

ψ ϕ

We now claim that in fact hϕ is an isomorphism. We’ll first do surjectivity. As B = ϕ(B) generates A, andϕ(B) = ψ(hϕ(B)), then hϕ(F ) contains B, so that hϕ(F ) = A since B is a basis for A. Next, we’ll do injectivity.

Let f ∈ ker(hϕ). Since ψ(B) generates F , then we get f =∑b∈B

mbψ(b) for some mb ∈ Z. Now,

0 = hϕ(f) = hϕ

(∑mbψ(b)

)=∑

mbhϕ(ψ(b)) =∑

mbϕ(b),

which means that mb = 0 for all b ∈ B, and hence f = 0, so that hϕ is injective. Applying Lemma 4 now completesthe proof.

B Lemma 5: If we have a free abelian group (A,ϕ : X → A) and a set Y with a bijection β : X → Y , then (A,ϕ ◦ β−1)is a free abelian group on Y .24

B Corollary: Let A be an abelian group. Let X be a set with a map ϕ : X → A. then (A,ϕ) is a free abelian groupif and only if ϕ(X) is a basis for A and ϕ is injective.25

B Fact: If X is an infinite set, and Fin(X) is the set of finite subsets of X, then |X| = |Fin(X)|.

3 October 2011 Free Groups, Group Actions

B Theorem: Let (Ai, ϕi : Xi → Ai) be free abelian groups for i = 1, 2. Then A1∼= A2 if and only if |X1| = |X2|.

24The proof here is “easy” if you draw the pictures.25The proof here is an application of the lemmas and the proposition.


Proof. (⇐) Let ψ : X1 → X2 be a bijection. Since (A2, ϕ2) is free abelian on X2, then (A2, ϕ2 ◦ ψ) is free abelianon X1 by Lemma 5. Thus, A1

∼= A2 by the universal property of their both being free on X1.(⇒) Now, assume A1

∼= A2, and let h : A1 → A2 be an isomorphism. Set 2Ai = {2a|a ∈ Ai} and note that 2Ai CAi.Set Ai = Ai/2Ai. We claim now that h induces an isomorphism h : A1 → A2. So set h(a) = h(a). We first show thisis well defined. Suppose a = b ∈ A1. Then a− b ∈ 2A1. Note that h(2A1) = 2h(A1) ⊆ 2A2 so that h(a)−h(b) ∈ 2A2

which gives h(a) = h(b). Next, we must show that h is an isomorphism. In fact, we have the following commutativediagram:

A1 A2

A1 A2

h

−

h

−− ◦ h

Since h and − are surjective, then so is − ◦ h. We claim then that ker(−◦h) = 2A1 and apply the first isomorphism

theorem.26 Note that (−◦h)(2a) = 2h(a) = 0 in A2 so that 2A1 ⊆ ker(−◦h). Now, suppose (−◦h)(a) = 0. Then

we have h(a) = 0 so that h(a) ∈ 2A2. So let h(a) = 2b for some b ∈ A2. But h is an isomorphism, so there existssome β ∈ A1 such that h(β) = b. Thus, h(a) = 2h(β) = h(2β) so that a = 2β ∈ 2A1 since h is an isomorphism.Thus, h is an isomorphism.Now, for i = 1, 2 we have the following:

Ai ∼=

⊕x∈Xi

Z

2⊕x∈Xi

Z∼=

⊕x∈Xi

Z⊕x∈Xi

2Z∼=⊕x∈Xi

Z/2Z,

which gives the isomorphism⊕x∈X1

Z/2Z ∼=⊕x∈X2

Z/2Z. Now, if |X1| <∞, then we get

2|X1| =

∣∣∣∣∣⊕x∈X1

Z/2Z

∣∣∣∣∣ =

∣∣∣∣∣⊕x∈X2

Z/2Z

∣∣∣∣∣ = 2|X2|

which means that |X1| = |X2|. If, on the other hand, X1 (and hence also X2) is infinite, then we have

|X1| = |Fin(X1)| =

∣∣∣∣∣⊕x∈X1

Z/2Z

∣∣∣∣∣ =

∣∣∣∣∣⊕x∈X2

Z/2Z

∣∣∣∣∣ = |Fin(X2)| = |X2|

which completes the proof.

B Comment: We’re now going to move on from free groups and start discussing group actions. We’ll also briefly coverthe Sylow Theorems, and will probably cover semi-direct products.

B Definition: Let G be a group and X a set. An action of G on X is a map α : G×X → X (usually we write g · xor simply gx for the action of g on the set element x) such that

1. 1G · x = x for all x ∈ X, and2. g1 · (g2 · x) = (g1g2) · x for all x ∈ X and for all g1, g2 ∈ G.

B Definition: We say the orbit of an element x ∈ X is the set G · x = {g · x|g ∈ G}.B Definition: Let x ∈ X. Then the stabilizer of x is StabG(x) = {g ∈ G|g · x = x}. Notice here that the stabilizer

is a subgroup of G.

B Comment: Sometimes the notation Gx is used for the stabilizer of x, but that notation sucks given our notation forthe orbits!

B Definition: We say an action is transitive if it has a single orbit. Equivalently, if for some x ∈ X (but really itwill be true for all x ∈ X), we have G · x = X.

26Apparently the isomorphism theorems are due to Emmy Noether?


B Comment: Giving an action of G on X is equivalent to giving a homomorphism from G to the group of permutationson the set X. Note here that this group is really the group of bijections X → X where the group law is composition.We’ll call this group of permutations Perms(X).

5 October 2011 More Group Actions; Starting Semi-Direct Products

B Recall: We have two equivalent definitions of an action being transitive. The first is that there is a unique orbit.The second is that given any two elements x, y ∈ X, there is a group element g ∈ G such that g · x = y.

B Example: We discussed the action of S1 on C where S1 acts on C via regular multiplication in C. Here, this actionis not transitive, but each orbit is a circle centered about the origin.

B Theorem: (In particular a version of Cayley’s Theorem) Let ϕ : G×X → X be an action of a group G on a setX, let x ∈ X, and H = StabG(x).

1. Then ϕ induces a homomorphism Φ : G→ Perms(X) which is given by g 7→ πg where πg(y) = g · y.

2. If the action is transitive, then ker Φ ⊆ H and moreover, ker Φ =⋂g∈G

gHg−1 and lastly, ker Φ contains every

subgroup N ≤ G such that N CG and N ⊆ H.

Proof. We’ll show each part:

1. We first need to show that πg is actually a permutation for every g ∈ G. So note that given g ∈ G, then(πg ◦ πg−1)(x) = πg(g

−1 · x) = g(g−1x) = eGx = x and (πg−1 ◦ πg)(x) = g−1(gx) = eGx = x so that πg is abijection for each g and hence an element of Perms(X).Next, we need to show that Φ is a homomorphism. That is, we need to show that πg1g2 = Φ(g1g2) =Φ(g1) ◦ Φ(g2) = πg1 ◦ πg2 . So let x ∈ X. We’ll show that both permutations send x to the same place. Now,πg1g2(x) = (g1g2)x = g1(g2x) = πg1(g2x) = πg1(πg2(x)) = πg1 ◦ πg2(x). Thus, Φ is a homomorphism.

2. First, we’ll show that ker Φ ⊆ H . So let a ∈ ker Φ. Then πa = 1X so that πa(y) = a · y = y for any y ∈ X. Hence,a · x = x so that a ∈ StabG(x) = H.

Next, we need to show that ker Φ ⊆⋂g∈G

gHg−1. So let a ∈ ker Φ and let g ∈ G. It is clearly enough to show that

a ∈ gHg−1. If a ∈ ker(Φ) then a(gx) = gx so that g−1agx = x. this means that g−1ag ∈ H and so a ∈ gHg−1.

Next, we’ll show the other inclusion ker Φ ⊇⋂g∈G

gHg−1. So let b ∈⋂g∈G

gHg−1, and let y ∈ X. Then y = γx for

some γ ∈ G since the action is transitive. Also, b = γhγ−1 for some h ∈ H by choice of b. Thus, multiplying bothsides of these two equations together gives by = γhγ−1γx. Simplifying the right hand side gives by = γhx = γx = y.Thus, Φ(b) = πb = 1X so that b ∈ ker(Φ).Lastly, we must show that ker(Φ) is the largest normal subgroup of G which is contained in H . So let N CG with

N ⊆ H, and choose any g ∈ G. Then N = gNg−1 ⊆ gHg−1. Thus, N ⊆⋂g∈G

gHg−1 ⊆ ker(Φ) which completes

the proof.

B Corollary: If a group G acts transitively on a finite set X and if |G| > n! where n = |X| > 1, then G has anontrivial normal subgroup, N .

Proof. We didn’t prove this in class, but I’ll do it now. Since |G| > |X|! = |Perms(X)|, then the homomorphismΦ from part 1 of the theorem cannot be injective. This means that ker(Φ) {1G}. By part 2 of the theorem above,we know that H = StabG(x) contains ker(Φ) for any x ∈ X. Now, since the action is transitive, and X has morethan one element, then there must be at least one g ∈ G such that gx 6= x. Hence, H � G, so that ker(Φ) is a propersubgroup of G as we now have {1G} � ker(Φ) ⊆ H � G. Since the kernel of any group homomorphism is a normalsubgroup, then we have that ker(Φ) is a non-trivial normal subgroup.

B Semi-Direct Products: We can do this in a more general setting, but we’ll give an example first to illustrate theidea.

B Example: Let G be a group, with subgroups N,H such that N ∩H = {1G} and N CG. Then, NH = {nh : n ∈N,h ∈ H} is a subgroup of G. We let H act on N by conjugation, and can refer to NH as an internal semi-directproduct.


7 October 2011 Semi-Direct Products & Back to Group Actions

B Definition: Let N,H be groups and Φ : H → Aut(N) where Aut(N) is the set of automorphisms of N . Note thatAut(N) ≤ Perms(N). We define the semi-direct product, N oΦ H, as follows. As a set N oH is just the setN ×H. We define the group operation27 as

(n1, h1) · (n2, h2) =(n1

(Φ(h1)(n2)

), h1h2

).

B Claim: N ∼= N × {1H}CN oH and H ∼= {1N} ×H ≤ N oH.28

B Recall: If N = Z/pZ where p is a prime, then Aut(N) ∼= Z/(p− 1)Z, and in general, Aut(Z/nZ) is of order φ(n)where φ is the Euler φ function.

B Example: Let N = Z/7Z and H = Z/3Z. Then, Aut(N) ∼= Z/6Z. Let Φ : H → Aut(N) be an injectivehomomorphism given by [1]3 7→ [2]6. Now, N o H has order 21, but is not an abelian group. You can see thisby considering the two elements ([1]7, [0]3) and ([0]7, [1]3) which do not commute since Φ([1]3) is not the identityautomorphism. On the other hand if we were to try and construct a semi-direct product of the groups in the otherorder, H oN , then we must use the zero map Φ : N → Z/2Z and so H oN = H ×N .

