Posts Tagged ‘Skolem-Noether theorem’

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).

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Corollary 1. If k is a field, then every automorphism of M_n(k) is inner.

Proof. Let f : M_n(k) \longrightarrow M_n(k) be an automorphism. Now let R=S=M_n(k) and apply the lemma we proved in part (1)\Box

We now extend the lemma in part (1) to any finite dimensional central simple k-algebra S.

Lemma. Let A be a finite dimensional central simple k-algebra and let B be a simple k-subalgebra of A. If g: B \longrightarrow A is a k-algebra homomorphism, then there exists an invertible element v \in A such that g(b)=vbv^{-1} for all b \in B.

Proof. We proved in here that A \otimes_k A^{op} \cong M_n(k), for some integer n. Let S = A \otimes_k A^{op} and let R=B \otimes_k A^{op}. By this corollary,  R is a simple k-subalgebra of S. Now let

f = g \otimes_k \text{Id}_{A^{op}}.

So f : R \longrightarrow S is an k-algebra homomorphism and thus, by the lemma in part (1), there exists an invertible element u \in S such that f(r)=uru^{-1}, for all r \in R. Let a \in A^{op} and choose r = 1 \otimes_k a. Then

r = 1 \otimes_k a = g(1) \otimes_k \text{Id}_{A^{op}}(a) =f(1 \otimes_k a)=f(r)=uru^{-1}.

So ru=ur. Thus u is in the centralizer of 1 \otimes_k A^{op} and hence u \in A \otimes_k 1 by the second part of this lemma.  Therefore

u =v \otimes_k 1,

for some v \in A. Note that since u is invertible, v is invertible too and u^{-1}=v^{-1} \otimes_k 1. Therefore for every b \in B we have

g(b) \otimes_k 1 = f(b \otimes_k 1)=u(b \otimes_k 1)u^{-1}=vbv^{-1} \otimes_k 1.

Thus g(b)=vbv^{-1}. \ \Box

Corollary 2. (Skolem-Noether, 1929) Let A be a finite dimensional central simple k-algebra. Every automorphism of A is inner. Also, if B is a simple k-subalgebra of A and f,g : B \longrightarrow A are k-algebra homomorphisms, then there exists an invertible element v \in A such that f(b)=vg(b)v^{-1} for all b \in B.

Proof. The first part is clear from the lemma. For the second part, we know from the lemma that there exist invertible elements v_1,v_2 \in A such that f(b)=v_1bv_1^{-1} and g(b)=v_2bv_2^{-1} for all b \in B. So if we let v=v_1v_2^{-1}, then f(b)=vg(b)v^{-1} for all b \in B. \ \Box

Throughout this two-part note k is a field and A is a finite dimensional central simple k-algebra. That means, as we mentioned in Remark 1 in this post, A=M_n(D) for some finite dimensional central k-division algebra D, i.e. the center of D is k and [D:k] < \infty. We are going now to prove a result due to Skolem and Noether: every automorphism of A is inner. That means if f: A \longrightarrow A is a k-algebra isomorphism, then there exists a unit u \in A such that f(a)= uau^{-1} for all a \in A. Note that if R is a simple k-algebra and S is any k-algebra, then any nonzero k-algebra homomorphism f: R \longrightarrow S is injective. The reason is that \ker f is an ideal of R and since R is simple, we must either have \ker f=(0) or \ker f = R. Since f is nonzero, \ker f \neq R and so \ker f = (0), i.e. f is injective. In particular, if R is a finite dimensional simple k-algebra, then {\rm{End}}(R)={\rm{Aut}}(R).

Remark 1. Let R \cong M_r(D), where D is a division ring. By the Wedderburn-Artin theorem, R has only one simple R-module, say M, up to isomorphism. This M is nothing but D^r, the set of all r \times 1 vectors with entries in D. It is also true that D \cong \text{End}_R(M). Since R is semisimple, every R-module is a finite direct product of M. Thus if V is any R-module, then V \cong M^n, for some integer n.

Remark 2. Let S = M_n(k). Since M_n(k) \cong End_k(k^n), we may look at an element s \in S as both an n \times n matrix and a k-linear transformations s: k^n \longrightarrow k^n. We will use both of them in the proof of the following lemma.

Lemma. Let S=M_n(k) and suppose that R is any simple k-subalgebra of R. If f: R \longrightarrow S is a k-algebra homomorphism,  then there exists an invertible element u \in S such that f(r)=uru^{-1}, for all r \in R.

Proof. Clearly R is Artinian because it is finite dimensional over k. Let M be the unique simple R-module (see Remark 1). Let V =k^n. Clearly V is a left S-module, and hence a left R-module, where the multiplication of an element of S by an element of V is just the multiplication of an n \times n matrix by an n \times 1 vector. So by Remark 1,

V \cong M^p, \ \ \ \ \ \ \ \ \ \ \ \ (1),

for some integer p. We can also define the multiplication of an element of r \in R by an element of v \in V in this way:

r \cdot v = f(r)v. \ \ \ \ \ \ \ \ \ \ \ \ (2)

With this definition V will have a second structure as an R-module because f is a ring homomorphism. Let’s put V_1=(V,+, \cdot). So, by Remark 1,

V_1 \cong M^q, \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)

for some integer q. Note that as k-modules, V and V_1 have the same structure because if r \in k, then f(r)=r and hence by (2), r \cdot v = rv. Thus \dim_k V = \dim_k V_1. But by (1), \dim_k V = p \dim_k M and by (3), \dim_k V_1 = q \dim_k M. Hence p=q and so, as R-modules, V \cong V_1. Let u: V \longrightarrow V_1 be an R-module isomorphism. Then obviously u is also an isomorphism between k-vector spaces and so u is an invertible element of S = M_n(k) \cong End_k(V). Finally, let r \in R and v \in V. Let u(v)=v_1. Then

uru^{-1}(v_1)=u(rv)=r \cdot u(v)=r \cdot v_1 = f(r)(v_1).

So uru^{-1}=f(r). \ \Box