modules and ideals of algebras of associative type

ISSN 1066-369X, Russian Mathematics (Iz. VUZ), 2008, Vol. 52, No. 8, pp. 20–27. c© Allerton Press, Inc., 2008.Original Russian Text c© N.A. Koreshkov, 2008, published in Izvestiya Vysshikh Uchebnykh Zavedenii. Matematika, 2008, No. 8, pp. 25–34.

Modules and Ideals of Algebras of Associative Type

N. A. Koreshkov1*

1Kazan State University, ul. Kremlyovskaya 18, Kazan, 420008 RussiaReceived March 09, 2006

Abstract—In this paper, we study some properties of algebras of associative type introduced inprevious papers of the author. We show that a finite-dimensional algebra of associative type over afield of zero characteristic is homogeneously semisimple if and only if a certain form defined by thetrace form is nonsingular. For a subclass of algebras of associative type, it is proved that any moduleover a semisimple algebra is completely reducible. We also prove that any left homogeneous ideal ofa semisimple algebra of associative type is generated by a homogeneous idempotent.

DOI: 10.3103/S1066369X08080033

Key words: algebra of associative type, homogeneously semisimple algebra, module, ideal,homogeneous idempotent.

Lie type algebras introduced in [1, 2] are natural generalizations of Lie algebras, associative algebras,Lie superalgebras with Z2-grading, and of algebras of some other classes. In [3] and [4], we studied aspecial case of algebras of Lie type called algebras of associative type. In the above mentioned papers,we gave examples of such algebras related to the notions of a group algebra and an algebra of Cartantype. Recall the definition of an algebra of associative type.

Let G be a semigroup, T a finite subset in G, A = ⊕α∈T

Aα a G-graded algebra over a field k, i.e.,

for any α, β ∈ T , AαAβ ⊂ Aα·β if α · β ∈ T and AαAβ = 0 if α · β /∈ T . An algebra A is said to be analgebra of associative type if, for any α1, α2, α3 ∈ T , there exists λ = λ(α1, α2, α3) ∈ k, λ �= 0, such that(aα1aα2)aα3 = λaα1(aα2aα3) for any aαi ∈ Aαi , i = 1, 2, 3.

In what follows we consider only finite-dimensional algebras of associative type. In [3], it hasbeen proved that such an algebra has a greatest homogeneous nilpotent ideal, which will be calledthe homogeneous radical of the algebra. (Here and in what follows a subspace M in a G-gradedalgebra A = ⊕

α∈TAα, T ⊂ G, is called homogeneous if M = ⊕

α∈TM ∩ Aα, T ⊂ G.) This fact gives rise

to homogeneously semisimple algebras of associative type, namely: an algebra A of associative type iscalled homogeneously semisimple if it has no nontrivial homogeneous nilpotent ideals.

The property of an algebra to be nilpotent depends, generally speaking, on the arrangement ofbrackets in the definition of powers of an algebra. But, as was shown in [3], if a homogeneoussubalgebra B of an algebra A of associative type is such that the product of any k elements b1, . . . , bk

of B is equal to zero for some arrangement of brackets, then the product of these elements is equal tozero for any arrangements of brackets. Thus, the arrangement of brackets has no influence on nilpotencyof homogeneous subalgebras in an algebra of associative type.

In addition, in [3], it has been shown that any homogeneously semisimple algebra of associativetype A is a direct sum of two-sided ideals Bi, i = 1, . . . , r, each of which is a homogeneously simplealgebra (i.e., an algebra with no nontrivial two-sided homogeneous ideals). What is more, each ideal Bi isa sum of homogeneous minimal left ideals Lij , j = 1, . . . , si, and any two ideals Lij and Lik, for a given i,are isomorphic as A-modules. In what follows the ideals B1, . . . , Br will be called simple componentsof A. We will also use the following two definitions.

*E-mail: [email protected].