B Lemma: Let G be a finite group, acting on a finite set X and choose any x ∈ X. Then |G| = |Gx| · |G · x|.Proof. This is mostly just an application of Lagrange’s theorem, and noting that there is a bijection between elementsof the orbit G · x and the left cosets of StabG(x) = Gx in G.

B Orbit Decomposition Theorem: Let G be a finite group acting on a finite set X. Let X1, . . . , Xr be the distinctorbits in X for the action. Choose xi ∈ Xi for each i. Then

|X| =r∑i=1

|G/Gxi | =r∑i=1

[G : Gxi ].

Proof. This proof is a simple application of the lemma and Lagrange’s theorem.

10 October 2011 More Group Actions

B Proposition: Let G be a finite group acting on a finite set X. Let n be the number of orbits. Let f : G→ Z be a

map where f(g) is the number of fixed points of g, i.e. f(g) = |{x ∈ X : gx = x}|. Then n|G| =∑g∈G

f(g).29

Proof. Let X1, . . . , Xn be the distinct orbits, and set fi(g) = |{x ∈ Xi : gx = x}|. Note then that f(g) =

n∑i=1

fi(g).

Let Γi be a “correspondence” in G × Xi. In particular, let Γi = {(g, x) ∈ G × Xi|gx = x}. We then have thefollowing diagram where π1 and π2 are the usual projection maps:

Γi ⊆ G×Xi G

Xi

π1

π2

Consider π2|Γi : Γi → Xi. Note that this map is surjective as (eG, x) ∈ Γi for all x ∈ Xi. Also, the fibers are

(π2|Γi)−1({x}) = {(g, x) : gx = x} = StabG(x)× {x}, and

(π1|Γi)−1({g}) = {(g, x) : x ∈ Xi and gx = x},27We didn’t actually show this operation turned N ×H into a group, but it isn’t hard to show, and some people even did it as homework.

The only tricky part is determining that (n, h)−1 = (Φ(h−1)(n−1), h−1).28This seems not too hard to show and part of it may have been done by some people for homework.29This was actually stated the previous day, but proved this day, so it’s listed here.


which is the set of fixed points of G in Xi and so |(π1|Γi)−1({g})| = fi(g). The fibers of a map are always disjoint,

and their union is the domain; hence,∑g∈G

fi(g) = |Γi| =∑x∈Xi

|StabG(x)|. Recall that |G| = |StabG(x)| · |G · x|.

Hence,∑g∈G

fi(g) = |Γi| =∑x∈Xi

|G|/|Xi| = |G|. Now, we add these up over i to get∑g∈G

f(g) =∑g∈G

n∑i=1

fi(g) =

n∑i=1

∑g∈G

fi(g) =

n∑i=1

|G| = n|G|.

B Corollary: Say G is a finite group acting transitively on a finite set X with |X| > 1. Then there exists an elementg ∈ G which is fixed-point free, that is, so that gx 6= x for all x ∈ X. In the language of the previous proposition,this says there is an element g ∈ G such that f(g) = 0.

Proof. Suppose not, then for each g ∈ G there exists an xg ∈ X such that g · xg = xg. Or equivalently, f(g) ≥ 1

for every g ∈ G. Since the action is transitive, |G| =∑g∈G

f(g). This means that f(g) = 1 for every g ∈ G, but this

is a contradiction since f(1G) = |X| ≥ 2.

B Corollary: Let G be a group, H a subgroup such that H � G. If G is a finite group, then⋃g∈G

gHg−1 ( G.

Proof. We know G acts on the left cosets of H by left multiplication. Set X = {gH}g∈G. Since H � G, then wehave that |X| > 1. Also, note that this action is transitive, and choose any a ∈ G. Then a ∈ gHg−1 for some g ∈ Gif and only if g−1ag ∈ H , which happens if and only if ag ∈ gH , or equivalently if and only if a is in the stabilizer of

gH under the group action, i.e. a · gH = gH . Now, let S =⋃g∈G

gHg−1. Then a ∈ S if and only if f(a) ≥ 1, where

as before, f(a) is the number of fixed points of a under the action, or equivalently, f(a) is the number of conjugatesgHg−1 which contain a. However, by the previous corollary, there is some b ∈ G such that f(b) = 0. Hence, b /∈ S.

12 October 2011 More Group Actions & the Class Equation

B We now provide an alternate proof for the Corollary proved last time:

Proof. Alternate Proof: Let G act on the set of conjugates of H by conjugation, and set X = {gHg−1}.30

Note that StabG(H) = {g ∈ G : gHg−1 = H} = NG(H).31 Recall that |gHg−1| = |H| for all g ∈ G, and that1G ∈ gHg−1 for all g ∈ G. Let n be the number of conjugates of H , i.e. n = |X| = |G|/|StabG(H)| = |G|/|NG(H)|.If n = 1, then H CG and for all g ∈ G, we have gHg−1 = H, so clearly

⋃g∈G

gHg−1 � G. So suppose n 6= 1. We

have the following computation:∣∣∣∣∣∣⋃g∈G

gHg−1

∣∣∣∣∣∣ ≤ n (|H| − 1) + 1 =|G|

|NG(H)|(|H| − 1) + 1 =

|G||NG(H)|

|H| − |G||NG(H)|

+ 1.

Now, NG(H) � G, |H|/|NG(H)| ≤ 1, and |G|/|NG(H)| = n ≥ 2, and so from the computation we get that∣∣∣∣∣∣⋃g∈G

gHg−1

∣∣∣∣∣∣ ≤ |G| − 2 + 1 = |G| − 1 < |G|.

B Class Equation: Let G act on itself via conjugation g ·h = ghg−1. This gives us a homomorphism Φ : G→ Aut(G)with kernel the center of the group, which we recall is denoted Z(G). The image of Φ is a subgroup of Aut(G) whichis known as the group of inner automorphisms, and is sometimes denoted Inn(G). For h ∈ G, we have that StabG(h)is {g ∈ G : ghg−1 = h} = CG(h), the centralizer of h in G. More generally, the centralizer of a subset S ⊆ G isCG(S) = {g ∈ G|gs = sg for all s ∈ S}. Note here that CG(G) = Z(G).We now assume that G is a finite group, and that X1, . . . , Xn are the disjoint orbits of the group under the conjugationaction. For each i, choose any xi ∈ Xi. Then we have the class equation:

|G| =n∑i=1

|Xi| =n∑i=1

[G : CG(xi)].

30This IS an action.31Recall that NG(H) is the normalizer of H in G.


However, this is more commonly written as follows and separates out the singleton orbits:

|G| = |Z(G)|+∑

[G : CG(xi)]

where now the sum is taken over those i’s such that CG(xi) � G.

B Proposition: Let G be a finite group, and H a subgroup of index 2. Then H CG.

Proof. Let g ∈ G. If g ∈ H , then obviously gH = Hg. However, if g /∈ H , then gH and Hg are both disjoint from H ,and since the index is 2, we must have that gH = Hg. Hence, gH = Hg for all g ∈ G, and so H is normal in G.32

14 October 2011 Group Actions & Sylow Theorems & Nilpotent Groups33

B Corollary: If G is a finite group and has an element with exactly two conjugates, then G has a non-trivial, proper,normal subgroup.34

Proof. Let g ∈ G be an element with exactly two conjugates, then under conjugation of the group on itself, StabG(g)has index 2. So by the proposition, StabG(g) is a proper normal subgroup of G. It remains to show that StabG(g)is a non-trivial, proper subgroup. However, this is easy since g ∈ StabG(g) and g 6= 1G since 1G has only itself asa conjugate, and so StabG(g) is a non-trivial, proper, normal subgroup.

B Proposition: Let G be a finite group, and let p be the smallest prime such that p divides the order of the group. IfG has a subgroup, H, of index p, then H CG.

Proof. We didn’t prove this in class, but I’ll do it here anyways. Let X be the set of left cosets of H in G. Then, Gacts on H by left multiplication, and this action is clearly transitive. Then the homomorphism, Φ, given by Cayley’stheorem has kernel ker(Φ) ⊆ H as we have that H = StabG(H) since gH = H if and only if g ∈ H.35 Our goalis to get that ker(Φ) = H . Now, we know that Perms(X) is a group of size p!, and so we must have that |G/ ker(Φ)|divides p!. However, if r divides the index of ker(Φ) in G, then r divides the order of G since the index divides theorder of the group. Since p is the smallest prime that divides the order of the group, and no number larger thanp divides G/ ker(Φ), then we must have that |G/ ker(Φ)| is either 1 or p. We now do the following computation:

|G/ ker(Φ)| = [G : ker(Φ)] = [G : H][H : ker(Φ)] = p[H : ker(Φ)] ≥ p.

Thus, we must have that [G : ker(Φ)] = p, so that ker(Φ) = H since we already knew that ker(Φ) ⊆ H. Hence His normal in G.

B Definition: Let p be a prime. A finite group, G, is called a p-group if |G| = pn for some n ∈ N.

B “Trivial” Example: The trivial group is a p-group for all primes p.36

B Lemma: Let G be a non-trivial p-group for some prime p. Then |Z(G)| > 1.

Proof. We apply the class equation. The size of the group G is the size of its center plus the sum of the order ofthe orbits of elements having conjugates other than themselves. Note that p divides the order of the group, andp divides the order of each orbit of elements having conjugates other than themselves. Thus, p must divide the orderof the center of G, and so |Z(G)| is at least p, and hence greater than 1.

B Cauchy’s Theorem: Let G be a finite group, and p a prime such that p divides the order of G. Then G has anelement of order p.

Proof. From class notes: “A beautiful proof can be given using Z/pZ acting via cyclic permutations on a subsetof G× . . .×G︸︷︷︸

p times

.” –B.H.37

B Definition: Let p be a prime. A subgroup of order pn of a finite group, G is called a p-subgroup.

B Definition: Let G be a finite group of order pn ·m where p is prime and (m, p) = 1. Then a p-subgroup of G thathas order pn is called a Sylow p-subgroup.

B Sylow’s Theorem(s): Let G be a finite group of order pn ·m where p is prime and p - m.

1. G has a Sylow p-subgroup.

32This proof is different from the one given in class, but makes more sense to me.33Oh My!34Groups with no non-trivial, proper, normal subgroup are called simple, so we could just say that then G is not a simple group.35The subgroup H and the stabilizer H from the theorem are now the same thing.36Hahaha!37This seems to indicate that we don’t need to know the proof, but at some point, I think I should understand and type up a real proof.


2. If Q is a p-subgroup of G and P is a Sylow p-subgroup, then Q ⊆ gPg−1 for some g ∈ G. In particular, anytwo Sylow p-subgroups of a group are conjugate.

3. If np denotes the number of Sylow p-subgroups, then np ≡ 1 (mod p) and np|m.38

B Definition: A sequence of subgroups G1 ≤ G2 ≤ . . . ≤ Gr = G of a group G is called an abelian tower ifGi CGi+1 and Gi+1/Gi is abelian and both statements hold for all i.

B Comment: Nilpotent and Solvable groups be defined based upon special abelian towers.

