Symmetric Groups and Iwahori–Hecke Algebras
4.5 Semisimple algebras and modules
4.5.7 Modules over a semisimple algebra
Corollary 4.51.LetJbe a two-sided ideal of a semisimple algebraA. ThenJ and the quotient algebraA/J are semisimple algebras.
Proof. Consider the splitting A =
λ∈Λ Aλ of A as in Theorem 4.49. By Example 4.47 (ii), eachAλ is semisimple. We have seen in the proof of Theo- rem 4.49 that there is a setΛ0⊂Λsuch thatJ =
λ∈Λ0 Aλ. By Lemma 4.48,
J =
λ∈Λ0
Aλ and A/J ∼=
λ∈Λ−Λ0
Aλ.
We conclude by using Example 4.47 (ii), which tells us that finite products of
semisimple algebras are semisimple.
We now summarize the representation theory of a (finite-dimensional) semisimple algebraA. Let{Aλ}λ∈Λ be the set of all nonzero two-sided ideals ofAthat are simple as algebras. This set is finite. For eachλ∈Λ, there is a unique up to isomorphism simpleAλ-moduleVλ. We viewVλas anA-module byAμVλ = 0 for μ=λ. Then the A-modules {Vλ}λ∈Λ are simple, and eve- ry simpleA-module is isomorphic to exactly one of them. Moreover, for any finite-dimensionalA-moduleM, there are a unique functiondM :Λ→Nand an isomorphism ofA-modules
M ∼=
λ∈Λ
VλdM(λ). (4.30)
The functiondM is called thedimension vector ofM.
LetDλ be the division ring EndA(Vλ). The following theorem is a conse- quence of Corollary 4.40, Lemma 4.48, and Theorem 4.49.
Theorem 4.54.With the notation above, A∼=
λ∈Λ
EndDλ(Vλ) and
dimKA=
λ∈Λ
(dimKVλ)2 dimKDλ
.
Corollary 4.55.If the ground fieldKis algebraically closed, then the algebra homomorphism
A→
λ∈Λ
EndK(Vλ)
obtained as the product overΛof the algebra homomorphismsA→EndK(Vλ) induced by the action ofAon Vλ is an isomorphism. Moreover,
dimKA=
λ∈Λ
(dimKVλ)2.
Proof. Applying Proposition 4.30 (c) to the simple A-module Vλ, we ob- tain dimKDλ = 1. Thus, Dλ = K. The corollary is then a reformulation
of Theorem 4.54.
Exercise 4.5.1.LetAbe a finite-dimensional algebra with radicalJ =J(A).
(a) Show that the quotient algebraA/J is semisimple.
(b) Prove that 1 +xis invertible in Afor anyx∈J.
Exercise 4.5.2.Let V be a finite-dimensional vector space over a field K.
Show that everyK-linear automorphism of the algebra A= EndK(V) is the conjugation by an element ofA. (Hint: An automorphism ofA defines a new A-module structure onV; now use the fact thatAhas only one isomorphism class of simple modules.)
Exercise 4.5.3.Let A be a semisimple algebra and M a finite-dimensional A-module. Show that there is an algebra isomorphism
EndA(M)∼=
λ∈Λ(A)
MdM(λ)
EndA(Vλ) .
Exercise 4.5.4.Let K be an algebraically closed field and A a semisimple K-algebra. Prove that there is an isomorphism ofA-modules
A∼=
λ∈Λ(A)
Vλdλ, wheredλ= dimKVλ.
Exercise 4.5.5.LetD be a division ring. For 1≤i, j≤n, letEi,j ∈Mn(D) be the matrix whose entries are all zero except the (i, j) entry, which is 1.
(a) Verify that the trace of the right multiplication by Ei,j in the matrix algebraMn(D) isnifi=j and is 0 otherwise.
(b) Prove that the trace form ofMn(D) is given for alla, b∈Mn(D) by a, b=nTr(ab).
