$\mathbf{Z}(M_n(F))\cong F$ for all fields $F$

Question

I'm studying Isaacs' Algebra: A Graduate Course. This is Exercise 12.17 in the book:

If $R$ is a simple ring with unity, show that $\mathbf{Z}(R)$ is a field. In particular, if $R=M_n(F)$, where $F$ is a field, show that $\mathbf{Z}(R)\cong F$.

I solved this exercise. For reference, here's my solution for the second part:

Solution for second part:

Let $R=M_n(F)$ and let $z\in\mathbf{Z}(R)$. The goal is to show that $z$ is a scalar matrix ($z=\lambda I$ for some $\lambda\in F$).

Let $e_{ij}$ denote the matrix wherein the $(i,j)$ entry is $1$ and every other entry is $0$. Then we can write $z=\sum_{k,\ell}z_{k\ell}e_{k\ell}$ for some $z_{k\ell}\in F$.

Comparing coefficients on both sides of the equation $ze_{ij}=e_{ij}z$, we get:

$z_{ki}\delta_{\ell j}=z_{j\ell}\delta_{ik}$ for all $i,j,k,\ell$.

Setting $i=k, j=\ell$, we get $z_{ii}=z_{jj}$ for all $i,j$. Let $\lambda=z_{ii}$.

Setting $i=k$, we get that for all $j\neq\ell$, $z_{j\ell}=0$. So $z$ has no off-diagonal elements.

We can conclude that $z=\lambda I$. $\blacksquare$

Question: Is there a more "conceptual" proof of this? Here are some thoughts:

If $F$ is algebraically closed, then $K=\mathbf{Z}(R)$ is a field that is also a finite-dimensional vector space over $F$. It follows that $K$ is an algebraic extension of $F$. Since $F$ is algebraically closed, this means $K\cong F$.

Alternatively, if $F$ is algebraically closed and $0\neq z\in\mathbf{Z}(R)$, then $z$ has a non-zero eigenvalue $\lambda\in F$. This means that $z-\lambda I$ is not invertible. But $z-\lambda I$ is an element of $\mathbf{Z}(R)$, which is a field, so $z-\lambda I=0$, and $z=\lambda I$.

Answer 1

Once you prove that $F^n$ is a simple module over $M_n(F)$, it follows by Schur's lemma that $Z(M_n(F))$ is a division algebra over $F$, which solves the problem if $F$ is algebraically closed. But $F^n$ is not only simple, it is absolutely simple; that is, it is simple after extension of scalars to the algebraic closure. As a general fact, if $A$ is an $F$-algebra we have

$$Z(A) \otimes_F \overline{F} \cong Z(A \otimes_F \overline{F})$$

so $Z(M_n(F))$ is a division $F$-algebra which, after extension of scalars to $\overline{F}$, is isomorphic to $\overline{F}$, and hence which must itself be isomorphic to $F$. More generally, if $M$ is a simple module of an $F$-algebra, then $\text{End}(M) \cong F$ iff $M$ is absolutely simple.

Alternatively, and more generally, you can develop some of the basic theory of Morita equivalence, and show that

1. the center of a ring is a Morita invariant (e.g. because it can be defined directly in terms of $\text{Mod}(R)$ as the ring of endomorphisms of the identity functor), and

2. $R$ is Morita equivalent to $M_n(R)$.

This proves conceptually the more general fact that $Z(M_n(R)) \cong Z(R)$.

Answer 2

If $A$, $B$ commute, then $B v = 0$ implies $B A v = A Bv=A 0 =0$. Hence $A$ invariates the kernel of $B$. Now if $B$ can be anything, so can its kernel. Therefore $A$ invariates every subspace ( $1$ dimensional is enough). From here to prove that $A$ is a scalar operator is not far.