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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?

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  • $\begingroup$ I'm still trying to figure out what your second paragraph means... can you formalize it a bit more please :) $\endgroup$
    – rschwieb
    Commented Apr 29, 2012 at 1:20
  • $\begingroup$ @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 $\psi$? $\endgroup$
    – joriki
    Commented Apr 29, 2012 at 4:41
  • $\begingroup$ @rschwieb Sorry, I admit I'm having a somewhat hard time expressing what I mean :(. I will try to fix it. $\endgroup$
    – Nastassja
    Commented Apr 30, 2012 at 4:17
  • $\begingroup$ In your statement about your "gut feeling" you never use $\rho$. What did you intend to say? $\endgroup$ Commented Apr 30, 2012 at 4:59
  • $\begingroup$ @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. $\endgroup$ Commented Apr 30, 2012 at 5:57

3 Answers 3

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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 $R$.

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  • $\begingroup$ 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. $\endgroup$
    – Nastassja
    Commented May 2, 2012 at 21:54
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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.

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  • $\begingroup$ 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. $\endgroup$
    – rschwieb
    Commented May 2, 2012 at 13:57
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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.

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