Why $J(M_n(R))=M_n(J(R))$ for any ring, where $J$ is the Jacobson radical of $R$?

Question

I found this statement in an exercise:

Show that $J(M_n(R))=M_n(J(R))$ for any ring $R$.

This doesn't look like a trivial thing to me at all. The definition of Jacobson radical that my book uses defines the Jacobson radical of a ring $R$ as the intersection of its left/right ideals. Equivalently, this is the same as the intersection of all annihilators of simple right/left $R$-modules.

The problem is that I know nothing about left/right ideals of $M_n(R)$ and I'm not sure if it's possible to find all left/right ideals of $M_n(R)$ for an arbitrary ring. Is there something I'm missing?

I'm not looking for a complete solution, please give me hints first.

Answer 1

Hints.

1. If $S$ is a simple left $R$-module, then $S^n$ is a simple left $M_n(R)$-module. (Note that any $M_n(R)$-submodule of $S^n$ has the form $S_1^n$ for some $R$-submodule $S_1$ of $S$.)

2. $\operatorname{ann}_{M_n(R)}S^n=M_n(\operatorname{ann}_RS)$.

3. The map $X\to X^n$ induces a bijective correspondence between the isomorphism classes of $R$-modules, respectively $M_n(R)$-modules. (The inverse is given by $Y\to e_{11}Y$.)

Answer 2

So, I'm going to verify the first claim suggested by @user26857:

1. If $S$ is a simple left $R$-module, then $S^n$ is a simple left $M_n(R)$-module.

We're going to show that $S^n$ has no submodules.

If we think of elements of $S^n$ as column vectors with entries in $S$, then $M_n(R)$ acts on $S^n$ by matrix multiplication from left.

If $S'$ is a submodule of $S$, then $(S')^n$, that is the set of all columns vectors with entries in $S'$ is a submodule of $S^n$. This is immediately seen from matrix multiplication.

On the other hand, if $X$ is a submodule of $S^n$, then using row operations on vectors in $S^n$, we can see that all of its rows must come from the same $R$-submodule of $S$.

Now we're going to verify the second claim:

2. $\operatorname{ann}_{M_n(R)}S^n=M_n(\operatorname{ann}_RS)$

It is immediately clear from matrix multiplication that if $A$ is a matrix with entries in $\operatorname{ann}_RS$, then $AX = 0$ for all $X \in S^n$. Therefore, $M_n(\operatorname{ann}_RS) \subseteq \operatorname{ann}_{M_n(R)}S^n$

For the other direction, if $A$ is a matrix with entries in $R$, then the left multiplication by $(s,0,\dots,0), \dots ,(0,0,\dots,s)$ for all $s \in S$ show that all entries must fall in $\operatorname{ann}_RS$. Therefore $\operatorname{ann}_{M_n(R)}S^n \subseteq M_n(\operatorname{ann}_RS)$.