20

MODULES AND IDEALS OF ALGEBRAS 21

Let G be a semigroup, M a G-set. If B = ⊕α∈T

Bα, T ⊂ G, |T | < ∞, is a G-graded algebra over a

field k, V = ⊕γ∈S

Vγ , S ⊂ M , |S| < ∞, a linear space over k, then a linear mapping f : B → Endk(V ) is

called a representation of associative type if

1) f(Bα)Vγ ⊂ Vα·γ when α · γ ∈ S or f(Bα)Vγ = 0 when α · γ /∈ S, where α ∈ T , γ ∈ S;

2) f(bαbβ)vγ = λα,β,γf(bα)f(bβ)vγ when bα ∈ Bα, bβ ∈ Bβ , vγ ∈ Vγ , and λα,β,γ ∈ k, λα,β,γ �= 0,where α, β ∈ T , γ ∈ S.

In this case, the space B will be called a B-module of associative type, and relations 1) and 2) will bewritten in the form

1)′ BαVγ ⊂ Vα·γ if α · γ ∈ S, BαVγ = 0 if α · γ /∈ S;

2)′ (bαbβ)vγ = λα,β,γbα(bβvγ) for any α, β ∈ T , γ ∈ S.

If M = G, V = B, and f(b) = Lb is the operator of left multiplication for b ∈ B, then B is an algebraof associative type.

Let G be a semigroup, M and M ′ two G-sets, and π : M → M ′ a G-mapping. Let V = ⊕γ∈S

Vγ ,

S ⊂ M , |S| < ∞, and V ′ = ⊕γ′∈S′

Vγ′ , S′ ⊂ M ′, |S′| < ∞, be two A-modules of associative type over

an algebra of associative type graded by G, A = ⊕α∈T

Aα, T ⊂ G, |T | < ∞. Then a linear mapping

ϕπ : V → V ′ will be called a homomorphism of an A-module V to an A-module V ′ if

1) π(S) ⊂ S′ and, for any γ ∈ S, ϕπ(Vγ) ⊂ V ′πγ ;

2) for any α ∈ T , β ∈ S, there exists a nonzero element λα,β ∈ k such that ϕπ(ax) = λα,βaϕπ(x)when a ∈ Aα, x ∈ Vβ .

If ϕπ is a bijection, then ϕπ is an isomorphism of A-modules of associative type.An algebra of associative type B regarded as a module over itself will be called the left regular B-

module. We denote this module by BB.A subspace U = ⊕

γ∈SUγ , Uγ = U ∩ Vγ , is called a B-submodule of a module V if U is a B-module in

the sense of the above given definition.We say that a B-module V of associative type is completely reducible if, for any B-submodule of

associative type W , there exists a B-submodule of associative type U such that V = W ⊕ U .By means of the standard arguments, one can easily prove that this definition is equivalent to the

decomposability of a B-module of associative type into a sum (not necessarily direct) of irreduciblesubmodules of associative type. Thus, the minimal left homogeneous ideals in an algebra of associativetype A are exactly the irreducible submodules of the left regular A-module of associative type AA.Therefore, the theorem on the decomposition of a homogeneous semisimple algebra proved in [3] impliesthe following theorem.

Theorem 1. If A is a finite-dimensional homogeneously semisimple algebra of associative type,then the left regular A-module AA is completely reducible.

Using this result, we prove the following statement.

Theorem 2. Each nonzero homogeneous two-sided ideal B of a homogeneously semisimplefinite-dimensional algebra of associative type A is the direct sum of some components of A.

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 52 No. 8 2008

22 KORESHKOV

Proof. An ideal B contains a homogeneous minimal left ideal L of A. Then Lx ⊂ B for any homoge-neous element x ∈ A. (An element is called homogeneous if it belongs to one of the components Aα.)In [3], it has been proved that minimal homogeneous left ideals L and L′ are isomorphic (as A-modules)if and only if L′ = La′ for some homogeneous element a′ ∈ L′. Therefore, B contains the sum BL of allhomogeneous minimal left ideals isomorphic to L. As has been shown in [3], BL is one of the simplecomponents of A.

Denote by B′ the sum of all simple components of A contained in B. Since the left regular module AAis completely reducible, we have B = B′ ⊕ B′′, where B′′ is an A-module or a left homogeneous ideal.If B′′ �= 0, then it contains a nonzero minimal homogeneous left ideal L′′. Consequently, according towhat has been said above, BL′′ ⊂ B′. On the other hand, B′ ∩ B′′ = 0, which leads to a contradiction.Therefore, B′′ = 0 and B is representable as the direct sum of some components of A.

