# Rewriting automorphism of matrix algebra in terms of automorphisms of the underlying ring?

I've used the following idea as a black box for some time now, but it occurred to me I don't fully understand why it's true.

Suppose $A=M_n(R)$ is the algebra of square matrices over some division ring $R$. Then for any $\phi\in\operatorname{Aut}(A)$, we can actually write $\phi$ as the composition of an automorphism induced by an automorphism $\psi$ of $R$ and the conjugation by some unit of $A$. More explicitly, for $\psi\in\operatorname{Aut}(R)$, this induces an automorphism $\tilde{\psi}$ of $A$ by applying $\psi$ to each of the entries in the matrix, for example, $$M=\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix} \mapsto \tilde{\psi}(M)\begin{pmatrix} \psi(a_{11}) & \psi(a_{12})\\ \psi(a_{21}) & \psi(a_{22}) \end{pmatrix}$$ and then we can conjugate by an invertible matrix in $A$, say $N$, to get $N\tilde{\psi}(M)N^{-1}$. I don't think the order of applying $\tilde{\psi}$ or conjugating matters, since if I conjugate first, then I could apply a different $\tilde{\psi}$. So the composition would be something like $\phi=\varphi_N\circ\tilde{\psi}$ where $\varphi_N$ is the conjugation by $N$ map.

My question is, why can any automorphism $\phi$ of $A$ actually be decomposed in this way?

• I'm still trying to figure out what your second paragraph means... can you formalize it a bit more please :) Commented Apr 29, 2012 at 1:20
• @Nastassja: It think it should say "conjugation by" instead of "conjugation of"? By the automorphism induced by some $\psi\in\operatorname{Aut}(R)$ I presume you mean the componentwise application of $\psi$? Commented Apr 29, 2012 at 4:41
• @rschwieb Sorry, I admit I'm having a somewhat hard time expressing what I mean :(. I will try to fix it. Commented Apr 30, 2012 at 4:17
• In your statement about your "gut feeling" you never use $\rho$. What did you intend to say? Commented Apr 30, 2012 at 4:59
• @Nastassja: if $\rho$ is an automorphism of $A$ then it isn't an element of $A$, and I'm not sure what you mean by conjugation here. Perhaps you mean the following: giving $N$ an $A$-module structure means specifying a ring map $A \to \text{End}(N)$, and any endomorphism $\rho : A \to A$ (no invertibility necessary) defines a new ring map $A \to A \to \text{End}(N)$ by composition. Commented Apr 30, 2012 at 5:57

A special case: Let $\phi:M_n(R)\to M_n(R)$ be an automorphism. It restricts to an automorphism of the center of $M_n(R)$, which is the same as the center $K=Z(R)$ of $R$, which is a field. If we suppose that this restriction $\phi|_K$ is the identity and that $R$ is finite, then the Nother-Skolem theorem tells us that $\phi$ is inner, that is, by conjugation by an invertible element of $M_n(R)$.

The general case is stated in Algebra IX: Finite Groups of Lie Type, Finite-dimensional Division Algebras, by A. I. Kostrikin and I. R. Shafarevich, in chapter II, section 3. They see $M_n(R)$ as an algebra over a field and look for automorphisms which are algebra automorphisms: but you can always take the ground field to be the prime field of the center of $R$.

• Thanks Mariano. I found a copy of Algebra IX, and Chapter II, Section 3 is about the Brauer Complex. Are you referring to section 2 on Semisimple Conjugacy Classes of $G^F$? Sorry if I'm wrong, I don't understand much in this book. Commented May 2, 2012 at 21:54

I think what you're describing is basically the contents of the Skolem-Noether theorem which states that a central simple algebra $A$ with center $Z$ has the property that every $Z$-algebra automorphism is inner.

So in the case of a matrix ring over a field $\mathbb{F}$, all $F$-linear automorphisms are inner.

However if your automorphism may not be $R$-linear and it may not be central if you use a division ring $R$, that is, the map doesn't fix $R$. Here you could say: "Well, $\phi(R)$ is isomorphic to $R$, so why don't I just identify $R$ with $\phi(R)$ and pretend the original map is $R$ linear?"

This sounds a little bit like what you described, although it's beyond what I've seen stated with the Skolem-Noether theorem.

• The quoted text is sleep-deprived speculation. While central simple algebras have only inner automorphisms, I doubt you can say the same for general ring automorphisms. Commented May 2, 2012 at 13:57

Suppose $$D$$ is a division ring, and $$\varphi$$ is a ring automorphism of $$M_n(D)$$. As usual, identify $$D$$ with the subring of scalar matrices in $$M_n(D)$$. Note that every automorphism $$\psi$$ of $$D$$ extends to an automorphism $$\psi\colon(d_{ij})\mapsto(\psi(d_{ij}))$$ of $$M_n(D)$$. Then there exists a matrix $$A\in M_n(D)$$ and an automorphism $$\psi$$ of $$D$$ such that $$\varphi(B)=A\psi(B)A^{-1}$$ for every $$B\in M_n(D)$$.

Elementary linear algebra proof: Identify $$M_n(D)=\rm{End}_D(V)$$, where $$V=D^n$$ is the right $$D$$ vector space of column vectors, and $$M_n(D)$$ acts by matrix multiplication from the left. Let $$\{e_i\}\subset V$$ and $$\{e_{ij}\}\subset M_n(D)$$ denote the standard bases, so that $$e_{ij}e_k=\delta_{jk}e_i$$ (Kronecker delta). Fix $$u\in V$$ such that $$v_1=\varphi(e_{11})u\neq 0$$, and set $$v_i=\varphi(e_{i1})v_1$$ for $$2\leq i\leq n$$. Since $$\varphi(e_{1i})v_i=v_1$$, $$v_i\neq 0$$. Then since $$\varphi(e_{ii})v_i=v_i$$ for all $$i$$ and $$\varphi(e_{jj})v_i=0$$ for all $$j\neq i$$, $$\{v_i\}$$ is a basis for $$V$$. Define $$A\in M_n(D)$$ by $$Ae_i=v_i$$. For every $$i,j,k$$ we have $$\varphi(e_{ij})Ae_k=\varphi(e_{ij})v_k=\varphi(e_{ij}e_{k1})v_1 =\delta_{jk}v_i=Ae_{ij}e_k,$$ and hence $$\varphi(e_{ij})A=Ae_{ij}$$. Conclude that $$\varphi(e_{ij})=Ae_{ij}A^{-1}$$ for all $$i,j$$.

Now suppose $$d\in D\subset M_n(D)$$. Then $$\varphi(d)\varphi(e_{ij})=\varphi(e_{ij})\varphi(d)$$, and hence $$(A^{-1}\varphi(d)A)e_{ij}=e_{ij}(A^{-1}\varphi(d)A)\quad \text{for all}\ i,j.$$ Conclude that $$A^{-1}\varphi(d)A\in D$$ is a scalar matrix, and hence that $$\psi\colon d\mapsto A^{-1}\varphi(d)A$$ is a well defined automorphism of $$D$$.

We've verified that $$\varphi(B)=A\psi(B)A^{-1}$$ for $$B=e_{ij}$$ and for $$B\in D$$. Since $$(d_{ij})=\sum d_{ij}e_{ij}$$ for every $$(d_{ij})\in M_n(D)$$, this establishes the claim.