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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.

user26857
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math.n00b
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2 Answers2

5

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 $Y\to e_{11}Y$.)

user26857
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  • Yes, $M_n(R)$ acts on $S^n$ by matrix multiplication. – user26857 Dec 05 '14 at 12:40
  • In general, there is a bijective correspondence between the $R$-submodules of $S$ and the $M_n(R)$-submodules of $S^n$ given by $N\mapsto N^n$. – user26857 Dec 05 '14 at 13:23
  • If $X\le S^n$, isn't this obvious that $X=\pi_1(X)^n$? (Here $\pi_1:S^n\to S$ is the projection on the first component.) For instance, $x=(x_1,\dots,x_n)\in X$ can be written as $(\pi_1(x),\pi_1(e_{12}x),\dots,\pi_1(e_{1n}x))$ where $e_{ij}$ is the matrix having $1$ in the $(i,j)$ entry and $0$ otherwise. Conversely, if $x_i=\pi_1(X_i)$ with $X_i\in X$, then $(x_1,\dots,x_n)=e_{11}X_1+\cdots+e_{n1}X_n\in X$. – user26857 Dec 05 '14 at 14:38
  • Oh! I can't believe it was this easy and I didn't see it on my own. Thanks.. (+1). The second part becomes immediately clear with this description now. How should we proceed now? I mean now we have established a one-to-one correspondence between $R$-submodules of $S$ and $M_n(R)$-submodules of $S^n$, but we need simple right/left $R$-modules and simple right/left $M_n(R)$ submodules. How do we proceed now? – math.n00b Dec 05 '14 at 15:57
  • I did, but we've found $M_n(R)$-submodules of $S^n$. We need to find the simple ones. No? – math.n00b Dec 05 '14 at 16:56
  • @math.n00b I've added more details to my answer. – user26857 Dec 05 '14 at 17:03
  • Yeah, you're right. I should've explained that my level is basic. I didn't think this question was so tricky, I thought it was much easier than this. About your edit, I still don't get it. Does this bijective correspondence send simple $R$-modules to simple $M_n($)-modules? If yes, why? – math.n00b Dec 05 '14 at 17:09
  • Of course! For $S$ simple implies $S^n$ simple. (In fact, the correspondence shows that every simple $M_n(R)$-module is (isomorphic to) $S^n$ for some simple $R$-module $S$.) – user26857 Dec 05 '14 at 17:13
  • Isn't $S\oplus 0 \oplus \cdots \oplus 0$ an R-"sub-module" of $S^n$?! (sorry, I corrected it now) – math.n00b Dec 05 '14 at 17:21
  • Oh! I see it now.. Thank you for your patience.. So, the fact that $S^n$ is simple as a $M_n(R)$ module comes from what? I don't think it's immediately clear. Is it? – math.n00b Dec 05 '14 at 17:24
  • $S$ is simple, as a left $R$-module. But you said, few comments above that "For $S$ simple implies $S^n$ simple.". What did you mean by this? – math.n00b Dec 05 '14 at 17:28
  • Please read that as follows: $S$ $R$-simple implies $S^n$ $M_n(R)$-simple. (How could be a direct sum of modules a simple module???) – user26857 Dec 05 '14 at 17:29
  • Yes, that is the claim #1 that I understood its proof. Now I see why this implies that the correspondence between $R$-submodules of $S$ and $M_n(R)$-submodules of $S^n$ send simple $R$-modules to simple $M_n(R)$-modules.. Thanks.. Everything is clear now I guess. I'd write an expanded answer on this thing here and make it community wiki for future reference. Please check if I've finally got it correctly. – math.n00b Dec 05 '14 at 17:33
  • Now please pay attention: the correspondence in 3. isn't that obvious. You have to prove that $e_{11}X^n\simeq X$ for any $R$-module $X$ (and this is easy, indeed), and $(e_{11}Y)^n\simeq Y$ for any $M_n(R)$-module $Y$, but this is not so easy. – user26857 Dec 05 '14 at 17:33
  • Please check if my explanations are correct so far. – math.n00b Dec 05 '14 at 17:55
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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:

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

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

user26857
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math.n00b
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  • @user26857: Is it correct so far? – math.n00b Dec 05 '14 at 17:53
  • Because I have to leave for a while, let me give you the promised link (containing an alternative proof): http://books.google.ro/books?id=2T5DAAAAQBAJ&pg=PA57&lpg=PA57&dq=Vn+simple+left+Mn%28R%29-module&source=bl&ots=PVqAcvGGgO&sig=a4UXfNoSdTHkDy7tMIbc4msiVIY&hl=en&sa=X&ei=3ayBVIDuBMrEPbv_gEg&ved=0CCsQ6AEwAw#v=onepage&q&f=false – user26857 Dec 05 '14 at 18:00
  • Thanks a lot for all your help. I'll try to complete my answer by then. Please check it later when you're back. Have a nice time. – math.n00b Dec 05 '14 at 18:15