Posts Tagged ‘splitting field’

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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Throughout, k is a field and A is a finite dimensional central simple k-algebra of degree n.

If K is a splitting field of A, then, by definition of splitting fields, there exists a K-algebra isomorphism f: A \otimes_k K \longrightarrow M_n(K). Now let a \in A and put q(x) = \det(xI - f(a \otimes_k 1)) \in K[x], i.e. the characteristic polynomial of f(a \otimes_k 1) in M_n(K). The goal is to prove that q(x) \in k[x] and q(x) does not depend on K or f. We will then call q(x) the reduced characteristic polynomial of a.

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. We now define the map \phi_p : F[x] \longrightarrow E[x] by \phi_p(\sum_{i=0}^ra_ix^i)=\sum_{i=0}^r \phi (a_i)x^i. Clearly \phi_p is a k-algebra injective homomorphism.

Lemma 1. Let F/k and E/k be splitting fields of A with a k-algebra homomorphism \phi : F \longrightarrow E. Suppose that f : A \otimes_k F \longrightarrow M_n(F) is an F-algebra isomorphism. There exists an E-algebra isomorphism g : A \otimes_k E \longrightarrow M_n(E) such that \det(xI - g(a \otimes_k 1))=\phi_p(\det(xI-f(a \otimes_k 1))).

Proof. We will apply the notation and Theorem 2 in this post. By that theorem there exists an E-algebra isomorphismg : 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. Thus

\det(xI - g(a \otimes_k 1)) = \det(xI - \phi_m(f(a \otimes_k 1)))=\phi_p(\det(xI - f(a \otimes_k 1))). \ \Box

Lemma 2. Let K be a splitting field of A and a \in A. Let f : A \otimes_k K \longrightarrow M_n(K) be a K-algebra isomorphism and let g: A \otimes_k K \longrightarrow M_m(K) be any K-algebra homomorphism. Then m=rn for some integer r \geq 1 and \det(xI - g(a \otimes_k 1))=(\det(xI - f(a \otimes_k 1)))^r for all a \in A.

Proof. Since A \otimes_k K is simple, g is injective and so g(A \otimes_k K) \cong A \otimes_k K is a central simple K-subalgebra of M_m(K). Thus, by the theorem in this post, there exists a central simple K-algebra C such that

M_m(K) \cong g(A \otimes_k K) \otimes_K C.

Let \dim_K C = r^2. Then, since \dim_K g(A \otimes_k K) = \dim_K A \otimes_k K = \dim_k A = n^2, we’ll get from the above isomorphism that m^2 = \dim_K M_m(K) = n^2r^2 and so m=nr. Now, let’s define a map h : A \otimes_k K \longrightarrow M_m(K) by

h(s) = \begin{pmatrix} f(s) & 0 & 0 & \ldots & 0 \\ 0 & f(s) & 0 & \ldots & 0 \\ 0 & 0 & f(s) & \ldots & 0 \\ . & . & . & \ldots & . \\ . & . & . & \ldots & . \\ . & . & . & \ldots & . \\ 0 & 0 & 0 & \ldots & f(s) \end{pmatrix}, \ \ \ \ \ \ \ \ (*)

for all s \in A \otimes_k K. Note that f(s) is repeated r times in h(s) because m=nr. Clearly h is an K-algebra homomorphism because f is so.  Thus, by the Skolem-Noether theorem (see Corollary 2), there exists an invertible matrix u \in M_m(K) such that g(t)=uh(t)u^{-1}, for all t \in A \otimes_k K. Thus if a \in A, then g(a \otimes_k 1) and h(a \otimes_k 1) are similar and so their characteristic polynomials are equal. It now follows from (*) that \det(xI - g(a \otimes_k 1)) = \det(xI - h(a \otimes_k 1))=(\det(xI - f(a \otimes_k 1)))^r. \ \Box

Corollary. Let K be a splitting field of A and a \in A. If f,g : A \otimes_k K \longrightarrow M_n(K) are K-algebra isomorphisms, then \det(xI - f(a \otimes_k 1))=\det(xI - g(a \otimes_k 1)). \ \Box

To be continued in part (2).