19 October 2011 Sylow Theorems & Nilpotent Groups

B Useful Fact: Let P be a Sylow p-subgroup of a finite group G. Then NG(NG(P )) = NG(P ).

Proof. We have that P CNG(P )CNG(NG(P )) but remember that normality is not a transitive relation. So, weknow that P is the unique Sylow p-subgroup of NG(P ) by part 2 of the Sylow Theorem(s). We easily get the inclusionNG(P ) ⊆ NG(NG(P )), so we need to show the other. Let g ∈ NG(NG(P )). Then gPg−1 C gNG(P )g−1 = NG(P )by choice of g. Thus, P = gPg−1 since P is the unique Sylow p-subgroup of NG(P ). In particular, this means thatg ∈ NG(P ) and so we get the other inclusion.

B Lemma: Let G be a finite group. Assume the Sylow subgroup of G are all normal. Then G is the direct product ofits Sylow subgroups.

Proof. Let |G| = pr11 · . . . · prss where pi are distinct primes and assume ri > 0 for all i. By part 1 of Sylow’sTheorem(s), there exists a Sylow pi-subgroup, Pi for each i. Then we need to show that G ∼= P1 × . . . × Ps. Ifs = 1, then the statement is trivial since G = P1 is a p1-group, so assume s > 1. Define θ : P1 × . . .× Ps → G by(g1, . . . , gs) 7→ g1 · . . . · gs. We need to check three things:θ is a Homomorphism: Let (g1, . . . , gs) and (h1, . . . , hs) be elements of P1 × . . .× Ps. Then,

θ((g1, . . . , gs) · (h1, . . . , hs)) = θ((g1h1, . . . , gshs)) = g1h1 · . . . · gshs.

Also, θ((g1, . . . , gs)) · θ((h1, . . . , hs)) = (g1 · . . . · gs) · (h1 · . . . · hs). We get that this is a homomorphism once weshow that gh = hg whenever g ∈ Pi, h ∈ Pj , and i 6= j. Note that ghg−1h−1 = (ghg−1)h−1 ∈ Pj since h ∈ Pj CG.Also, ghg−1h−1 = g(hg−1h−1) ∈ Pi since g ∈ Pi CG. However, we have that the orders of the subgroups Pi andPj are coprime, and so we must have that the order of the element ghg−1h−1 is 1, and hence ghg−1h−1 = eG = eso that gh = hg.Injectivity: Suppose that (g1, . . . , gs) ∈ ker(θ). Then we have g1 · . . . · gs = eG = e. Let mi = |G|/prii so thatmi and prii are coprime. Thus, there exist x, y ∈ Z such that 1 = xmi + yprii . Now, e = (e)xmi = (g1 · . . . · gs)xmi =gxmi1 · . . . · gxmis = gxmii since if i 6= j, then the order of gj divides mi. Thus, we get

e = gxmii = g1−yprii = gi ·

(gpriii

)−y= gi · (1G)−y = gi.

This means that (g1, . . . , gs) = (e, . . . , e) and so θ is injective.Surjectivity: This is clear since the orders of the two groups are the same and θ is injective.

B Definition(s): Let G be a group. Consider the following sequence of quotients:

G = G0 � G1 � G2 � . . .

where Gi+1 = Gi/Z(Gi). We get a sequence then of surjective homomorphisms σj : G→ Gj . Let

C0(G) := (eG), and

Cj(G) := ker(σj).

T hen C0(G) < C1(G) < C2(G) < . . . is called the ascending central series. We’ll show next time that it isan abelian tower.39 The group G is nilpotent if G = Ci(G) for some i ∈ N. Equivalently, we can say that G isnilpotent if |Gi| = 1 for some i ∈ N.

B Example: Any abelian group is nilpotent. This is obvious since G/Z(G) = G1 is the trivial group.

38This second bit is usually (in my oh-so-humble experience) stated that np divides the order of the group, but given the first part,the two are equivalent.

39We did a terrible job doing it this day, but next time it actually makes sense!


21 October 2011 Nilpotent Groups

B Note: The following are fairly easy facts to show about the set of group homomorphisms f : A→ B and g : B → C.

1. f(A) ∼= A/ ker(f)2. ker(f) ⊆ ker(g ◦ f)3. f(ker(g ◦ f)) ∼= ker(g ◦ f)/ ker(f)4. ker(g ◦ f) = f−1(ker(g))5. If f is surjective, then f(ker(g ◦ f)) = f(f−1(ker(g))) = ker(g) by applying fact 4 from this list. Applying this

to the 3rd fact gives usker(g) ∼= ker(g ◦ f)/ ker(f) (1)

B Proposition: The ascending central series defined in the last class is an abelian tower.

Proof. Let f : G → Gi, and g : Gi → Gi+1 be the given surjections. Then ker(f) = Ci(G), ker(g) = Z(Gi),and ker(g ◦ f) = Ci+1(G) all by definition or simple observation. Plugging these into equation (1) gives usZ(Gi) ∼= Ci+1(G)/Ci(G). In particular, this means that the ascending central series is an abelian tower.

B Facts:

– If p is prime, then any finite p-group, G, is nilpotent.

Proof. If |Gi| > 1, then since Gi is a p-group, we have that |Z(Gi)| > 1 so |Gi+1| = |Gi/Z(Gi)| = |Gi|/|Z(Gi)|is less than |Gi|. Hence, we muse eventually get that Gt is trivial.

– A product of nilpotent groups is nilpotent.

Proof. Let B and C be nilpotent groups, and set G = B × C. Note that Z(G) = Z(B)× Z(C), so that we have

G/Z(G) =B × C

Z(B)× Z(C)∼= B/Z(B)× C/Z(C). In general, we get Gi ∼= Bi × Ci and so eventually we get both

Bi and Ci are trivial, and hence Gi is trivial as well.

– We can easily extend this to finite products, but it is not true for infinite products.– A group G = B(1)× . . .×B(r) where each B(i) is a finite pi-group for primes pi is nilpotent.

B Lemma: Let G be a nilpotent group, and let H � G. Then H � NG(H).

Proof. Let n be the largest index such that Cn(G) ⊆ H. We know such an n exists since G is nilpotent. Thenchoose any b ∈ Cn+1(G) \H. Since Cn+1(G)/Cn(G) ∼= Z(Gn) ∼= Z(G/Cn(G)), then bCn(G) ∈ Z(G/Cn(G)). Leth ∈ H . Then bhCn(G) = (bCn(G))(hCn(G)) = (hCn(G))(bCn(G)) = hbCn(G) and this all takes place in the groupG/Cn(G). Thus, there is some a ∈ Cn(G) ≤ H so that bha = hb. Thus, as h and a are both elements of H, weget, b−1hb = ha ∈ H. Thus, we get b ∈ NG(H) but we also had b /∈ H, and so H � NG(H) as desired.40

B Theorem: A finite group G is nilpotent if and only if it is the product of its Sylow subgroups.

Proof. (⇐) If G is a product of its Sylow p-subgroups for primes p which divide the order of the group, then Gis a finite product of nilpotent groups, and hence is nilpotent.(⇒) Say G is a finite nilpotent group. It is enough to show that each Sylow subgroup is normal in G, because ifso, then G is a product of its Sylow subgroups (see the first Lemma from 19 Oct). If G is a p−group, then we’reclearly done. So, we assume otherwise, and let P � G be a Sylow p−subgroup of G. From the Useful Fact atthe beginning of the 19th of October, we know that NG(NG(P )) = NG(P ). Also, by the Lemma just previous tothis, we know that if NG(P ) � G, then we’d have NG(P ) � NG(NG(P )) which would be a contradiction, and sowe must have that NG(P ) = G. Thus, P CG and that completes the proof.

24 October 2011 Solvable Groups

B Definition: A group G is solvable if it has an abelian tower of the following form:

(eG) = Gm CGm−1 C . . .CG0 = G,

where Gi/Gi+1 is abelian for all i. A series of this form is called a solvable series for G.

B Note: It is clear that nilpotent groups are all solvable. We’ll see later that not all solvable groups are nilpotent.

40I’ve modified this proof a bit from the one given in class so that I understand it better.


B Feit-Thompson Theorem: (Conjectured by Burnside) Every finite group of odd order is solvable.41

B Derived Series: Let G(1) (also written G′) be the commutator subgroup of G, [G,G], and call it the first derivedsubgroup of G. Recursively, G(i+1) is called the i+ 1st derived subgroup, and is the first derived subgroup ofthe ith derived subgroup, that is, G(i+1) = (G(i))′. Note that G ⊇ G(1) ⊇ G(2) ⊇ . . ., and that G(i)/G(i+1) is abelian(we know this from the homework). This sequence of subgroups is an abelian tower known as the derived series.

B Note: If there is an m such that G(m) = (eG), then G is solvable.

B Proposition: Let G be a group. Then G is solvable if and only if the derived series terminates at the identity.

Proof. (⇐) This is clear!(⇒) Assume G is solvable. Then there is an abelian tower Gm = (eG)CGm−1 C . . .CG0 = G. Since, G0/G1 isabelian, then G1 ⊇ G(1). We can inductively show that G(i) ⊆ Gi. Then we get G(m) ⊆ Gm = (eG), which completesthe proof since Gi/Gi+1 being abelian means that G′i ⊆ Gi+1 for all i. Thus, G(i+1) = (G(i))′ ⊆ G′i ⊆ Gi+1.

B Useful Fact: Let G be a group, and N CG, then G is solvable if and only if N and G/N are.

Proof. (⇒) Assume G is solvable, then N (i) ⊆ G(i), so if G(m) = (1G), then so does N (m), and so G being solvableimplies that N is solvable. Next, if f : G→ H is any group homomorphism, then f(G′) ⊆ H ′, and if f is surjective,then f(G′) = H ′. Thus, under the quotient homomorphism q : G → G/N , we get q(G(i)) = (G/N)(i), so that Gbeing solvable implies that G/N is also solvable.(⇐) We’ll build an abelian tower for G out of towers for N and G/N . First, let Nm = (eG)CNm−1 C . . .CN0 = N ,be a solvable series for N , and let Gs = (eG/N ) C Gs−1 C . . . C G0 = G/N be a solvable series for G/N . Set

Gi = q−1(Gi). Then Gi/Gi+1∼= (Gi/N)/(Gi+1/N) ∼= Gi/Gi+1 is abelian. Note that G0 = G, and Gs = N . Then

we have the following as a solvable series for G, which completes the proof:

(eG) = Nm C . . .CN0 = N = Gs C . . .CG0 = G.

B Corollary: If G is solvable, then any subgroup of G and every homomorphic image of G is solvable.42

B Example: Not every solvable group is nilpotent. The group S3 is solvable. This is because A3 C S3, and A3 isabelian, and every abelian group is solvable. Also, S3/A3

∼= Z/2Z which is abelian and hence also solvable. Thus bythe Useful Fact we’ve got that S3 is solvable. However, S3 is not nilpotent! Note here that Z(S3) = (1S3

), and sothe ascending central series is constant. Alternately, the sylow subgroups of S3 are of order 3 and 8, but S3 has order6, and so it cannot be the direct product of its Sylow subgroups.