(c) Deduce thatMn(D) is semisimple if and only ifn is invertible inD.
Exercise 4.5.6.Let K be a field of characteristic p > 0 and G the cyclic group of order p. Show that (g−1)p = 0 ∈ K[G] for all g ∈ G. Deduce that the group algebra K[G] contains a nonzero nilpotent ideal and is not semisimple.
Exercise 4.5.7.LetAbe a finite-dimensional algebra over a fieldKof char- acteristic zero. Prove that all elements of the radical ofAare nilpotent. (An elementaofAisnilpotent ifaN = 0 for some integerN ≥1.)
Solution.Setd= dimKA. For eacha∈J(A) andn≥1, Tr((Ra)n) = Tr(Ran) =a, an−1= 0.
Ifλ1, . . . , λdare the eigenvalues of Rain an algebraic closure ofK, the previous equalities imply that
λn1+· · ·+λnd = 0
for all n ≥ 1. By Newton’s formulas (which require the ground field to be of characteristic zero), all elementary symmetric polynomials in λ1, . . . , λd vanish. This implies that the characteristic polynomial of Ra is a monomial of degreedand hence (Ra)d= 0. Therefore,ad= (Ra)d(1) = 0.
Exercise 4.5.8.Let K be a field of characteristic zero. Prove that a finite- dimensionalK-algebraA that does not contain nonzero nilpotent left ideals is semisimple.
Solution.It suffices to prove that J =J(A) = 0. Assume thatJ = 0 and pick a nonzero left idealI⊂J of minimal dimension overK. By assumption, the idealI is not nilpotent, and in particular,I2 = 0. Hence, there isx∈I such thatIx= 0. By the minimality of I and the inclusionIx⊂I, we have Ix=I. Hence, there is e∈I such thatex=x. It follows that
x=ex=e(ex) =e2x . We thus obtain (e−e2)x= 0. The left ideal
I={y∈I|yx= 0}
is a proper subideal of I, since Ix = 0. By the minimality of I, we must haveI= 0. Since e−e2∈I, we havee=e2. Hence,
e=e2=e3=· · ·.
Now, by Exercise 4.5.7, the elemente∈I⊂J is nilpotent (this is where we use the characteristic-zero assumption). From these two facts we deduce that e = 0. Hence, x = ex = 0 and Ix = 0, which contradicts the choice of x.
Therefore,J= 0.
Exercise 4.5.9.An elementeof an algebraAis anidempotent ife=e2. (a) Show that ife∈Ais an idempotent, then so isf = 1−e.
(b) Suppose that an idempotente∈Ais central, that is,ecommutes with all elements ofA. Setf = 1−e. Prove thatAe andAf are two-sided ideals ofA, that viewed as algebrasAeandAf haveeandf as respective units, and that the mapAe×Af →A,(a, b)→a+bis an algebra isomorphism.
(c) Show that the unique nonzero central idempotent of a simple algebra is its unit.
Exercise 4.5.10.A nonzero central idempotenteof an algebraAisprimitive if it not expressible as a sum of two nonzero central idempotents whose product is zero.
(a) Prove that ife is a primitive central idempotent ofA, then there are no algebrasA1,A2such thatAe∼=A1×A2.
(b) LetAbe a product ofr <∞simple algebras. Show that there is a unique set {e1, . . . , er} of primitive central idempotents ofA such thateke = 0 for all distinctk, ∈ {1, . . . , r} ande1+· · ·+er= 1.
Exercise 4.5.11.LetAbe a finite-dimensional algebra. Prove the following:
(a) The sum of two nilpotent left ideals ofA is a nilpotent left ideal.
(b) Any nonnilpotent left ideal ofAcontains a nonzero idempotent.
(c) The sum J of all nilpotent left ideals ofAis a two-sided ideal.
(d) If the ground field has characteristic zero, thenJ is the radical ofA.