Scalar extensions

Proof. By Skolem-Noether (Theorem 2.1.29) any automorphism of a central simple algebra is inner. LetA ∼=Q

A_iwithA_i simple. Lete_ibe the element ofAthat is the identity ofA_i. All thee_i lie in the centre ofAand hence are left unchanged under the application ofφ. Thus,φdescends to an automorphism ofA_i. NowA_i can be considered as an algebra over its centre, as such it is a central simple algebra. Hence, every restriction ofφis inner, and, consequently, so isφ. 2 Corollary 2.1.31 Letkbe a field andr ≥1an integer. Then every automorphism ofMat_r(k)comes

from conjugation by an invertible matrix. 2

2.2 Scalar extensions

Definition 2.2.1 LetK/kbe a field extension,Aa finite dimensionalk-algebra andV anA-module.

The scalar extension ofA byK is defined as the K-algebraAK := K ⊗_k Aand the one ofV as the A_K-moduleV_K := K ⊗_kV. In terms of the representationA → End_k(V), this gives rise to A_K →End_K(V_K).

Remark 2.2.2 Let Gbe a (profinite) group,ka field andV₁ andV₂ twok[G]-modules. The tensor product representation ofV1andV2is defined asV1⊗_kV2with thek[G]-action given byg(v1⊗v2) = gv₁⊗gv₂.

In terms of representations ofk[G]-algebras we obtain ak-algebra homomorphism k[G]−−−−−→^g7→(g,g) k[G×G]∼=k[G]⊗_kk[G]→End_k(V₁⊗_kV₂).

IfV2is a1-dimensional representation, then the tensor product representation is called a twist.

Proposition 2.2.3 LetAbe a finite dimensional semi-simplek-algebra. Then for any finite separable extensionK/k, theK-algebraK⊗_kA=A_Kis semi-simple.

Proof. (See [Lang], Theorem XVII.6.2.) By Lemma 2.1.17 we may assume thatK/kis a Galois extension and we letG= Gal(K/k). We will show thatN := Jac(K⊗_kA)is zero.

We letGact onK⊗_kAbyσ(x⊗a) =σ(x)⊗a. We see thatσN = N for anyσ ∈G, asN is maximal nilpotent by Proposition 2.1.13. Letz∈ N be any element. It can be written in the form z=P_n

i x_i⊗a_iwith{a₁, . . . , a_n}forming a basis ofAandx_i∈K. Now we use the traceTr_K/kto make an element inJac(A). Lety∈Kbe any element. We have

Tr_K/k(yz) = Tr_K/k(X This element is still inN, but also inA, and hence inJac(A)by Lemma 2.1.17. It follows that it is equal to0, whenceTr_K/k(yx_i) = 0for alliand ally. Since by separabilityTr_K/kis a non-degenerate bilinear form, it follows thatxi= 0for alli, whencez= 0, as desired. 2

28 CHAPTER 2. GENERAL REPRESENTATION THEORY

IfAis ak-algebra andK/ka field extension, then

K⊗_kMat_n(A)∼= Matn(K⊗_kA) (see Exercise 13).

Theorem 2.2.4 Let R be a commutative ring, letA andB be R-algebras and letV andW be A-module. Suppose thatBis flat overRandV a finitely presentedA-module. Then the natural map

B⊗_RHom_A(V, W)→Hom_B⊗RA(B⊗_RV, B⊗_RW) is anR-isomorphism.

Proof. (See [Karpilovsky], Theorem 3.5.2.) Omitted. 2

Lemma 2.2.5 Letkbe a field andAak-algebra. LetK/kbe a field extension. LetW be a simple A_K-module. Then there exists a simpleA-moduleV such thatW occurs as a composition factor of VK.

Proof. We know thatW occurs as a composition factor of AK. LetVi be a composition series ofA. Then a composition series ofA_K is obtained by taking the composition factors of every(V_i)_K. 2

Lemma 2.2.6 Letkbe a field, Aak-algebra andV₁ andV₂ twoA-modules of finitek-dimension.

LetK/kbe a field extension. IfV₁andV₂have a common composition factor, then so do(V₁)_Kand (V2)_K. Conversely, if(V1)_K and(V2)_Kare semi-simple and have a common composition factor, then so doV₁andV₂.

Proof. (See [CurtisReiner], 29.6.) Suppose first that V₁ andV₂ have a common composition factorS, i.e. a simple module occuring in the composition series. Then all the composition factors of S_Koccur in the composition series of both(V₁)_Kand(V₂)_K.

Conversely, if(V₁)_K and (V₂)_K are semi-simple and have a common composition factor, then by Schur’s lemmaHomA_K((V1)K,(V2)K)is non-zero. Note thatV1 andV2 are also semi-simple by Lemma 2.1.17. From Theorem 2.2.4 it follows thatHom_A(V₁, V₂)is also non-zero, implying thatV₁

andV₂have a common composition factor. 2

We now come to the concept of Galois conjugate modules.

