Splitting fields

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.

2.3 Splitting fields

We draw the attention to Exercise 15.

Definition 2.3.1 LetRbe a ring andT ⊂Rbe a subring. We define the centralizer ofT inRas Z_R(T) :={r∈R |rt=tr∀t∈T}.

For important properties of the centralizer see Exercise 15. We include the following proposition, although it will not be needed for the subsequent proofs.

2.3. SPLITTING FIELDS 31

Proposition 2.3.2 Letkbe a field and considerk-algebrasA, A^′, B, B^′ such thatB ⊆AandB^′ ⊆ A^′. Then

Z_A⊗kA^′(B⊗_kB^′) =Z_A(B)⊗_kZ_A^′(B^′).

Proof. Exercise 16. 2

Proposition 2.3.3 Arbitrary scalar extension of a central simple algebra is central simple.

Proof. (See [Kersten], Satz 5.10.) Omitted. 2

Proposition 2.3.4 Letkbe a field and Aa finite dimensional central simplek-algebra. LetB ⊂ A be a simple subalgebra withdim_kB =n. Then there is ak-algebra homomorphism

Z_A(B)⊗_kMatn(k)∼=A⊗_kB^opp.

Proof. (See [Kersten], Korollar 8.4 (i).) Omitted. 2

Theorem 2.3.5 Letkbe a field andDa division algebraDoverk. LetKbe a subfield ofD.

ThenDis split byK, i.e.D_K∼= Matr(K)for somer≥1, if and only ifKis a maximal subfield.

In that case,ris equal to[K:k]and is called the index ofD.

Proof. This is a direct consequence of Proposition 2.3.4. For, we obtain isomorphisms Z_D(K)⊗_kMat_r(k)∼=D⊗_kK^opp ∼=D_K.

IfKis a maximal subfield, thenZ_D(K) =Kby Exercise 15. IfKis properly contained in a maximal subfieldL, thenZ_D(K)properly containsL, and is hence not a field. 2 Lemma 2.3.6 Letkbe a field andDa finite dimensional division algebra overk. Suppose that any maximal subfield ofDis equal tok. ThenD=k.

In particular, there is no finite dimensional division algebra over an algebraically closed field other than the fieldkitself.

Proof. Letd∈ Dbe any element. Inside ofDconsiderk(d), i.e. the smallest ring containingk andd. This ring is commutative, askis in the centre ofD(by the definition of ak-algebra). Sok(d)

is a field extension ofkand thus equal tok. 2

Corollary 2.3.7 LetDbe a division algebra over a fieldk. All its maximal subfields are isomorphic.

Proof. This follows from Theorem 2.3.5. 2

Corollary 2.3.8 (Wedderburn) A finite division ring is a finite field.

32 CHAPTER 2. GENERAL REPRESENTATION THEORY

Proof. (See [Bourbaki], 11.1.) Let D be a finite division ring with centre k and let K be a maximal subfield ofD. By Corollary 2.3.7 and basic algebra, every other maximal subfield is of the formxKx⁻¹. As every element ofDis contained in some maximal subfield, it follows that

D^×= [

x∈D^×

xK^×x⁻¹.

Notice that forx^′ =xtwitht∈Kwe havex^′K^×x^′−1 =xK^×x⁻¹. The number of distinctxK^×x⁻¹ is, thus, at most equal to the number of elements inD^×/K^×. Moreover, the number of elements of xK^×x⁻¹ is always equal to the number of elements ofK^×. Consequently, all distinctxK^×x⁻¹ are pairwise disjoint. Since they all contain the unit element, the number of distinctxK^×x⁻¹ has to be

one. 2

Corollary 2.3.9 (Wedderburn) Letkbe either an algebraically closed or a finite field. LetAbe a finite dimensional semi-simplek-algebra. ThenAis the direct product of matrix algebras.

Proof. By Lemma 2.3.6 and Corollary 2.3.8 we know that the only finite dimensional division algebras overkare fields. Hence, the corollary is a consequence of Corollary 2.1.27. 2 Definition 2.3.10 Letkbe a field andAak-algebra.

(a) An irreducibleA-moduleV is called absolutely irreducible (or geometrically irreducible) if for every extensionK/kthe moduleV_K =K⊗_kV is an irreducibleA_K =K⊗_kA-module.

(b) A field extensionK/kis called a splitting field ofAif every irreducibleA_K-module is absolutely irreducible.

Theorem 2.3.11 An irreducibleA-moduleV is absolutely irreducible if and only if End_A(V)∼=k,

i.e. the onlyA-endomorphisms ofV are left multiplications by an element ofk.

Proof. Since the statement is about endomorphism rings, by Exercise 6 we may assume thatV is a faithful module. This implies thatAis a simple ring withV the only simple module (Lemma 2.1.22).

Let us now first assume thatV is an absolutely irreducibleA-module. LetD = End_A(V)and let K be a field splitting field D. By Theorem 2.1.26 we know that A ∼= Matr(D^opp) and that V ∼= (D^opp)^r. For theK-dimensions we obtain

n:= dim_KV_K =rdim_kD^opp. Using Exercise 13 we have

Matr(K⊗kD^opp)∼=K⊗kMatr(D^opp)∼=K⊗kA∼=AK ∼= Matn(K).

2.3. SPLITTING FIELDS 33

TheK-dimension is thus

n²=r²dim_kD.

Comparing with the above yieldsdim_kD= (dim_kD)², whencedim_kD= 1.

Now we assume conversely thatEnd_A(V) =D=k. ThenA ∼= Matn(k)for some integern≥ 1. Hence, under this isomorphism V is isomorphic to the simple modulekⁿ, which is absolutely

irreducible by Lemma 2.1.18 (a). 2

Corollary 2.3.12 LetAbe a simplek-algebra. Its simple module is absolutely irreducible if and only ifA∼= Matr(k)for somer.

Proof. This follows from Theorem 2.3.11 and Wedderburn’s Theorem 2.1.26, which statesA ∼= Matr(D^opp)withD= End_A(S)withS the only simpleA-module. 2 Proposition 2.3.13 Letkbe a field and letDbe a finite dimensional divisionk-algebra. Then any maximal subfield ofDis a separable extension ofk.

Proof. (See [Kersten], Theorem 10.2.) Omitted. 2

Corollary 2.3.14 Letk be a field andAa finite dimensionalk-algebra. There is a finite separable splitting field ofA.

Proof. Let S_i be the finitely many simple A-modules and D_i = End_A(S_i) the corresponding division algebras. For eachDi, letKi be the field provided by Theorem 2.3.5 such that (Di)_Ki = Mat_si(K_i). The fieldK_iis finite and separable (by Proposition 2.3.13), hence the compositeKof all theK_i is also a finite separable extension ofk.

As the statement is about simple modules, we can work with B := A/Jac(A). Since K is separable,B_K =A_K/(K⊗_kJac(A))is also semi-simple by Proposition 2.2.3. We haveB =Q

iB_i and eachBiis of the formMatri(D_i^opp)by Wedderburn’s Theorem 2.1.26. Consequently,

BK =Y by Lemma 2.1.18 (d). Hence, every simple module of B_K is absolutely simple. Since the simple modules ofB_Kare the same as those ofA_K(asK⊗_kJac(A)⊆Jac(A_K)acts trivially), the corollary

follows. 2

Remark 2.3.15 We should mention that the theory we are exposing is about the Brauer group, which, however, we do not wish to define.

At the end of this section, we draw the reader’s attention to Exercise 17, in which a simple but not absolutely simple module should be exhibited.

34 CHAPTER 2. GENERAL REPRESENTATION THEORY