In what follows we assume that all algebras of associative type and their modules are finite-dimensional vector spaces over a field k of zero characteristic.

Following [5] (P. 31), we will say that a homogeneous element a is properly nilpotent if ab and ba arenilpotent elements for any homogeneous element b.

The condition of proper nilpotency of an element can be described in terms of a certain form.

Let k[x] be the ring of polynomials in one variable. Denote A[x] = k[x]⊗k

A, V [x] = k[x]⊗k

V , where

V = ⊕γ∈S⊂M

Vγ is a module of associative type, A = ⊕α∈T⊂G

Aα. We make V [x] = ⊕γ∈S⊂M

Vγ [x], Vγ [x] =

k[x]⊗k

Vγ into a module of associative type over A[x] = ⊕α∈T⊂G

Aα[x], Aα[x] = k[x]⊗k

Aα by letting

(f ⊗ a)(g ⊗ v) = fg ⊗ av, a ∈ Aα, v ∈ Vγ , f, g ∈ k[x]. Then the linear mapping ρ : A → Endk(V )defining the structure of A-module of associative type on V extends to the linear mapping ρx : A[x] →k[x]⊗

kEndk(V ) acting as follows: ρx(f ⊗ a) = f ⊗ ρ(a), f ∈ k[x], a ∈ Aα.

Denote by δf the leading coefficient of a polynomial f(x) ∈ k[x]. Define a mapping SV from A[x] ×A[x] to k by

SV (a, b) = δ tr(Xρx(a)ρx(b)

), a, b ∈ A[x],

X =

[X1 0

. . .0 Xs

]

, Xi = xiEi, Ei =[ 1 0

. . .0 1

], ni = dimVγi is the dimension of Ei, s is the number of

elements in S.If a, b ∈ A, then SV (a, b) = δ tr

(Xρ(a)ρ(b)

), where ρx(1 ⊗ a) is identified with ρ(a). Any matrix

ρ(a), a ∈ Aα, can be represented in the form [ρij(a)], i, j = 1, . . . , s, where ρij(a) ∈ Endk(Vγj , Vγi).

Choosing a basis in V that agrees with the grading of V , one can rewrite the second relation in thedefinition of a module of associative type in the form

ρ(ab) = ρ(a)ρ(b)Λ, ρ(a), ρ(b), ρ(ab),Λ ∈ Endk(V ), (1)

where a, b are homogeneous elements of A, a ∈ Aα, b ∈ Aβ , and Λ =

[Λ1 0

. . .0 Λs

]

, Λi = λi(α, β)Ei,

λi(α, β) ∈ k, λi(α, β) �= 0, Ei =[ 1 0

. . .0 1

], ni = dim Vγi is the dimension of Ei.

By analogy with associative algebras, an A-module V of associative type will be called exact if fromthe fact that av = 0 for any v ∈ V and some a ∈ A it follows that a = 0.

Proposition 1. Let V = ⊕γ∈S⊂M

Vγ be an exact A-module of associative type over a graded algebra

A = ⊕α∈T⊂G

Aα. Then a homogeneous element a is properly nilpotent if and only if SV (a, b) =

SV (b, a) = 0 for any homogeneous element b ∈ A.



Proof. Let z be a homogeneous nilpotent element of A and ρ(z) = [ρij(z)], i, j = 1, . . . , s, the matrixof z with respect to the representation ρ in V . Then tr ρii(z) = 0, i = 1, . . . , s. In fact, since ρ(ab) =ρ(a)ρ(b)Λ (see (1)) for homogeneous elements a and b, we have

(ρ(z)

)n = ρ(z . . . (zz) . . .︸︷︷︸

n

)Λ=0, where n

is the nilpotency degree of z, i.e., the operator ρ(z) is nilpotent.If z is a homogeneous element of A, then, for any j = 1, . . . , s, the matrix ρ(z) = [ρij(z)] has no more

than one nonzero block ρij(z). In the case when ρii(z) = 0, we also have tr ρii(z) = 0. If ρii(z) �= 0,then, for any n,