26 October 2011 Finish Solvable Groups & Start Rings (Fields)

B Corollary: Sn is not solvable for n ≥ 5.

B Corollary: An is not solvable for n ≥ 5.

B Definition: A ring is an abelian group (R,+) with a binary operation ∗ called multiplication such that

– ∗ is associative– R contains a multiplicative identity, denoted 1R or 1.43

– Multiplication distributes over addition on both the left and the right. That is, for a, b, c ∈ R, we havea(b+ c) = ab+ ac and (a+ b)c = ac+ bc.

B Note: If we also have the ∗ is commutative, then we say that R is a commutative ring.

B Example: If 0 = 1, then R = {0}.B Facts: Let R be a ring, and b ∈ R. Then the following are easy to show from the properties.

– For any b ∈ R, we get −b = (−1)b = b(−1).– For any b ∈ R, we get b0 = 0b = 0.

B Definition: We say a ring R is a division ring if 1 6= 0 and for any b ∈ R \ {0}, there exists b−1 ∈ R such thatbb=1 = b−1b = 1.

B Definition: A commutative division ring is a field.

B Example: The real quaternions are a division ring. As a set they are Q = {a + bi + cj + dk : a, b, c, d ∈ R}, where 1is a multiplicative identity for Q, addition works like a vector space, and ij = k, jk = i, ki = j, ji = −k, kj = −i,ik = −j, and i2 = j2 = k2 = −1. We also have that multiplication is associative and distributes over addition. It ispossible in this ring to find inverses of any non-zero element in a manner similar to the field C, but in this ring,multiplication is not commutative, so it is not a field.

41Thompson got a fields medal for this!42The proof of this is embedded in the proof of the Useful Fact.43This is not always part of the definition, but it will be for our class.


B Wedderburn’s Little Theorem: A finite division ring is a field.

B Definition: Let R be a commutative ring, and let a ∈ R be nonzero. We say a is a zero-divisor of R if there is anonzero b ∈ R such that ab = 0.

B Definition: A commutative ring R where 1 6= 0 which has no zero-divisors is called an integral domain, or simplya domain.

B Example: Z is a domain, as is any field.

B Proposition: Let R be a finite domain. Then R is a field.

Proof. Let b ∈ R be a nonzero element. Then {b, b2, . . .} is a finite subset of R, which does not contain 0. Thus,there exists i < j such that bi = bj . We can then cancel to get 1 = bj−i = b · bj−i−1.

B Fact: Let F be a finite field, and let P be the prime field, that is, the field generated by 1 or equivalently thesmallest subfield of F which contains 1. Then |P | = p for some prime p, and the additive order in (F,+) of everynon-zero element is p.

B Definition: Let F be a field. The smallest positive integer such that mb = 0 for all b ∈ F is called the characteristicof F , and is denoted char(F ) = m if it exists. If no such integer exists, we say char(F ) = 0.

B Example: Let F = Z/pZ, where p is a prime, then P = F , and char(F ) = p. If F is a subfield of C, then charF = 0.

B Fact: If F is a field and char(F ) = p > 0, then p is a prime, and the prime subfield of F is isomorphic to Z/pZ.

28 October 2011 Field Extensions

B Terminology: Let F ⊆ E be fields. We say F is a subfield of E or that E is an extension field of F .

B Note: If F ⊆ E is a field extension, then E is an F−vector space.

B Definition: We say F ⊆ E is a finite extension if the vector space dimension of E over F is finite. Otherwise, wesay E is an infinite extension. Either way, we write [E : F ] = dimF (E) and call it the degree of the extension.

B Fact: If F ⊆ E ⊆ K are fields, then [K : F ] = [K : E][E : F ].

B Definition: Let R, S be rings. We say a map h : R→ S is a ring homomorphism if the following hold:

1. h(0R) = 0S2. h(1R) = 1S3. h(a+ b) = h(a) + h(b) for all a, b ∈ R4. h(ab) = h(a)h(b) for all a, b ∈ R

B Aside: The composition of homomorphisms is also a homomorphism. The only homomorphism out of the stupidring, {0} is the trivial map to the stupid ring.

B Example: Let F ⊆ E be a field extension and let α ∈ E. Consider the polynomial ring F [x], and let evα : F [x]→ Ebe the map given by evaluating the polynomial at α. Then evα is a homomorphism.

B Notation: If ker(evα) 6= 0, then Lang denotes the unique monic generator of the ideal by Irr(α, F, x) and calls it theirreducible polynomial for α over F .

B Note: If ker(evα) 6= 0, then Irr(α, F, x) is an irreducible polynomial since F [x]/(Irr(α, F, x)) is isomorphic to asubring of E, but E is a domain, and hence F [x]/(Irr(α, F, x)) is also a domain, so Irr(α, F, x) cannot have non-trivialfactors.

B Definition: Let F ⊆ E be fields, and let α ∈ E. We say α is algebraic over F if ker(evα) 6= 0. That is, if there isa non-trivial polynomial f ∈ F [x] such that f(α) = 0. We say α is transcendental if it is not algebraic over F . Wesay a field extension F ⊆ E is algebraic if every element of E is algebraic over F .

B Criterion: Let F ⊆ E be a field extension. If [E : F ] <∞, then the extension is algebraic. The converse is falsesince Q is algebraic over Q but has infinite degree.

Proof. If [E : F ] < ∞, then dimF (F [x]) = ∞ so any homomorphism h : F [x] → E has a nontrivial kernel. Inparticular, for any α ∈ E, we get that ker(evα) 6= 0. Thus, α is a root of some non-trivial polynomial, Irr(α, F, x).

B Comment: Let F ⊆ E be a field extension, and let α ∈ E. If α is algebraic, then F (α) is the field generated by Fand α inside E, and is also the image of evα, or equivalently, F [α]. That is, when α is algebraic over F , we getF [α] ∼= F (α). This is false if α is transcendental.

31 October 2011 Algebraic Extensions

B Note: We started today by proving the comment from the previous day.


B Fact: If α is algebraic over a field F , then

[F (α) : F ] = [F [α] : F ] = dimF (F [α]) = dimF

(F [x]

(Irr(α, F, x))

)= deg(Irr(α, F, x)) <∞.

However, if α is transcendental, then [F (α) : F ] =∞.

B Question: Why is the set of elements in a field E which are algebraic over a subfield F actually a subfield of E?44

B Theorem: Let k be a field, L an algebraically closed field such that there is an embedding σ : k ↪→ L. Let E be analgebraic extension field of k. Then there is a homomorphism σ′ : E → L such that the diagram below commutes.45

E

k L

σ′

σ

B Corollary: Any two algebraic closures, A,B, of a field k are isomorphic over k; that is, there exists an isomorphismφ : A→ B such that φ|k = idk.46

B Notation: Lang denotes the algebraic closure of a field k by ka. Note that ka can technically be any algebraicallyclosed field containing k which is algebraic over k. The point of the corollary is that the algebraic closure of a field isunique up to isomorphism.

2 November 2011 More Algebraic Extensions

B Recall: A field K is algebraically closed if every non-constant polynomial f ∈ K[x] has a root in K. We say analgebraically closed field K which contains k is an algebraic closure of k if K is algebraic over k.47

B Comments:

– The only algebraic extension of an algebraically closed field, k, is the field k itself.– Every field has an algebraic closure.

B Definition: We say a field extension k ⊆ K is an algebraic closure if K is algebraically closed, and the extensionis algebraic.

B Example: C is algebraically closed, but is not the algebraic closure ofQ. Instead, Qa = {z ∈ C : z is algebraic over Q}.B Definition: Let S ⊆ k[x] be a set of non-constant polynomials. Then K is a splitting field for S if every element

of S splits (factors) into linear factors in K[x] and if K is generated over k by the roots of elements of S.

B Examples:

1. C is a splitting field for x2 + 1 over R.2. Q[

√2] is a splitting field over Q for x2 − 2.

3. Q[ 3√

2] is not a splitting field over Q for x3 − 2 since it’s missing the complex roots.4. For any field k, ka is the splitting field over k for the set S = {all positive degree polynomials in k[x]}.

B Definition: A field extension k ⊆ K is normal if K is a splitting field of some set of polynomials of positive degreein k[x].

B Theorem:48 If A and B are extension fields of k, and A,B are splitting fields for the same set S of polynomials,then A ∼= B over k. That is, there is an isomorphism φ : A→ B such that φ|k = idk.

B Theorem:49 Consider fields k ⊆ K ⊆ ka. The following three statements are equivalent:

1. K is a normal extension of k.2. Every homomorphism h : K → ka such that h|k = 1k is an automorphism of K.3. Every irreducible polynomial in k[x] which has a root in K actually splits (factors linearly) in K.

4 November 2011 Field Homomorphisms & Separability

B Example: The extension Q ⊆ Q( 3√

2) is a finite algebraic extension, but it is not a normal extension since not all of

the roots of Irr( 3√

2,Q, x) are in Q( 3√

2). We also explained that Q[x]/(x3 − 2) ∼= Q[ 3√

2] ∼= Q[ 3√

2eiπ3 ] ∼= Q[ 3

√2e

2iπ3 ].

Hence, there are at least three homomorphisms over Q from Q[ 3√

2] to Qa.

44We justified this, but I won’t here.45Cited as review.46We partially proved the corollary given the theorem, but it is one of those non-interesting proofs where you just follow your nose.47The algebraic closure of a field k is frequently denoted k.48This was cited from Lang, and is review.49Also cited and review.


B Proposition: Let k be a field, and let α ∈ ka. Let f(x) = Irr(α, k, x). Then the number of distinct homomorphismsfrom k[α] to ka over k is equal to the number of distinct roots of f(x) in ka.

Proof. This is just a sketch! It is easy to show that if h is such a homomorphism, then f(h(α)) = 0 so that h(α)is also a root of f(x). Since h is determined by its value on α, then it is clear that the number of homomorphismsover k is no larger than the number of distinct roots of f in ka. Also, you can define a homomorphism by sendingα to any of the other roots, so there are exactly as many homomorphisms over k as there are distinct roots of theirreducible polynomial.

B Corollary: The number of k−homomorphisms of k[α]→ ka is no larger than deg Irr(α, k, x) if α ∈ ka.

B Example: Let k be a field, and y and indeterminate, then x2 − y ∈ (k(y))[x] is irreducible. If k has characteristic 2,then x2 − y = x2 + y = (x+

√y)2 ∈ (k(y))a[x] and so x2 − y only has a single root in k(y)a.

B Definition: Let k be a field. We define a linear operator (known as the derivative) by ddx : k[x] → k[x] via

ddxax

n = anxn−1 and by extending linearly.

B Comments:

– The derivative satisfies the Liebnitz rule:

d

dx(fg) =

df

dxg + f

dg

dx

– If k ⊆ E is an algebraic field extension, then ddx on k[x] is the restriction of d

dx on E[x].– We can use derivatives to detect polynomials with multiple roots since the linear factors aren’t unique and due

to the Liebnitz rule! In particular, α is a multiple root of f(x) ∈ k[x] if and only if the power of (x− α) in thefactorization of f(x) is strictly greater than one.

