Let $Q$ be a division ring, and let $M_n(Q)$ be the ring of $n \times n$ matrices over $Q$. If $Q' \subset M_n(Q)$ is the subring of scalar matrices, is it generally true that every ring automorphism of $M_n(Q)$ restricts to an automorphism of $Q'$? If $Q$ is a field, the answer is yes, for then $Q'$ is merely the center of $M_n(Q)$. But I'm not sure where to go when commutativity of $Q$ is taken away.

  • 2
    $\begingroup$ When you say "diagonal matrices," you really appear to have scalar matrices in mind. $\endgroup$
    – user49640
    Commented Feb 19, 2017 at 7:03
  • $\begingroup$ I had the same thought. I think you mean constant diagonal matrices. Because the ring of diagonal matrices is not preserved even if the scalars are a field. $\endgroup$
    – rschwieb
    Commented Feb 19, 2017 at 9:05
  • $\begingroup$ @user49640 Yeah, I realized that a bit too late. Edited. $\endgroup$
    – Feryll
    Commented Feb 19, 2017 at 13:30

1 Answer 1


No. The subring of scalar matrices is not automorphism-invariant. In fact, the subgroup of $\mathrm{GL}(n,R)$ of scalar matrices may not be inner-automorphism-invariant, i.e. normal.

For example, let $Z$ be the subgroup of $G=\mathrm{GL}(2,\mathbb{H})$ of scalar quaternionic matrices. Given an arbitrary $\lambda\in Z$ we can explicitly compute $N_{\lambda}=\{g\in G:g\lambda g^{-1}\in Z\}$ (which is a subset but not a subgroup). Indeed, writing out $g\lambda =\mu g$ gives

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} =\begin{pmatrix} \mu & 0 \\ 0 & \mu \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$

$$ \iff \begin{pmatrix} a\lambda & b\lambda \\ c\lambda & d\lambda \end{pmatrix} =\begin{pmatrix} \mu a & \mu b \\ \mu c & \mu d \end{pmatrix} $$

If all $a,b,c,d$ are nonzero this gives four equations which, after conjugating, gives $\mu=$

$$ {}^a\lambda={}^b\lambda={}^c\lambda={}^d\lambda.$$

(If any of $a,b,c,d$ are $0$ then we simply omit them from the equality. Note ${}^px:=pxp^{-1}$.)

This is, in turn, equivalent to $a,b,c,d$ being equal modulo $C_G(\lambda)$. Without loss of generality we may assume $\lambda$ is not central, so $C_G(\lambda)\cong\mathrm{GL}(2,\mathbb{C}_\lambda)$ where $\mathbb{C}_\lambda\cong\mathbb{C}$ is the complex subalgebra of $\mathbb{H}$ generated by $\lambda$ (i.e. $\mathrm{span}\{1,\lambda\}$).

Now we may compute the normalizer $N_G(Z)$. It is the intersection of all $N_\lambda$, so if $g\in N_G(Z)$ then it must be in $x\mathrm{GL}(2,\mathbb{C}_u)$ (where $\mathbb{C}_u:=\mathbb{R}\oplus\mathbb{R}u$ and $x$ is one of the nonzero entries of $g$) for all imaginary unit quaternions $u$, or in other words $x^{-1}g$ is in $\bigcap_u \mathrm{GL}(2,\mathbb{C}_u)=\mathrm{GL}(2,\mathbb{R})$. Therefore, $N_G(Z)=Z\cdot\mathrm{GL}(2,\mathbb{R})$ which is proper in $G$, hence $Z$ is not a normal subgroup.

In fact this argument generalizes to any dimension,

$$ N_{\mathrm{GL}(n,\mathbb{H})}(\mathbb{H}^\times\cdot I)=\mathbb{H}^\times\cdot\mathrm{GL}(n,\mathbb{R}). $$


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