(ρ(z)n

)ii

=(ρii(z)

)n.Thus, for a homogeneous nilpotent element z, the block ρii(z) is also nilpotent. In particular,

tr ρii(z) = 0.Let a be a homogeneous properly nilpotent element. Then, for any homogeneous element b, the prod-

uct ab is nilpotent. Therefore, tr ρii(ab)=0, i=1, . . . , s. But ρ(ab) = ρ(a)ρ(b)Λ, i.e., tr(ρ(a)ρ(b)

)ii

= 0,i = 1, . . . , s. Therefore,

SV (a, b) = δ tr(Xρ(a)ρ(b)

)= δ

s∑

i=1

xi tr(ρ(a)ρ(b)

)ii

= 0.

Let SV (a, b) = 0 for any homogeneous element b and some element a. Since we have that SV (a, b) =δ tr

(Xρ(a)ρ(b)

)= δ tr

(Xρ(ab)Λ

), it follows tr

(ρ(ab)

)= 0. Let us replace b by b1 = b(ab). By the

assumption, SV (a, b1) = 0. But

SV (a, b1) = δ tr(Xρ(a)ρ(b(ab))

)= δ tr

(Xρ(a(b(ab)))P

)

= λ−1a,b,abδ tr

(X(ρ(ab)2)P

)= λ−1

a,b,abδ tr(X(ρ(ab))2P ′),

where P , P ′ are block-diagonal matrices with nonzero diagonal coefficients, and, in each block, the co-efficient is constant. Consequently, tr

(ρ(ab)

)2 = 0. Repeating the procedure, we obtain tr(ρ(ab)

)n=0for any natural n. Since the basic field is of zero characteristic, ρ(ab) is a nilpotent operator, i.e.,(ρ(ab)

)m = 0 for some natural m. Using one more time the relation(ρ(ab)

)m = ρ((ab)[m]

)Λ, where

(ab)[m] =((ab)(. . . ((ab)(ab)) . . . )︸︷︷︸

m

), we obtain ρ

((ab)[m]

)= 0. Since ρ is an exact representation,

(ab)[m] = 0.The above discussion can be repeated for the element ba, which proves the proper nilpotency of a.

Let M = G, V = A, and let ρ be the regular representation defined by the operators of left multipli-cation Lx, x ∈ A. All the arguments of Proposition 1 are also valid in this case except for the transitionfrom ρ

((ab)[m]

)= 0 to (ab)[m] = 0, where the exactness of ρ is used. In this case, we apply the relation

L(ab)[m] = 0 to ab. Then (ab)[m+1] = 0, which proves the following proposition.

Proposition 2. If A is an algebra of associative type, then a homogeneous element a ∈ A isproperly nilpotent if and only if SA(a, b) = SA(b, a) = 0 for any homogeneous element b ∈ A.

Let I be a subspace in an algebra A. Denote I⊥L,V = {a ∈ A, SV (a, b) = 0, b ∈ I}, I⊥R,V = {a ∈ A,

SV (b, a) = 0, b ∈ I}, where V is an A-module of associative type. If I = A, then A⊥L,V will be called the

left kernel of SV and denoted by KerL SV . Respectively, A⊥R,V will be called the right kernel and denoted

by KerR SV .if G is a group, we will say that a G-set M has no fixed points when gm = m ⇔ g = 1.

Proposition 3. Let A = ⊕α∈T⊂G

Aα, and let G be a group, V = ⊕γ∈S⊂M

Vγ an A-module of associa-

tive type, M a G-set with no fixed points. If I is a homogeneous two-sided ideal in A, then I⊥L,V ,

I⊥R,V are also homogeneous two-sided ideals and I⊥L,V = I⊥R,V .


24 KORESHKOV

In particular, from Proposition 3 it follows that KerL SV = KerR SV . We will denote this set byKer SV and call it the kernel of SV .