– If k is a field of characteristic p > 0, then ddxx

p = pxp−1 = 0.

B Definition: Let k ⊆ E be a field extension, and let α ∈ E be algebraic over k. We say α is separable over k ifIrr(α, k, x) does not have any multiple roots in ka. An extension k ⊆ E is separable if E is algebraic over k andevery element of E is separable over k.

B Definition: A finite algebraic extension is Galois if it is both normal and separable.

7 November 2011 Characterizing Separable Extensions

B Frobenius: Let k be a field of characteristic p > 0. We get a homomorphism F : k → k given by a 7→ ap, which iscalled the Frobenius map. This is because in a field of characteristic p, we have (a+ b)p = ap + bp and in anycommutative ring (ab)p = apbp.

B Example: Let k be a field, f ∈ k[x] where d = deg f ≥ 1. If f ′(x) = 0 (since d > 1, this means k is a field ofpositive characteristic, say p), then f is actually a polynomial in xp and p|d. If we work over ka, then we can takepth roots of elements of k, and so f ′(x) = 0 gives that f(x) = (h(x))p where h ∈ ka[x].

B Corollary: Let k be a field. Let α be algebraic over k, and let f = Irr(α, k, x). The following are equivalent:

– α is not separable over k,– f ′(α) = 0, and– f ′(x) = 0.50

B Corollary: Let k ⊆ K be an algebraic field extension, and assume that the characteristic of k is zero, (and henceso is the characteristic of K). Then K is a separable extension of k.51

B Proposition: Let k ⊆ K be finite fields. Then the extension is separable and normal.

Proof. Let p > 0 be the characteristic of both fields. Then the prime field is Z/pZ. This means K is a Z/pZvector space of dimension dimZ/pZK = m. That is, K ∼= Z/pZ⊕ . . .⊕ Z/pZ︸︷︷︸

m times (copies)

. This gives that |K| = pm, and

[K : Z/pZ] = m. Let K∗ = K \ {0} be the multiplicative group of K. Then |K∗| = pm − 1 and by Lagrange’sTheorem, α|K

∗| = 1K for all α ∈ K∗. Thus, xpm−1 − 1 is a polynomial of degree pm − 1 with pm − 1 distinct roots.

Hence, xpm−1 − 1 and xp

m − x are separable and the roots of the second polynomial are the pm distinct elementsof K. Now, for any α ∈ K, we get that Irr(α, k, x) divides xp

m − x, and so Irr(α, k, x) cannot have any multipleroots, so that α is separable over k so that the extension is normal (since α was arbitrary). Moreover, the extension isnormal since xp

m − x splits in K[x] and the set of roots of xpm − x is precisely K, and so it generates K over k.

50The proof is not really difficult and I’ve seen it before, so I’m omitting it here.51This is almost entirely a direct application of the previous corollary.


B Comment: If K is an algebraic extension of k, then if E is the set of all elements of K which are separable over k,then E is an intermediate field, k ⊆ E ⊆ K.52

9 November 2011 Separable Degree

B Definition: Let K be a finite algebraic extension of a field k so that k ⊆ K ⊆ ka. We define the separabledegree of K over k, denoted [K : k]s to be the cardinality of the set of k−homomorphisms K → ka.

B Note/Example: For simple algebraic extensions, the separable degree [k(α) : k]s is the number of distinct roots ofIrr(α, k, x). If α is separable, then the separable degree and the degree [k(α) : k] are equal. If α is not separable, thenwe know char(k) = p > 0, and if f(x) = Irr(α, k, x), then f ′(x) = 0, so that f(x) = (h(x))p for some h(x) ∈ ka[x]. Infact, there is an m > 0 such that f(x) = (g(x))p

m

where g has no multiple roots. In this case, [k(α) : k]s = deg g(x)

and [k(α) : k] = deg f(x) = (deg g(x)) · pm = pm[k(α) : k]s. We refer to [k(α):k][k(α):k]s

as the inseparable degree and

denote it [k(α) : k]i.

B Theorem: Let k ⊆ E ⊆ K ⊆ ka be fields and assume K is a finite extension of k. Then the separable degree ismultiplicative and finite. Moreover, [K : k]s divides [K : k]. Here, we call [K : k]/[K : k]s the inseparable degreeand denote it by [K : k]i.

Proof. Let SK/k be the set of k−homomorphisms K → ka. So |SK/k| = [K : k]S . Similarly, SK/E is the set ofE−homomorphisms K → ka and SE/k is the set of k−homomorphisms E → ka. Thus, |SK/E | = [K : E]s and|SE/k| = [E : k]s. For each k−homomorphism γ : E → ka pick an extension γ : ka → ka so that γ|E = γ. We’llnow define a map

ϕ : SK/k → SK/E × SE/kvia

h 7→((

h|E)−1

◦ h, h|E).

Showing that this map is bijective will complete the proof. We’ll do this by finding an inverse to ϕ. Let

λ : SK/E × SE/k → SK/k

be given by (h, g) 7→ g ◦ h. Note that h : K → ka is a E−homomorphism (and hence a k−homomorphism) andg : ka → ka is a k−homomorphism and so their composition exists as written and is a k−homomorphism from Kto ka, and hence an element of SK/k as desired. We now must show that compositing these two maps is the identityin both directions. We have

(λ ◦ ϕ)(h) = λ (ϕ(h)) = λ

(((h|E)−1

◦ h, h|E))

= h|E ◦((

h|E)−1

◦ h)

= h

and similarly since h|E = idE and g|E = g, we get

(ϕ ◦ λ)((h, g)) = ϕ(λ((h, g)))

= ϕ(g ◦ h)

=

((˜(g ◦ h) |E

)−1

◦ (g ◦ h), g ◦ h|E)

=

((˜g|E)−1

◦ g ◦ h, g|E)

= (h, g).

Thus, we have that |SK/k| = |SK/E | · |SE/k| as desired to give that separable degree is multiplicative.Now, since K is a finite extension of k, we can pick a finite set of elements α1, . . . , αr so that

k ⊆ k(α1) ⊆ k(α1, α2) ⊆ . . . ⊆ k(α1, . . . , αr) = K.

Using what we learned about simple extensions in the Note/Example, we have that

|SK/k| = |SK/k(α1,...,αr−1)| · |Sk(α1,...,αr−1)/k(α1,...,αr−2)| · . . . · |Sk(α1)/k|= [K : k(α1, . . . , αr−1)] · [k(α1, . . . , αr−1) : k(α1, . . . , αr−2)] · . . . · [k(α1) : k].

52We’ll justify this first thing on the 14th of November using the idea of separable degree.


As each of the factors53 if finite, then so is their product, |SK/k|. That is, [K : k]s <∞. The fact that separabledegree in general divides the regular degree is a combination of the facts that degree is multiplicative, separabledegree is multiplicative, and that for a simple extension separable degree divides the regular degree.

11 November 2011 Separability & Characteristic p

B Definition: Let k ⊆ E be a field extension. We define Autk(E) := {f ∈ Aut(E) : f |k = idk}.B Note: Every field homomorphism is injective!5455 Also note that for any field extension k ⊆ E, the set Autk(E) is

actually a group under composition.

B Comment: Given any field k, the group Autk(ka) acts on the roots of any polynomial f ∈ k[x]. By lettingσ ∈ Autk(ka), we get the following commutative diagram where the two vertical(ish) maps are the obvious inclusionmap(s):

ka ka

k

σ

This map σ can then be extended to an automorphism of ka[x] so that the similar diagram commutes where againthe vertical(ish) maps are the obvious inclusion map(s):

ka[x] ka[x]

k[x]

σ

By definition if f =

n∑i=1

aixi ∈ ka[x], then σ(f) =

n∑i=1

σ(ai)xi. Because σ is a homomorphism, we have σ(f + g) =

σ(f)+σ(g) and σ(fg) = σ(f)σ(g). So if f ∈ k[x] is monic, then in ka[x] we have that f(x) = (x−α1)·. . .·(x−αm) andso f = σ (f(x)) = (x− σ(α1)) · . . . ·(x− σ(αm)) which means that σ permutes the roots of f , and this gives the action.

B Note: The action need not be transitive; however, if f ∈ k[x] is irreducible, then the action will be transitive. Weshow that now: If α1, α2 are roots in ka of an irreducible polynomial f , then k[α1] ∼= k[x]/(f) ∼= k[α2] where themaps are given by x 7→ αi and called τ and σ respectively. So there is a k−isomorphism στ−1 : k[α1]→ k[α2] givenby α1 7→ α2, and this can be extended to all of ka to get a k−homomoprhism γ : ka → ka such that γ|k[α1] = στ−1.Thus, γ(α1) = α2 and the action is transitive as desired.

B Corollary: If f ∈ k[x] is a (monic),56 irreducible polynomial that factors over ka as f(x) = (x−α1)m1 ·. . .·(x−αr)mr .Then m1 = . . . = mr.

Proof. Choose any αi, αj roots of the polynomial f . For simplicity, make them α1 and α2 in the expansion of fin ka[x]. Then, there is some k−homomorphism γ : ka → ka such that γ(α1) = α2 by the transitivity of the actionas explained above. Then γ(f) = f as f ∈ k[x] and γ is a k−homomorphism. However, we also see that

(x− α1)m1 · . . . · (x− αr)mr = f = γ(f) = (x− α2)m1(x− γ(α2))m2 · . . . · (x− γ(αr))mr .

As the power of (x− α2) is the same in each expansion of f we must have that m1 = m2 since γ is bijective andk[x] is a UFD.

Note: If char(k) = p > 0, then p divides the multiplicity of each root (and they’re all the same by the corollary).In fact, each multiplicity must be a power of p. We’ll show that now: Let f be a monic, irreducible polynomialin k[x]. Then over ka[x], f factors as f(x) = (x− α1)m · . . . · (x− αr)m by the above corollary. Now, Let m = psn

53The correct word is not productands...idiot!54This is because fields have no nontrivial ideals, the kernel must be an ideal, and not everything can be mapped to 0.55This means that we never need to show that field homomorphisms are injective on the homework (or Friday night activities)! Ever!56Monic just makes the proof easier, but isn’t necessary!


where p - n. Then using the fact that the Frobenius map is in fact a homomorphism we get

f(x) =(

((x− α1) · . . . · (x− αr))ps)n

=(

(xps

− αps

1 ) · . . . · (xps

− αps

r ))n

= g(xps

)

where g is some polynomial in k[x]. Moreover, g is irreducible with roots αps

i for each i. Since g(x) =((x− αp

s

1 ) · . . . · (x− αpsr ))n

then if n > 1 we get that g is irreducible with multiple roots, but p - n which

is a contradiction, and so we must have n = 1 so that m = ps.

B Corollary: Let f ∈ k[x] be irreducible with a field of characteristic p > 0. Then there exists an s ≥ 0 such thatf(x) = g(xs) and g is an irreducible, separable polynomial.