Definition 2.2.7 LetK/kbe a Galois extension andAak-algebra.

(a) The Galois groupG= Gal(K/k)acts on the set ofA_K-modules from the left as follows:

LetW be anA_K-module. For anyσ∈Gal(K/k)we let^σW be theA_K-module whose underlying K-vector space is equal toW equipped with theA_K-action

(x⊗a)._σw:= (σ⁻¹(x)⊗a)._oldw for allx∈Kand alla∈A.

2.2. SCALAR EXTENSIONS 29

(b) Given anAK-moduleW, the decomposition groupDW(K/k) =DW is defined as the stabilizer ofW, i.e. as the subgroup consisting of those σ ∈ Gsuch that ^σW ∼= W. For some reason, [Karpilovsky] calls this the inertia group.

Remark 2.2.8 (i) It is easy to check that the action is indeed a left action.

(ii) IfW is simple, then so is^σW.

(iii) Ifσ ∈Gal(K/k)andV is anA-module, we define the isomorphism (ofk-vector spaces) σ:V_K x⊗v7→σ(x)⊗v

−−−−−−−−→V_K.

It is easy to check that for a submodule W ⊂ V_K the mapσ defines an isomorphism ofA_K -modules from^σW toσ(W).

(iv) Suppose thatW is a matrix representation ofA_K, i.e.A_K → End_K(W) = Mat_n(K). Then the matrix representation for^σW is obtained from the one ofW by applyingσ⁻¹ to the matrix entries.

Lemma 2.2.9 Letkbe a field andAak-algebra. LetK/kbe a Galois extension. LetV be a simple A-module. If the simpleA_K-moduleW occurs in the composition series ofV_K with multiplicitye, then so does^σW for allσ ∈Gal(K/k).

Proof. Note that^σVK ∼=VKasAK-modules. Thus^σW, which naturally occurs in the decompo-sition series of^σV, is isomorphic to a composition factor ofV. The statement on the multiplicities

follows also. 2

Lemma 2.2.10 LetK/kbe a finite Galois extension. Denote byW_Athe moduleW considered as an A-module (rather than as anA_K-module). Then there is an isomorphism ofA_K-modules

(W_A)_K ∼= M

σ∈Gal(K/k) σW.

Proof. We give the map: x ⊗v 7→ (. . . , x._σv, . . .) = (. . . , σ⁻¹(x)v, . . .). It is an A_K -module homomorphism. That it is an isomorphism can be reduced to the isomorphismK⊗_kK ∼= Q

σ∈Gal(K/k)σK,which is easily checked. 2

Proposition 2.2.11 LetK/kbe a finite Galois extension. LetV be indecomposable and letW be an indecomposable direct summand ofV_K. Then there exists an integeresuch that

VK = M

σ

σWe

,

whereσruns through a system of coset representatives forGal(K/k)/DW.

30 CHAPTER 2. GENERAL REPRESENTATION THEORY

Proof. (See [Karpilovsky], Theorem 13.4.5.) The proof is based on the uniqueness of a decompo-sition into indecomposables (Krull-Schmidt theorem, Theorem 2.1.9). DecomposeV_K into different indecomposables

V_K∼=V₁ⁿ¹⊕V₂ⁿ² ⊕ · · · ⊕V_r^nr withV₁ =W. We obtain

V^[K:k]∼= (V_K)_A∼= (V₁)ⁿ_A¹⊕(V₂)ⁿ_A²⊕ · · · ⊕(V_r)ⁿ_A^r asA-modules. As a consequence,(V₁)_A∼=V^sfor somes. Lemma 2.2.10 yields

V_K^s ∼= ((V₁)_A)_K ∼= M

σ∈Gal(K/k) σV₁,

whence for everyithere isσ_i ∈Gal(K/k)such thatV_i = ^σiV₁. From Lemma 2.2.9,n₁ =n_i =:e for allifollows. We now rewrite the first displayed equation:

V_K∼=W^e⊕(^σ2W)^e⊕ · · · ⊕(^σrW)^e.

By assumption, all the ^σiW are different. Lemma 2.2.9 implies that all the Galois conjugates occur.

This finishes the proof. 2

Corollary 2.2.12 LetV be simple and letW be a simple composition factor ofVK. Then there exists an integeresuch that

V_K = M

σ

σWe

,

whereσruns through a system of coset representatives forGal(K/k)/D_W.

Proof. SinceV is simple, it is a simpleB :=A/Jac(A)-module. Note thatB_K ∼=A_K/(K⊗_k Jac(A))(using thatK/kis flat) is also semi-simple (Proposition 2.2.3). Proposition 2.2.11 implies thatV_K = L

σσWe

with some indecomposable, and due to the semi-simplicity, hence simple,B_K -moduleW. We note thatW is also a simpleAK-module. The isomorphism is also an isomorphism ofA_K-modules, sinceK⊗_kJac(A)⊆Jac(A_K)acts trivially. 2

We draw the attention to Exercise 14.