Proof. Note first that I⊥L,V (respectively, I⊥R,V ) is a subspace in A. In fact, if SV (a, b) = 0 and

SV (a′, b) = 0, then tr(Xρ(a)ρ(b)

)= 0 and tr

(Xρ(a′)ρ(b)

)= 0, where ρ is the representation corre-

sponding to the module V . Since the trace function and the representation ρ are linear mappings, wehave tr

(Xρ(αa + α′a′)ρ(b)

)= 0, α,α′ ∈ k. Consequently, SV (αa + α′a′, b) = 0. If the latter relation

holds for any b ∈ I, then αa + α′a′ ∈ I⊥L,V .

In addition, I⊥L,V and I⊥R,V are homogeneous spaces. By virtue of the homogeneity of I, it sufficesto verify the orthogonality condition for b = bβ ∈ I ∩ Aβ . Let SV (a, bβ)=0 and a =

∑

α∈Saα. If α · β �=1,

then all the diagonal blocks of the matrix ρ(aα)ρ(bβ) are zero since the G-set M has no fixed points.

Consequently, in these cases, tr(Xρ(aα)ρ(bβ)

)= 0. Therefore, tr

(Xρ

(a −

∑

α�=β−1

aα

)ρ(bβ)

)= 0.

Consequently, SV (aβ−1 , bβ) = 0, i.e., all components of a are orthogonal to bβ and, therefore, belongto I⊥L,V .

Let a∈Aα, b∈Aβ , and let SV (a, b) = 0. Then tr(Xρ(a)ρ(b)

)= 0, i.e.,

s∑

i=1

s∑

j=1xi tr

(ρij(a)ρji(b)

)= 0,

where s is the number of components of the grading in the module V . If ρji(b) �= 0 for some i, then thenumber j is determined uniquely since each i-th block column can contain only one nonzero block.Therefore, the coefficient of xi in the sum under consideration equals tr

(ρij(a)ρji(b)

). Thus, in the case

when ρji(b) �= 0, we also have tr(ρij(a)ρji(b)

)= 0.

Consider C =s∑

k=1

s∑

t=1xk tr

(ρkt(b)ρtk(a)

). If ρkt(b) �= 0 for some k, then tr

(ρtk(a)ρkt(b)

)= 0. But

tr(ρtk(a)ρkt(b)

)= tr

(ρkt(b)ρtk(a)

). Consequently, C = 0. Moreover, SV (b, a) = δC = 0.

Thus, I⊥L,V = I⊥R,V . We will denote this set by I⊥V . In particular, KerL SV = KerR SV = Ker SV .

Let us verify that I⊥V is a two-sided ideal. For this, we show that if a ∈ Aα, b ∈ Aβ , and c ∈ Aγ , thenSV (ab, c) = 0 ⇔ SV (a, bc)=0. Let us compute each of the indicated values of the function SV . We have

SV (ab, c) = δ

s∑

i=1

s∑

j=1

xi tr(ρij(ab)ρji(c)

)= δ

s∑

i=1

s∑

j=1

s∑

k=1

λixi tr

(ρik(a)ρkj(b)ρji(c)

)

since δ tr(Xρ(ab)ρ(c)

)= δ tr

(XΛρ(a)ρ(b)ρ(c)

). (On condition that the G-set M has no fixed points,

any matrix ρ(a) = [ρij(a)] has no more than one nonzero block ρij(a) for each i. Therefore, ρ(ab) =ρ(a)ρ(b)Λ̃ = Λρ(a)ρ(b), where the block-diagonal matrix Λ is obtained from the matrix Λ̃ by a permu-tation of diagonal blocks.) Respectively,

SV (a, bc) = δ

s∑

i=1

s∑

k=1

xi tr(ρik(a)ρki(bc)

)= δ

s∑

i=1

s∑

k=1

s∑

j=1

λ′ix

i tr(ρik(a)ρkj(b)ρji(c)

)

since δ tr(Xρ(a)ρ(bc)

)= δ tr

(Xρ(a)ρ(b)ρ(c)Λ′).

If SV (ab, c) = 0, thens∑

k=1

s∑

j=1tr

(ρik(a)ρkj(b)ρji(c)

)= 0 for any i = 1, . . . , s since all λi are nonzero.

But then, from the expression for SV (a, bc), we have SV (a, bc) = 0, which proves the above formulatedcondition of “partial” invariance.

