Posts Tagged ‘degree of central simple algebras’

Throughout this post, k is a field and A is a finite dimensional central simple k-algebra.

Definition. A field L is called a splitting field of A if k \subseteq L and A \otimes_k L \cong M_n(L), as L-algebras, for some integer n \geq 1.

Remark. By this lemma, the algebraic closure of k is a splitting field of any finite dimensional central simple k-algebra. Also, if f L is a splitting field of A, then \dim_k A = \dim_L A \otimes_k L = \dim_L M_n(L)=n^2 and so n = \deg A.

Theorem 1. Let A = M_n(D), where D is some finite dimensional central division k-algebra and let L be a field containing k. Then L is a splitting field of A if and only if L is a splitting field of D.

Proof. Let \dim_k D = m^2. Then \dim_k A = (mn)^2 and so \deg A = mn. If L is any field containg k, then

A \otimes_k L \cong M_n(D) \otimes_k L \cong M_n(D \otimes_k L). \ \ \ \ \ \ \ \ \ (*)

Now, suppose that L is a splitting field of A. Then A \otimes_k L \cong M_{mn}(L). Also, since D \otimes_k L is a finite dimensional central simple L-algebra, D \otimes_k L \cong M_r(D') for some division ring D' and some integer r. Therefore, by (*), we have M_{mn}(L) \cong A \otimes_k L \cong M_{rn}(D') and thus m=r and D' \cong L. Hence D \otimes_k L \cong M_m(L). Conversely, if L is a splitting field of D, then D \otimes_k L \cong M_m(L) and (*) gives us A \otimes_k L \cong M_{mn}(L). \ \Box

Corollary. There exists a finite Galois extension L/k such that L is a splitting field of A.

Proof. Trivial by Theorem 1 and the theorem in this post. \Box

Notation. Let F/k and E/k be field extensions and suppose that \phi : F \longrightarrow E is a k-algebra homomorphism. Note that, since F is a field, \phi is injective. Now, given an integer n \geq 1, we define the map \phi_m: M_n(F) \longrightarrow M_n(E) by \phi_m([u_{ij}]) = [\phi(u_{ij})] \in M_n(E). Clearly \phi_m is a k-algebra injective homomorphisms.

Theorem 2. Let F/k and E/k be field extensions and suppose that \phi : F \longrightarrow E is a k-algebra homomorphism. Suppose also that F is a spiltting field of A with an F-algebra isomorphism f : A \otimes_k F \longrightarrow M_n(F). Then E is a splitting field of A and there exists an E-algebra isomorphism g : A \otimes_k E \longrightarrow M_n(E) such that g(a \otimes_k 1) = \phi_m(f(a \otimes_k 1)) for all a \in A.

Proof. The map \phi_m: M_n(F) \longrightarrow M_n(\phi(F)) is an isomorphism and so

A \otimes_k \phi(F) \cong A \otimes_k F \cong M_n(F) \cong M_n(\phi(F)).

Therefore

A \otimes_k E \cong A \otimes_k (\phi(F) \otimes_{\phi(F)} E) \cong (A \otimes_k \phi(F)) \otimes_{\phi(F)} E \cong M_n(\phi(F)) \otimes_{\phi(F)} E \cong

M_n(\phi(F) \otimes_{\phi(F)} E) \cong M_n(E).

So we have an isomorphism g: A \otimes_k E \longrightarrow M_n(E). It is easy to see that g(a \otimes_k 1)=\phi_m(f(a \otimes_k 1)) for all a \in A. \ \Box

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For the definition of splitting fields of central simple algebras see here. Throughout, D is a division algebra with the center k and \dim_k D < \infty. For a subalgebra B of D, we will denote the centralizer of B in D by C_D(B).

Theorem. Let L be a subfield of D and suppose that k \subseteq L. Then D \otimes_k L \cong M_n(C_D(L)), where n = \dim_{C_D(L)} D = \dim_k L.

Proof. Let R = D \otimes_k L. As we saw in here, D has a structure of a (right) R-module and

End_R(D) \cong C_D(L). \ \ \ \ \ \ \ (1)

Note that R is a finite dimensional central simple L-algebra. In particular, it is primitive. We are now going to show that D is a faithful simple R-module. First, D is faithful because the annihilator of D in R is an ideal of R and, since R is a simple algebra, the annihilator is zero. Now, let I be a non-zero R-submodule of D. Let x \in D and y \in I. Then, since I is an R-module, we have yx = y(x \otimes_k 1) \in I. Thus I is a right ideal of D and so I=D. Hence D is a simple R-module. So, by (1) and the structure theorem for primitive rings, we have R \cong M_n(C_D(L)), where n = \dim_{C_D(L)} D. To complete the proof of the theorem, we only need to show that \dim_k L=n. Well, we have

\dim_k D = \dim_L R = \dim_L M_n(C_D(L))=n^2 \dim_L C_D(L)

and thus

(\dim_k L)^2(\dim_k D)=n^2 (\dim_k C_D(L))(\dim_k L)=n^2 \dim_k D. \ \ \ \ \ \ \ (2)

The last equality in (2) is true by the lemma in this post. Therefore \dim_k L = n. \ \Box

Corollary 1. If L is a maximal subfield of D, then L is a splitting subfield of D. In fact, D \otimes_k L \cong M_n(L), where n = \deg D = \dim_L D=\dim_k L.