B Corollary: Let α be algebraic over a field k, with char k = p > 0. Then αps

is separable for some s ≥ 0.

B Fact: Let k ⊆ K be a field extension, and let α1, . . . , αr ∈ K be algebraic and separable over k. ThenE = k(α1, . . . , αr) is a separable extension of k.

Proof. Let β ∈ E and assume that β is not separable over k. Then [k(β) : k]s < [k(β) : k]. Thus, [E : k]s = [E :k(β)]s · [k(β) : k]s < [E : k(β)] · [k(β) : k] = [E : k]. However, we also have that because simple algebraic extensionsare separable, then

[E : k]s = [E : k(α1, . . . , αr−1)]s · . . . · [k(α1) : k]s= [E : k(α1, . . . , αr−1)] · . . . · [k(α1) : k] = [E : k]

This is a contradiction, and so we must have that β is separable over k.

14 November 2011 Theorem of the Primitive Element

B Corollary: Let k ⊆ K be an algebraic extension of fields. Also, let E = {α ∈ K : α is separable over k}. Then Eis an intermediate field k ⊆ E ⊆ K.

Proof. It is enough to show that k(E) = E. It is clear that E ⊆ k(E), so let β ∈ k(E). Then there existα1, . . . , αr ∈ E such that β ∈ k(α1, . . . , αr) ⊆ k(α). But we saw last time that k(α1, . . . , αr) is separable over k.Thus, β is separable over k and so β ∈ E.

B Definition: Let k ⊆ F be an algebraic extension of fields, and let E = {α ∈ F : α is separable over k}. Then wecall E the separable closure of k in F .

B Comment: If F 6= E in the above definition, then char k = p > 0, and for each β ∈ F , there is an n (which dependson β) so that βp

n ∈ E. This is because Irr(β, k, x) = f(xpn

) where f is separable. Moreover, Irr(β, k, x) dividesxp

n − βpn = (x− β)pn

and so we have that Irr(β, k, x) = (x− β)m, and as we saw before, m must be a power of p.

B Definition: We say an algebraic extension k ⊆ F (where char k = p > 0) is purely inseparable if for eachelement β ∈ F there is an n such that βp

n ∈ k. If k = F , then regardless of characteristic, we also say F is a purelyinseparable extension of k.

B Corollary: Let k ⊆ F be an algebraic field extension. Let E be the separable closure of k in F . Then k ⊆ E isseparable and E ⊆ F is purely inseparable.

B Definition: We say an extension k ⊆ F is simple if there is an α ∈ F such that k(α) = F . If k(α) = F , we saythat α is a primitive element.

B The Theorem of the Primitive Element: Let k ⊆ E be a finite extension of fields.

1. If E is separable over k, then E is a simple extension, that is, there is an element δ ∈ E such that E = k(δ).2. The extension is simple if and only if there are only finitely many intermediate fields.

Proof. 1. First, suppose that |k| < ∞. Then E is also a finite field. But E∗ = E \ {0} is a cyclic group (undermultiplication), so let E∗ = (α). Then we get E = k(α). Next, suppose that k is an infinite field. It is enoughto show that k(α, β) = k(δ) for some δ ∈ k(α, β). Let σ1, . . . , σn be the distinct k−homomorphisms of E → ka.Note n = [E : k]s = [E : k]. So consider the polynomial

P (x) =∏i 6=j

(σi(α) + σi(β)x− (σj(α) + σj(β)x)) .


In particular, since σi 6= σj , then we have that P (x) 6= 0. Since |k| =∞, we can find a c ∈ k so that P (c) 6= 0.57

Thus, σi(α) + σi(β)c is different for all i = 1, . . . , n. Thus, σi|k(α+βc) is different for each i = 1, . . . , n. Thus,[k(α + βc) : k]s ≥ n. This gives then that n ≤ [k(α + βc) : k]s = [k(α + βc) : k] ≤ [k(α, β) : k] = n. Hence,k(α+ βc) = k(α, β).

2. First, suppose that |k| < ∞. Then for any finite extension k ⊆ E, we have that |E| < ∞ as well, and k ⊆ Eis simple, and so there are only finitely many intermediate fields. Next, suppose that k is an infinite field. We’llstart by assuming there are finitely many intermediate fields and showing that the extension is simple. Again,it is enough to show that for α, β ∈ K that k(α, β) is actually a simple extension. So there exist c1, c2 ∈ k sothat c1 6= c2 6= 0 and k(α + c1β) = k(α + c2β). This gives that (c1 − c2)β = (α + c1β)− (α + c2β) ∈ k(α + c2β).Thus, β ∈ k(α + c2β). Now using the fact that β ∈ k(α + c2β), we get that, α = α + c2β − c2β ∈ k(α + c2β).Thus, k(α, β) ⊆ k(α+ c2β) ⊆ k(α, β) and hence k(α+ c2β) = k(α, β).Now, assume k ⊆ E is a simple extension. In particular, let E = k(δ) for some δ ∈ ka = Ea. Our goal is to showthat there are only finitely many intermediate fields between k and E. We’ll show there is an injection from theset of intermediate fields to the set of monic factors of f = Irr(δ, k, x). Since the monic factors is clearly a finiteset, then the other set will be finite as well, and that will complete the proof. So let F be an intermediate field, i.e.k ⊆ F ⊆ E. Then F (δ) = E. So consider gF (x) = Irr(δ, F, x). Then clearly gF |f . We’re now ready to define a map

ϕ : {intermediate fields} → {monic factors of f}

by setting ϕ(F ) = gF . Instead of directly showing that ϕ is injective, we’ll actually show that we can recover F fromgF . Let KgF = k(a1, . . . , am) where gf = xm + a1x

m−1 + . . .+ am. It is clear that KgF ⊆ F as each ai ∈ F . Sowe must show the other inclusion. We’ll do this by looking at degrees. It is clear that [E : F ] = [F (δ) : F ] = deg gFas E = F (δ) and by the definition of gF . Similarly, it is clear that [E : KgF ] = deg(Irr(δ,KgF , x)) = deg(gF ) sinceE = KgF (δ) and Irr(δ,KgF , x) = gF . Thus, we have [E : F ] = [E : KgF ] = [E : F ] · [F : KgF ] so that [F : KgF ] = 1and hence F = KgF so that there is an inverse map, an hence ϕ is injective (and hence completing the proof).

16 November 2011 Proof Completion

B Today was short and we just finished the proof that I’ve put completely above for the sake of simplicity.

18 November 2011 None

B No class today. Brian Harbourne was out of town.

21 November 2011 Separable Degree & Galois Theory

B Recall: Let k ⊆ F be fields, and assume the extension is algebraic. Then we have k ⊆ E ⊆ F where E := {α ∈ F :α is separable} and E ⊆ F is purely inseparable.

B Recall: If k ⊆ F is a finite extension and E is the separable closure of k in F , then

[F : k] = [F : E] · [E : k],

[F : k]s = [F : E]s · [E : k]s, and

[F : k]i = [F : E]i · [E : k]i

since each type of degree is multiplicative (i.e. this holds for any intermediate field k ⊆ E ⊆ F ).

B Proposition: Let k ⊆ F be a finite field extension, and let E be the separable closure of k in F . Then[E : k]s = [E : k] = [F : k]s, [F : E]s = 1, [F : E] = [F : E]i, and [E : k]i = 1.

Proof. We’ll start by showing that [E : k]s = [E : k]. By the theorem of the primitive element, there is an α ∈ Esuch that E = k(α) (since E is a separable extension of k). However, [E : k] = deg(Irr(α, k, x)) and Irr(α, k, x)is separable. For each root, β, of Irr(α, k, x) we get a k−homomorphism k(α) → ka given by α 7→ β. Thus, thenumber of k−homomorphisms k(α)→ ka is at least the number of distinct roots of Irr(α, k, x), but that’s equal todeg Irr(α, k, x) = [E : k] since α is separable. Hence, [E : k]s ≥ [E : k] ≥ [E : k]s and so we get that [E : k]s = [E : k].Next, we’ll show that [F : E]s = 1. Once we’ve shown that, then we get [F : k]s = [F : E]s · [E : k]s = [F : E]s · [E :k] = [E : k]. So, let σ be an E−homomorphism σ : F → Ea = F a. Thus, σ|E = idE . Let β ∈ F . If char(E) = 0,then F is automatically a separable extension of k so that F = E and σ(β) = β. However, if char(E) = p > 0, then

57This is because the polynomial can have only finitely many roots.


βpn ∈ E for some n. Thus, σ(β)p

n

= σ(βpn

) = βpn

. This then gives that (σ(β) − β)pn

= σ(β)pn − βpn = 0 and

hence σ(β) = β. This gives that [F : E]s = 1.Now, we have that [F : E] = [F : E]s · [F : E]i by definition of the inseparable degree, since [F : E]s = 1, then weclearly have [F : E] = [F : E]i. Similarly, we have [E : k] = [E : k]s · [E : k]i = [E : k] · [E : k]i and so [E : k]i = 1.

B Note: We’ll now start Galois Theory!!!

B Remark: Let K be a field. We’ll associate a group to K, namely, we associate Aut(K) to it. For a field extensionk ⊆ K, we associate the group Autk(K) ≤ Aut(K). Similarly, given G ⊆ Aut(K), we can associate a subfield,KG := {α ∈ K : ∀g ∈ G, g(α) = α}. This is called the fixed field of the subgroup.

B Comment: These associations can both be described as contravariant functors. In particular, let K be a field. Let Gbe the functor from subfields of K to subgroups of Aut(K) where G(E) = AutE(K) for any subfield E ⊆ K. Thesecategories are equipped with morphisms of inclusion in both cases. Similarly, let F be the functor from subgroups ofAut(K) to subfields of K given by F(G) = KG for any G ≤ Aut(K). We should discuss the contravariance of thesefunctors. If H < G < Aut(K), then KG ⊆ KH since fewer things are fixed by all of G than are fixed by all of H.Similarly, if k ⊆ E ⊆ K, then AutE(K) < Autk(K) since if E is fixed then k is automatically fixed.

B Motivation: The idea behind Galois theory is to understand the composition of these two functors. In generalthey’re not inverse functors, but it would be awfully nice if they were.

B Example: Consider the field extension Q ⊆ Q( 3√

2). Then we want to figure out what G(Q) = AutQQ( 3√

2) is. It is

clear that AutQ( 3√2)Q( 3√

2) = {1Q( 3√2)}. Let σ : Q( 3√

2)→ Q( 3√

2) be a Q−homomorphism. Then we can extend this

to a Q−homomorphism σ : Qa → Qa. Here σ permutes the roots of Irr( 3√

2,Q, x) = x3 − 2. However, Q( 3√

2) ⊆ Rand so Q( 3

√2) contains only one root of x3 − 2. Thus, σ( 3

√2) = 3

√2 so that σ = 1Q( 3√2) and hence G(Q) = {1Q( 3√2)}

as well.