If G is a group, M = G, V = A, and ρ is the regular representation, then the assumptions ofProposition 3 hold, and we obtain the following statement.



Corollary. Let G be a group and A = ⊕α∈T⊂G

Aα an algebra of associative type. If I is a homogeneous

two-sided ideal in A, then I⊥L,A and I⊥R,A are also two-sided ideals and I⊥L,A = I⊥R,A.

In particular, KerL SA = KerR SA. This set (denoted by Ker SA) will be called the kernel of SA.By virtue of the relation (ab)c = λa(bc), λ �= 0, λ ∈ k, for homogeneous elements a, b, c of an

algebra A, by induction, one can easily verify that if I is a homogeneous two-sided ideal in A, then anyits power Ik is also a homogeneous two-sided ideal. Hence it follows that the sum of two homogeneousnilpotent ideals is also a homogeneous nilpotent ideal. Therefore, for a finite-dimensional algebra A,there exists a greatest homogeneous nilpotent ideal. We will denote it by Rqr(A).

Proposition 4. Let G be a group and A a G-graded algebra of associative type. Then Rqr(A) =Ker SA.

Proof. Let I be homogeneous nilpotent ideal in A. Then, for any homogeneous element x of I, we havexa ∈ I. Consequently, x is properly nilpotent when a is homogeneous. Therefore, from Proposition 2 itfollows that I ⊂ Ker SA. In particular, Rqr(A) ⊂ Ker SA.

Conversely, let y ∈ KerSA. By Proposition 2, y is a properly nilpotent element. In particular, y2 isnilpotent. But then y is also nilpotent. Using theorem 1 in [3], we conclude that Ker SA is nilpotent.Therefore, Ker SA ⊂ Rqr(A). Two inclusions obtained prove the equality of the proposition.

As was defined above, an algebra A of associative type is homogeneously semisimple if it has nohomogeneous two-sided nilpotent ideals, i.e., Rqr(A) = 0. Proposition 4 implies the following theorem.

Theorem 3. Let G be a group and A a G-graded algebra of associative type. Then A is homoge-neously semisimple if and only if the form SA is nondegenerate.

If A is semisimple, then it is homogeneously semisimple. Consequently, the following theorem holds.

Theorem 4. If A is a semisimple G-graded algebra of associative type, where G is a group, thenthe form SA is nondegenerate.

Theorem 5. If A = ⊕α∈T

Aα, T ⊂ G, |T | < ∞, is a semisimple algebra of associative type and G is

a group, then 1 ∈ T and A1 is a semisimple associative algebra.

Proof. In [3], it has been shown that a semisimple algebra of associative type A = ⊕α∈T

Aα, T ⊂ G,

contains a homogeneous idempotent e. If G is a group, then the idempotent e belongs necessarilyto A1, 1 is the unity of G, i.e., A1 �= 0. Substituting the idempotent e into the relation (ab)c = λa(bc),a, b, c ∈ A, we obtain λ = 1, i.e., A1 is an associative algebra.

For any homogeneous element a ∈ Aα, the matrix of the operator of left multiplication La in terms ofa basis agreeing with the grading of A is of the form [Lij(a)], i, j = 1, . . . , r, r = |T |, and Lij(a) is therestriction of La to the subspace Aαj with values in Aαi . As was mentioned in Proposition 1, for each j,there is no more than one nonzero block Lij(a). Since G is a group, it follows that, for any i, there is nomore than one nonzero block Lij(a). In particular, all diagonal blocks Lii(c) of the matrix Lc are nonzerowhen c ∈ Aα, α �= 1.

If a ∈ Aα, b ∈ Aβ , and α · β �= 1, then SA(a, b) = 0 since LaLb = LabΛ, Λ =

[Λ1 0

. . .0 Λr

]

, Λi = λiEi,

λi ∈ k, λi �= 0, Ei is the identity operator of the space Aαi , i = 1, . . . , r, and all diagonal blocks of Lab

are nonzero. Consequently, SA(A1, Aβ) = 0 when β �= 1.If z ∈ A1 is such that the matrix Lz is nilpotent, then all diagonal blocks Lii(z) are also nilpotent

matrices. Therefore, the trace of each diagonal block tr Lii(z) equals zero.Let I be a nilpotent ideal in A1. Then, for a ∈ I, b ∈ A1, the operator Lab is nilpotent by virtue

of the relation Lz[n] = (Lz)nΛ (see Proposition 1). Therefore, tr(XLaLb) =∑

λixi tr

(Lii(ab)

)=0.