Proof. If L is a maximal subfield, then k \subset L and C_D(L)=L. \ \Box

Corollary 2. If L is a splitting subfield of D, then L is a maximal subfield of D.

Proof. We only need to show that C_D(L)=L. By Theorem 2, D \otimes_k L \cong M_n(C_D(L)), where n = \dim_k L. Since L is a splitting subfield of D, we also have D \otimes_k L \cong M_m(L), where m = \deg D. Thus M_n(C_D(L)) \cong M_m(L). So m = n and C_D(L) \cong L. Thus \dim_k C_D(L)=\dim_k L and hence C_D(L)=L because L \subseteq C_D(L). \ \Box

Corollary 3. Let L be a subfield of D. Then the following statements are equivalent.

1) L is a maximal subfield of D.

2) L is a splitting subfield of D.

3) k \subset L and \dim_k L = \deg D.

4) C_D(L)=L.

Proof. Straightforward! \Box

As in part (1), k is a field, A is a finite dimensional central simple k-algebra and B is a simple k-subalgebra of A. We will also be using notations and the result in the lemma in part (1), i.e. R = A \otimes_k B^{op}, M is the unique simple R-module,  A \cong M^n, D = \text{End}_R(M) and C_A(B) \cong M_n(D). Finally, as usual, we will denote the center of any algebra S by Z(S). We now prove a nice relationship between dimensions.

Lemma. \dim_k C_A(B) \cdot \dim_k B = \dim_k A.

Proof. We have R \cong M_m(D) and M \cong D^m, for some integer m. Thus A \cong D^{mn}, as k-modules, and hence

\dim_k A = mn \dim_k D. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

We also have

\dim_k A \cdot \dim_k B = \dim_k R = \dim_k M_m(D)=m^2 \dim_k D \ \ \ \ \ \ \ \ \ \ \ \ \ (2)

and

\dim_k C_A(B)=\dim_k M_n(D)=n^2 \dim_k D. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)

Eliminating \dim_k D in these three identities will give us the result. \Box

Theorem. If B is a central simple k-algebra, then C_A(B) is a central simple k-algebra too and we have A =BC_A(B) \cong B \otimes_k C_A(B).

Proof. By the lemma in part (1), C_A(B) is simple and thus, since B is central simple, B \otimes_k C_A(B) is a simple algebra. Thus the map \phi : B \otimes_k C_A(B) \longrightarrow BC_A(B) defined by \phi(b \otimes_k c)=bc is a k-algebra isomorphism. Hence B \otimes_k C_A(B) \cong BC_A(B) and so \dim_k BC_A(B)=\dim_k B \otimes_k C_A(B)=\dim_k A, by the above lemma. Therefore BC_A(B)=A. Clearly if c is in the center of C_A(B), then c commutes with every element of both B and C_A(B). Hence  c commutes with every element of A and thus c \in k. So Z(C_A(B))=k, i.e. C_A(B) is a central simple k-algebra. \Box

The Double Centralizer Theorem. C_A(C_A(B))=B.

Proof. Obviously B \subseteq C_A(C_A(B)). So, to prove that B=C_A(C_A(B)), we only need to show that \dim_k B = \dim_k C_A(C_A(B)).  Applying the above lemma to C_A(B) gives us

\dim_k C_A(C_A(B)) \cdot \dim_k C_A(B) = \dim_k A. \ \ \ \ \ \ \ \ \ \ \ \ \ (*)

Plugging \dim_k C_A(B)=\frac{\dim_k A}{\dim_k B}, which is true by the above lemma, into (*) finishes the proof. \Box

Corollary. If B is a subfield of A, then \deg A = (\deg C_A(B))(\dim_k B).

Proof. We first need to notice a couple of things. First, since B is commutative, B \subseteq C_A(B). Since B is commutative, C_A(C_A(B))=Z(C_A(B)). Thus, by the above theorem, Z(C_A(B))=B and so C_A(B) is a central simple B-algebra. Now, by the lemma

(\deg A)^2=\dim_k A = (\dim_B C_A(B))(\dim_k B)^2=(\deg C_A(B))^2 (\dim_k B)^2. \ \Box