# automorphisms on direct sum of matrices

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

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 $$(2)$$.

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