Hence it follows that δ tr(XLaLb) = 0, a ∈ I, b ∈ A1, i.e., SA(I,A1) = 0. Using the above obtainedcondition SA(A1, Aβ) = 0 when β �= 1, we have SA(I,A) = 0. Since A is semisimple, the form SA isnondegenerate (Theorem 4). Consequently, I = 0, i.e., A1 is a semisimple associative algebra.


26 KORESHKOV

Theorem 6. Let A = ⊕α∈T

Aα, T ⊂ G, be a semisimple finite-dimensional algebra of associative

type over a field of zero characteristic, and let V = ⊕γ∈S

Vγ , S ⊂ M , be a finite-dimensional A-

module of associative type. If G is a group and a G-set M has no fixed points, then V iscompletely reducible.

Proof. Let ϕ be the representation of associative type corresponding to the module V = ⊕α∈S

Vα. Denote

by ϕ1 the restriction of the representation ϕ to the subalgebra A1. Each subspace Vα, α ∈ S, is a moduleof associative type over A1. Denote the corresponding representation of associative type by ϕ1,α. Thenthe mapping ψα = λ1,1,αϕ1,α, where λ1,1,α is the constant from item 2 of the definition of a module ofassociative type, gives the standard representation of the semisimple associative algebra A1 in Vα. Sinceevery finite-dimensional associative semisimple algebra has unity e1 ([6], P. 34), its image ψα(e1) isa projector. Therefore, the following decomposition takes place: Vα = V ′

α ⊕ V ′′α , where V ′

α = {vα ∈ Vα,ψα(e1)vα = vα}, V ′′

α = {vα ∈ Vα, ψα(e1)vα = 0}.

Let W ′′ = ⊕α∈S

V ′′α , W ′ = ⊕

α∈SV ′

α. Then V = W ′ ⊕W ′′ and W ′, W ′′ are A-submodules. In fact, since

ψα(e1)vα = 0 when vα ∈ V ′′α and ψα(e1) = λ1,1,αϕ1,α(e1), it follows that e1vα = 0. Then, obviously,

A1vα = 0. But, as has already been mentioned, A is the sum of left homogeneous ideals Aek generatedby idempotents ek belonging to A1. Therefore, AVα = 0, α ∈ S, i.e. AW ′ = 0.

Let vα ∈ V ′α. Assume that aβ ∈ Aβ exists such that aβvα = v′γ + v′′γ and v′′γ �= 0. Then e1(aβvα) =

e1v′γ ∈ V ′

γ since e1v′′γ = 0. On the other hand,

e1(aβvα) = ϕ(e1)ϕ(aβ)vα = λ−11,β,αϕ(e1aβ)vα = λ−1

1,β,αϕ(λ1,1,βaβ)vα = λ−11,β,αλ1,1,βaβvα,

where λ1,1,β is the constant from the definition of the regular representation. We used here the factthat the operator of left multiplication by e1 acts as a nonzero scalar on each subspace Aβ , β ∈ T . Infact, if B = 〈aβ ∈ Aβ , β ∈ T , e1aβ = 0〉 �= 0, then A1B = 0. Therefore, AB = 0 (see the correspondingdiscussion for the space V ′′

α ). Then B is a homogeneous two-sided ideal in A with the property B2 = 0,which contradicts the fact that A is a semisimple algebra. Thus, for any aβ ∈ A, we have aβvα ∈ V ′

γ

when vα ∈ V ′α, i.e., AW ′ ⊆ W ′. Therefore, we consider the image ψa(e1) of the unity element e1 of A1 to

be the identity operator on Vα. In particular, Vα = A1Vα. Consequently, V = AV .

