If I have a ring automorphism $f$ on ${M_n(R)}$ $\!\oplus$${M_m(R)}$ for some field $R$ and some $n,m$, then:

  • if $n\neq m$, then is $f$ necessarily inner?
  • and if $n=m$, then would $f$ either be inner or $f(x\oplus y)=g(y\oplus x)$ for some inner ring automorphism $g$, for all $x$ and $y$?

How do I generalise this phrasing, can I say something like: $f = g\circ p$ for some inner $g$ and permutation $p$ on the indices (or something)?

Taken as fact: any ring automorphism on $M_n(R)$ is inner


1 Answer 1


The centre $Z$ of $M_n(R)\oplus M_m(R)$ is $\def\espan{\operatorname{span}}Z=\espan\{I_n\oplus 0, 0\oplus I_m\}$. Because $f$ is an automorphism, $f(Z)=Z$ (proof at the end). As $f(0)=0$ and $f(I)=I$, we either have (proof at the end) $$\tag1 f(I_n\oplus 0)=I_n\oplus 0,\qquad\qquad f(0\oplus I_m)=0\oplus I_m, $$ or $$\tag2 f(I_n\oplus 0)=0\oplus I_m,\qquad\qquad f(0\oplus I_m)=I_n\oplus 0, $$ When $(1)$, occurs, let $\tilde f:M_n(R)\to M_n(R)$ be given by $\tilde f(A)=\pi_1(f(A\oplus 0))$, where $\pi_1(A\oplus B)=A$. Then $\tilde f$ is an automorphism of $M_n(R)$. This means that there exists $V\in M_n(R)$ such that $\tilde f(A)=VAV^{-1}$ for all $A$. Similarly, $\tilde{\tilde f}:M_m(R)\to M_m(R)$, given by $\tilde{\tilde f}(B)=\pi_2(f(0\oplus B))$ is given by $B\longmapsto WBW^{-1}$. We have $\def\abajo{\\[0.2cm]}$ \begin{align} f(A\oplus 0)&=f(A\oplus 0)=f((I_n\oplus 0)(A\oplus 0))=(I_n\oplus 0)f(A\oplus 0)=\tilde f(A)\oplus 0, \end{align} and similarly $f(0\oplus B)=0\oplus \tilde{\tilde f}(B)$. Hence \begin{align} f(A\oplus B)&=f(A\oplus 0)+f(0\oplus B)=VAV^{-1}\oplus WBW^{-1}\abajo &=(V\oplus W)(A\oplus B)(V\oplus W)^{-1}, \end{align} and $f$ is inner.

When $(2)$ occurs, we have that $f(M_n(R)\oplus 0)\subset 0\oplus M_m(R)$, which by looking at the dimensions implies that $n\leq m$. And $f(0\oplus M_m(R))\subset M_n(R)\oplus 0$ implies that $m\leq n$, so $m=n$. Now we can reason in a similar way as above, to conclude that $f(A\oplus B)=g(B\oplus A)$ where $g$ is inner.

Proof that $f(Z)=Z$. This is true for any automorphism of any algebra. Let $A$ be algebra with centre $Z$ and $f$ an automorphism of $A$. Let $z\in Z$, $a\in A$. As $f$ is surjective, there exists $b$ with $a=f(b)$. Then $$f(z)a=f(z)f(b)=f(zb)=f(bz)=f(b)f(z)=af(z).$$ So $f(Z)\subset Z$. Now let $w\in Z$. Then, as $f$ is surjective, $w=f(z)$ for some $z\in A$. For any $a\in A$, $$f(za)=f(z)f(a)=wf(a)=f(a)w=f(a)f(z)=f(az).$$ as $f$ is injective, $za=az$, and so $z\in Z$. That is, $Z\subset f(Z)$. So $f(Z)=Z$.

Proof that either $(1)$ or $(2)$ holds.

Write $Z_1=I_n\oplus 0$, $Z_2=0\oplus I_m)$. We have $Z_1+Z_2=I$, $Z_1Z_2=0$. Write $$ f(Z_1)=aZ_1+bZ_2,\qquad\qquad f(Z_2)=cZ_1+dZ_2. $$ Since $f$ is an automorphism and $Z_1+Z_2=I$, $Z_1Z_2=0$, we have $$ 0=f(Z_1)f(Z_2)=acZ_1+bdZ_2,\qquad\qquad I=f(Z_1+Z_2)=(a+c)Z_1+(b+d)Z_2. $$ So $ac=bd=0$, $a+c=b+d=1$. If $a\ne0$, then $c=0$ and $a=1$; this forces $d\ne0$ (because $f(Z_2)\ne0$) and so $b=0$ and $d=1$; this gives $(1)$. And when $a=0$ then $c\ne0$ so $c=1$. The condition $a=0$ forces $b\ne0$, so $d=1$ and this is case $(2)$.

  • $\begingroup$ The point of being an automorphism is to preserve structure. I've added details to the answer. $\endgroup$ May 16 at 17:21

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