## Reduced characteristic polynomials (1)

Posted: October 1, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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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 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 $a \in A.$ 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).

## Automorphisms of central simple algebras (2)

Posted: January 26, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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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 $b \in B. \ \Box$

## Automorphisms of central simple algebras (1)

Posted: January 26, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , ,

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 $uru^{-1}=f(r). \ \Box$