Since A is a finite sum of minimal left homogeneous ideals Aek generated by homogeneous

idempotents ek ∈ A1, it follows that V =∑

vα∈V

n∑

k=1

(Aek)vα. Obviously, (Aek)vα = ⊕β∈T

(Aβek)vα is an

A-submodule of the module V . Consider the homomorphism of A-modules Aek → (Aek)vα defined byaek → (aek)vα. If V ′ is an A-submodule in (Aek)vα, then it appears as V ′ = ⊕

β∈T(Bβek)vα, Bβ ⊂ Aβ

and (Bβek)vα ⊂ V ′. By virtue of the fact that the G-set M has no fixed points, the preimage of the A-module V ′ is the left homogeneous ideal B = ⊕

β∈TBβek. Since Aek is a minimal left homogeneous ideal

in A, the ideal B either equals zero or coincides with Aek . Consequently, the submodule (Aek)vα eitheris irreducible or equals zero. Thus, V is a sum of irreducible A-modules, i.e., V is completely reducible.

The following result demonstrates a certain structure similarity of associative algebras and algebrasof associative type.

Theorem 7. Let A be a semisimple algebra of associative type with grading defined by a groupG, and let I be a left homogeneous ideal in A. Then I = Ae for some homogeneous idempotent eof A.



Proof. Since A is semisimple, the ideal I cannot be nilpotent and, therefore, contains a nonzeroidempotent [3]. Since the set T of indices of the grading of A is a subset of the group G, all its idempotentsare contained in A1, 1 is the unity of G.

To each idempotent e ∈ I ∩ A1, let us assign the space A(e) = 〈xα ∈ I ∩ Aα, α ∈ T , xαe = 0〉.Then A(e) is a left homogeneous ideal in A. The set of left homogeneous ideals A(e) is nonempty.Consequently, it has a minimal element A(e0).

If A(e0) = 0, then from the equality xe0 − xe20 = 0 for any homogeneous element x ∈ I it follows that

(x − λxe0)e0 = 0, λ �= 0, λ ∈ k, i.e., x = λxe0. But then I = Ie0 ⊂ Ae0 ⊂ I, whence I = Ae0.Assume that A(e0) �= 0 and show that this assumption leads to a contradiction. Since A(e0) is a

nonzero left homogeneous ideal of a semisimple algebra A, it follows that A(e0) contains an idempotente1 ∈ A1. By definition, e1 ∈ I and e1e0 = 0. Consider the element e = e0 + e1 − e0e1. Using the factthat A1 is an associative algebra (Theorem 5), one can easily check that e is an idempotent. The relatione1e = e1 �= 0 implies e �= 0.

Let x ∈ A(e), and let x be a homogeneous element. Then xe = 0 and (xe)e0 = 0. But ee0 = e0.Consequently, xe0 = 0, i.e., x ∈ A(e0). Therefore, A(e) ⊆ A(e0). But e1 ∈ A(e0) since e1e0 = 0, ande1 /∈ A(e) since e1e = e1 �= 0. Therefore A(e) is strictly contained in A(e0), which contradicts theminimality of A(e0). Thus the case A(e0) �= 0 is impossible.

REFERENCES1. Y. Bahturin and M. Zaicev, “Identities of Graded Algebras,” J. Algebra. 205 (1), 1–12 (1998).2. Y. Bahturin, M. Zaicev, and S. K. Sehgal, “G-Identities of Nonassociative Algebras,” Matem. Sborn. 190 (11),

3–14 (1999).3. N. A. Koreshkov, “On Nilpotency and Decomposition of Algebras of Associative Type,” Izv. Vyssh. Uchebn.

Zaved. Mat., No. 9, 34–42 (2006) [Russian Mathematics (Iz. VUZ) 50 (9), 32–39 (2006)].4. N.A. Koreshkov, “A Class of Algebras of Associative Type,” Izv. Vyssh. Uchebn. Zaved. Mat., No. 3, 38–46

(2007) [Russian Mathematics (Iz. VUZ) 51 (3), 33–41 (2007)].5. N. G. Chebotaryov, Introduction to the Theory of Algebras (OGIZ, Gostekhizdat, Moscow–Leningrad,

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modules and ideals of algebras of associative type

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