28 November 2011 Artin’s Theorem

B Definition: Let k ⊆ K be an algebraic extension of fields. We say it is a Galois extension if it is normal andseparable.

B Fundamental Theorem of Galois Theory: Let k ⊆ K be a finite Galois extension of fields. Then F and G give abijection from the set of intermediate fields k ⊆ E ⊆ K to the set of subgroups of Autk(K).58 Moreover, an intermediatefield E is Galois over k if and only if AutE(K)CAutk(K), and in this case, Autk(E) ∼= Autk(K)/AutE(K).

B Artin’s Theorem: Let K be any field, and let G be a finite subgroup of Aut(K). Let k = KG. Then k ⊆ K is afinite Galois extension with Galois group G = Autk(K) and [K : k] = |G|.Proof. We start by claiming that k ⊆ K is separable. So let α ∈ K. It is enough to show that f(α) = 0 for somepolynomial f ∈ k[x] which does not have multiple roots. So let {α1, . . . , αr} = Gα be the orbit of α under the action ofG. Also, do this so that αi 6= αj whenever i 6= j. Then set f(x) = (x−α1) · . . . ·(x−αr). Note here that f ∈ K[x], andthat G acts on K, so that we have an action of G on K[x]. In particular, if g ∈ G, then gf = (x−g(α1))·. . .·(x−g(αr)).But we know that {g(α1), . . . , g(αr)} = {α1, . . . , αr} since G just permutes the elements of an orbit. Thus, gf = f .Hence the coefficients of f as a (non-factored) polynomial are fixed by the action of G, and so they’re elements ofk and f ∈ k[x]. But f is separably by construction, and so k ⊆ K is separable and algebraic.We now claim that k ⊆ K is a normal extension. Let α ∈ K and let fα be the f as defined in the previous part.Note that fα splits over K and the roots are all in K. Thus, K is the splitting field of {fα : α ∈ K}. Thus, k ⊆ Kis Galois and G ≤ Autk(K).Now, if [K : k] < ∞, then KG = k, and of course KAutk(K) = k. Thus, under the Galois bijection, both G andAutk(K) correspond to k. Thus, G = Autk(K). Then by the usual statement of the Fundamental Theorem of GaloisTheory, we have that |G| = |Autk(K)| = [K : k]. So consider the case when [K : k] =∞. Then there must be anintermediate field k ⊆ E ⊆ K so that |G| < [E : k] <∞.59 By choosing such an intermediate field E, we know thatE will be a finite extension of k. Then the theorem of the primitive element says that E = k(γ) for some γ ∈ E. Now,we claim that fγ where fγ is as before is actually the irreducible polynomial of γ over k. That is, fγ = Irr(γ, k, x).It is clear that fγ ∈ k[x] from our previous reasoning, and that fγ(γ) = 0 by definition. So thus, we get thatIrr(γ, k, x)|fγ . On the other hand, the group Autk(K) acts transitively on the roots of fγ , but preserves the setof roots of Irr(γ, k, x). Since g(Irr(γ, k, x)) = Irr(γ, k, x) for all g ∈ Autk(K), we see that if δ is a root of Irr(γ, k, x)then so is gδ for all g ∈ Autk(K). Thus, the set of roots of Irr(γ, k, x) are exactly (Autk(K)) · γ and so contains G · γ.Since Irr(γ, k, x)|fγ , then we get the other inclusion, and so the sets of roots are equal, and hence fγ = Irr(γ, k, x)as they’re both monic and separable. Now, we have |G| ≥ t = deg Irr(γ, k, x) = [k(γ) : k] = [E : k] > |G|. This isa contradiction, and so we can’t ever have that [K : k] =∞, and that completes the proof.

58We defined these functors on 21 November 2011.59This can be done by adding elements one at a time until you get that the index is sufficiently large.


B Aside: If k ⊆ K is a Galois extension, then |Autk(K)| = [K : k].

B Aside: Here, we see how to get extensions k ⊆ K with any Galois group, as long as G ≤ Aut(K).

B Noether’s Problem: What groups occur as Galois groups of finite Galois extensions of Q?60

30 November 2011 Galois Extensions & Finite Fields

B Aside: If k ⊆ F is a finite field extension, then what can we say about |Autk(F )| and [F : k]? If F/k is Galois, thenthey’re equal. More generally, for any finite extension F/k, then [F : k]s is equal to the number of k−homomorphismsF → F a. Thus, |Autk(F )| ≤ [F : k]s since for each σ ∈ Autk(F ) we have a k−homomorphism F → F a given bycomposition with the inclusion morphism F → F ↪→ F a. Here the image of the composition is all of F , but there canbe k−homomorphisms from F → F a whose image is not F . If σ is the k−homomoprhism of F → F , then we callthe composition ϕσ.

B Example: Consider the homomorphism ϕ : Q( 3√

2)→ Qa which is given by sending Q to itself identically and 3√

2 to

a nonreal root of x3 − 2. Then ϕ(Q( 3√

2)) 6= Q( 3√

2) since the nonreal root is not in Q( 3√

2).

B Comments:

– An extension F/k is normal if and only if for all α ∈ F , then all roots of Irr(α, k, x) are in F as well.– When F/k is normal, then for any k−homomorhism ϕ : F → F a, then ϕ(F ) = F .– Thus, ϕ = ϕσ for some σ ∈ Autk(F ).– So when F/k is finite and normal (but not necessarily separable) then [F : k]s = |Autk(F )|.– If F/k is not Galois, then [F : k] > [F : k]s and so [F : k] > |Autk(F )|.’

B In Conclusion: If F/k is finite, we have

|Autk(F )| ≤ [F : k]s ≤ [F : k]

where if the extension is normal then the first is an equality and if it is also Galois, then they’re both equalities. (i.e.the first is an equality when the extension is normal and the second is an equality when the extension is separable).

B Finite Fields: Let K be a finite field, and set p = char(K) > 0. Note that |K| = pn for some n ≥ 1. Also, let k bethe prime field, so k ∼= Z/pZ. Thus, [K : k] = n. We have the Frobenius automorphism, which we’ll call ϕ that sendselements of K to their pth powers. In particular if α ∈ k then ϕ(α) = αp = α so that ϕ ∈ Autk(K) ≤ Aut(K).

B Claim: Autk(K) is a cyclic group generated by the Frobenius map ϕ for any finite field K where k is the prime field,and hence isomorphic to Z/pZ for some prime p.

Proof. It is enough to show that the order of ϕ is n.61 So recall that K∗ is a cyclic group under multiplication,and let K∗ = (α). But the order of ϕ is the least m ≥ 1 such that βp

m

= ϕm(β) = β for all β ∈ K. However, thenumber of roots of xp

m − β is at most pm so the order of ϕ is at least n since we need pn roots to get all elementsof K as roots of the polynomial xp

m − β. On the other hand, the order of α in K∗ is pn − 1. So αpn−1 = 1 and

so (αpn−1)α = α or equivalently αp

n

= α. Hence, ϕn(α) = α so that ϕn(αi) = αi for all i. Since α generates K∗

then ϕn(β) = β for all β 6= 0 and so ϕn(β) = β for all β ∈ K. Thus, n is the order of ϕ.

B Comment: If k ⊆ E ⊆ K, then Autk(K) ⊇ AutE(K), so AutE(K) is generated by a power of ϕ.

2 December 2011 Finite Fields

B Last Time: We saw that if F is a finite field with prime field k = Z/pZ, then Autk(K) = (ϕ) where ϕ is theFrobenius homomorphism. If |F | = pn for some n ≥ 1 then the order of ϕ is n.

B Note: If F1 and F2 are finite fields with |F1| = |F2|, then F1∼= F2.62

B Galois Correspondence for Finite Fields: Say k is the prime field, and F is an extension field with |F | = pn.Consider an intermediate field k ⊆ E ⊆ F . Then these correspond to (respectively)

Autk(F ) ⊇ AutE(F ) ⊇ AutF (F ) = {idF }.

Here, |k| = p, |F | = pn and n = [F : k] = [F : E][E : k] so if we let [F : E] = r and [E : k] = s, then n = rs, and|E| = ps. Let ϕF be the Frobenius map on the field F . As a vector space E ∼= ks. Also Autk(E) is a cyclic group oforder s, and Autk(E) ∼= Autk(F )/AutE(F ). Since |AutE(F )| = [F : E] = r, it is clear that AutE(F ) is a group oforder r, but it is not clear which group of order r it is. Since |Autk(F )| = n and Autk(F ) = (ϕF ) and |AutE(F )| = r,

then AutE(F ) = (ϕsF ). Thus, we have k ⊆ E ⊆ F correspond to (ϕ[k:k]F ) ⊇ (ϕ

[E:k]F ) ⊇ (ϕ

[F :k]F ) = (1F ) respectively.

60This problem is still open and interesting.61This is sufficient because we know that the group has order n = [K : k] = |Autk(K)| so if it has an element of order n then it will

be cyclic of order n.62This is true because of the uniqueness of splitting fields, since if |F | = pn, then F is the splitting field of xpn − x over k ∼= Z/pZ.


B Example(s):

– For every integer n ≥ 1, there is a finite field, F of order pn where p is any prime. Namely, F is the splittingfield for xp

n − x over k = Z/pZ. If n is prime, then the only intermediate fields are F and k. (Due to the Galoiscorrespondence!)

– Let p = 31991. Note that this is prime. (We could use any other prime, but this one is large enough to be awkwardto raise to powers, and yet still relatively small in computer terms). We want to find all intermediate fields of thefield of order p12. Note here that the lattice of subfields only depends on the exponent (and not on the prime itself).This is because Autk(F ) depends only on n when k = Z/pZ and |F | = pn. First, we’ll find the lattice of subgroupsof the cyclic subgroup of order 12. For notational simplicity, we’ll denote the cyclic group of order n by Cn:

C1

C2 C3

C4 C6

C12

Thus, there are 6 intermediate fields, and their lattice is the same as for the lattice of subgroups, except thebigger fields are at the top, rather than at the bottom of the picture.

– How many elements of F generate F over k if F and k are fields, p is prime, |k| = p and |F | = p6? Here, the

goal is to count all α ∈ F so that k(α) = F . We have the following lattice of subfields, where Cn is the subfieldcorresponding to the cyclic group of order n:

C1 = F

C2 C3

C6 = k

Let S = F \(C2∪C3). Then we want to count the size of S.63 So, |S| = |F |−(|C2|+ |C3|)+ |C6| = p6−p2−p3 +p.

5 December 2011 Transcendence Degree & Transcendence Bases

B Definition: Let k ⊆ K be fields, and S a subset of K. We say that the elements of S are algebraicallyindependent over k if for any r and any subset {s1, . . . , sr} ⊆ S of r distinct elements, the k−homomorphismϕ : k[x1, . . . , xr]→ K given by xi 7→ si is injective.

B Note: If {s1, . . . , sr} ⊆ K is algebraically independent over k, then k(x1, . . . , xr) ∼= k(s1, . . . , sr) and the isomorphismcan be given by xi 7→ si, which is a k−homomoprhism.

B Definition: If S ⊆ K is algebraically independent over k for some subfield k ⊆ K, and if K = k(S), then we saythat k ⊆ K is a purely transcendental extension.

B Example: Let k = Q, and K = Q(x). Then k ⊆ K is a purely transcendental extension (this is easily seen by usingthe set S = {x}). However, x2 ∈ K, and so Q ( Q(x2) ( Q(x) so with respect to the set T = {x2} (which is also analgebraically independent set), it appears as if the extension is not purely transcendental since we get that k(T ) ( K.

B Definition: Let k ⊆ K be fields. Partially order the set of subsets S ⊆ K which are algebraically independent overk by inclusion. We say a subset B ⊆ K which is algebraically independent over k is a transcendence base if B isnot properly contained in any other subset S ⊆ K which is algebraically independent over k.

B Example: Let k = Q, K = Q(x). Then {x} is a transcendence base for k ⊆ K. Also, {x2} is a transcendence basefor the extension even though k(x2) ( K.

B Note: If S is a transcendence base for k ⊆ K, then k(S) ⊆ K is an algebraic extension.

B Comment: Sometimes it is impossible to pick a transcendence base S for a field extension k ⊆ K such that k(S) = K.

B Example: For example, Q ⊆ Q(√

2) is an algebraic extension and so its transcendence base is the empty-set. Also,

Q ⊆ Q(π) is purely transcendental, but Q ⊆ Q(π,√

2) is probably not purely transcendental.

63This will be done using inclusion exclusion.


B Theorem: Let k ⊆ K be fields. Then there is a transcendence base for this extension.64

B Key Lemma: Let k ⊆ K be fields. Let S ⊆ K be algebraically independent over k. Let u ∈ K. Then T = S ∪ {u}is algebraically independent over k if and only if u is transcendental over k(S).

Proof. We’ll prove this next time, but first state some corollaries now.

B Corollary 1: Let k ⊆ K be fields, with S ⊆ K an algebraically independent set over k. Then S is a transcendencebase for k ⊆ K if and only if k(S) ⊆ K is algebraic.

B Corollary 2: Let S1 and S2 be transcendence bases for k ⊆ K. If |S1| <∞, then |S2| <∞ and |S1| = |S2|.65

B Definition: We define the transcendence degree of k ⊆ K, denoted trdegk(K) to be |S| where S is atranscendence basis for k ⊆ K.

7 December 2011 Transcendence Bases

B Corollary 3: Let k ⊆ K be fields, and let S ⊆ K be a subset. If K is algebraic over k(S), then S contains atranscendence base for k ⊆ K.

B Corollary 4: Let k ⊆ K be fields. Let S ⊆ K be algebraically independent over k. Then there is a transcendencebase S′ for k ⊆ K so that S ⊆ S′.

B Corollary 5: Let k ⊆ K ⊆ L be fields. Then trdegk(K) + trdegK(L) = trdegk(L).

B Note: We’re now ready to prove things:– Proof. of Key Lemma: (⇒) We’ll prove this direction by contrapositive. So we’re assuming that u is

algebraic (i.e. not transcendental) over k(S) and we need to show that T = S ∪ {u} is not algebraically

independent over k. So let f = Irr(u, k(S), x) and write it as f(x) =

d∑i=0

gihixi where gi = gi(s1, . . . , sr) and

hi = hi(s1, . . . , sr) 6= 0 where gi, hi ∈ k[x1, . . . , xr] for some r ∈ N and s1, . . . , sr ∈ S. Then, let f =

d∑i=0

gi

hixi

and note that f ∈ k(x1, . . . , xr, x). Let H = h0 · h1 · . . . · hd. Then Hf is a polynomial, which we’ll call F .

Then F (s1, . . . , sr, u) = H(s1, . . . , sr)f(s1, . . . , sr, u) = H(s1, . . . , sr)f(u) = 0 as f(u) = 0. So F ∈ kerϕ where

ϕ : k[x1, . . . , xr, x]→ K is given by xi 7→ si and x 7→ u. But F is not the zero polynomial since f 6= 0, so that

f 6= 0, and each hi 6= 0 so each hi 6= 0. Thus, H 6= 0, and so F 6= 0. This means that kerϕ 6= 0 and so T is notalgebraically independent.(⇐) Now, we’ll assume that u is transcendental over k(S) and we need to show that T = S ∪ {u} is algebraicallyindependent over k. Let t1, . . . , tr ∈ T . If ti 6= u for any i, then ti ∈ S for all i and hence t1, . . . , tr arealgebraically independent, so that k[x1, . . . , xr]→ K given by xi 7→ ti is injective. So instead suppose that ti = ufor some i. Without loss of generality, we may assume u = tr. Then ϕ : k[x1, . . . , xr]→ K given by xi 7→ ti fori < r and xr 7→ u = tr can be written as the composition of two maps. Namely, ϕ1 : k[x1, . . . , xr]→ k(S)[xr]given by xi 7→ ti for i < r and xr 7→ xr and ϕ2 : k(S)[xr] → K given by xr 7→ u. Then ϕ1 is injective ast1, . . . , tr−1 are algebraically independent over k and ϕ2 is injective as u is transcendental over k(S). Thus,ϕ = ϕ2 ◦ ϕ1 is injective, so that T is algebraically independent over k.

– Proof. of Corollary 1: S is not a transcendence base if and only if there is some u ∈ K such that S ∪ {u} isalgebraically independent over k. By the Key Lemma, this is true if and only if u is transcendental over k(S)which is true if and only if K is not algebraic over k(S).

B Aside: If t1, . . . , tr ∈ S and S is algebraically independent over k, then k[x1, . . . , xr−1]→ k(S) is injective wherexi 7→ ti. So in the proof of the Key Lemma we use the easy fact that k[x1, . . . , xr−1][xr]→ k(S)[xr] is also injective.

B Comment: The empty-set is always algebraically independent.

B Comment: Any subset of an algebraically independent set is also algebraically independent.

B Comment: If α ∈ S with S a subset of K and k ⊆ K a subfield, then if S is algebraically independent over k, wemust also have that α is transcendental over k.

B Note: We didn’t prove the remaining Corollaries in class, but they’re not terribly difficult so I’ll include them here.– Proof. of Corollary 2: First suppose S1 = ∅. Then k(S1) = k so that K/k is an algebraic extension. Thus,K has no elements that are transcendental over k and hence S2 = ∅. In particular, this gives |S2| < ∞ and|S1| = |S2|.So now suppose that S1 = {s1, . . . , sr} with si 6= sj if i 6= j, for some r ≥ 1. Then K/k(s2, . . . , sr) is notalgebraic since s1 is transcendental over k(s2, . . . , sr). Thus, there is a transcendental element in S2, since

64The proof is just a simple application of Zorn’s lemma.65It is true in general that they have the same cardinality, but that’s significantly harder to prove, and requires transfinite induction...ick!


k(s2, . . . , sr)(S2) = K is not algebraic over k(s2, . . . , sr). So let σ1 ∈ S2 be transcendental over k(s2, . . . , sr). Wealso get that s1 is algebraic over k(σ1, s2, . . . , sr) since otherwise S1 would not be a transcendence base. Thus,k(σ1, s1, . . . , sr) is algebraic over k(σ1, s2, . . . , sr), and since K is algebraic over k(σ1, s1, . . . , sr), then K is alsoalgebraic over k(σ1, s2, . . . , sr), then we have that K is algebraic over k(σ1, s2, . . . , sr), so that {σ1, s2, . . . , sr}is a transcendence base for the extension K/k. By induction, we can replace all of the elements of S1 withelements of S2. If we run out of elements of S1 to replace, or if we run out of elements of S2 to replace elementsof S1 with, then we get a contradiction to one of the sets being a transcendence base. Thus, we must have that|S1| = |S2|.

– Proof. of Corollary 3: Let Σ be a maximal algebraically independent subset of S. Then K/k(S) is algebraicand k(S)/k(Σ) is algebraic, so that K/k(Σ) is algebraic. Thus, Σ is a transcendence base for K/k by the firstcorollary.

– Proof. of Corollary 4: Let Σ be a maximal algebraically independent subset of K which contains S. ThenK/k(Σ) is algebraic by the maximality, so that Σ is a transcendence base for K/k by corollary 1.

– Proof. of Corollary 5: Let S be a transcendence base for K/k and let T be a transcendence base for L/K.Noe that T is automatically algebraically independent over k. If |S| =∞, then L/k has a transcendence basethat contains S. If |T | =∞, then L/k has a transcendence base that contains T , and so if either S or T is aninfinite set, then trdegk(L) =∞ and the statement holds.So now we assume that both S and T are finite sets. Let S = {s1, . . . , sm} and T = {t1, . . . , tn}. If S ∪ T isnot algebraically independent over k, then there is a nonzero polynomial f ∈ k[x1, . . . , xm, y1, . . . , yn] such thatf(s1, . . . , sm, t1, . . . , tn) = 0. Since f ∈ k[x1, . . . , xm, y1, . . . , yn] ⊆ k(S)[y1, . . . , yn], and since T is algebraicallyindependent over K (and hence over k(S)), then we have that f(s1, . . . , sm, y1, . . . , yn) is the zero element ofk(S)[y1, . . . , yn]. However, since the original polynomial f was not the zero polynomial, then some monomial inthe variables y1, . . . , yn appears as a term in f with coefficient a nonzero element of k[x1, . . . , xm], which whenevaluated at s1, . . . , sm gives zero. But this contradicts the fact that the set S is algebraically independent over k,and hence S ∪ T is algebraically independent over k.It remains to show that L is algebraic over k(S ∪ T ). However, we know that K is algebraic over k(S), so thatK(T ) is algebraic over k(S ∪ T ), and we also have that L is algebraic over K(T ). Thus, L is algebraic overk(S ∪ T ), so that S ∪ T is a transcendence base for L/k and hence we get the result:

trdegk(K) + trdegK(L) = trdegk(L).

9 December 2011 topic here

B Hilbert & Rings of Invariants: Let G be a group, acting as k−automorphism on R = k[x1, . . . , xn] where kis a field. The big question here is whether the ring of invariants, RG is always finitely generated over k. Hilbertproved that the answer is yes for certain groups, called reductive groups. The key idea in the proof was thatR is Noetherian! Hilbert’s 14th problem was asking if the ring of invariants is always finitely generated. Zariskireformulated the problem in terms of an intermediate field, and Nagata found a counterexample. Apparently Reeswas also instrumental in solving the problem, but we weren’t told how in class.

B Fact: Let k ⊆ E ⊆ K be fields such that K = k(S) for some finite subset S ⊆ K. Then E is finitely generated overk.66

B Special Case: If k ⊆ E ⊆ K are fields and K is a purely transcendental extension of k with trdegkK <∞, and ifE/k is algebraic, then it can be shown that E is finitely generated over k.

66“It is a challenging, but doable problem to prove this fact.” - B.H.

math 901 fall 2011 course notess-kshulti1/fall2011/901.notes.pdf · 2014. 9. 18. · math 901 notes...

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