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This question is on a sample exam and I'm struggling.

Let $R=M_3(\mathbb{R})$ be the ring of 3x3 real matrices. Determine a composition series of $R$ as

(1) an $R$-module


(2) an $\mathbb{R}$-module.

So, in each problem, we need a strictly decreasing series of submodules of $R$ such that the factors are simple.

My wording is probably going to be off, because I'm new to the subject. My apologies.

In the first problem, we consider $R$ as an $R$ module. This doesn't give us any additional structure on $R$, so we can look at this as the problem of finding a decreasing sequence of subrings such that each subsequent subring is a maximal ideal in the previous. Then the factors would be simple and we would have ourselves a compositions series of $R$ as an $R$-module.

Do the definitions of "simple" for a ring and module coincide like this?

I don't really know how to proceed with this. Something about $R$ being matrices over $\mathbb{R}$ rather than $\mathbb{Z}$ is making it hard for me to find submodules/subring/ideals.

Any suggestions?

Thank you.


I posted this without having looked at the second question. I see now that $R$ as an $\mathbb{R}$-module is a vector space, because $R$ is a field. Thus, it has a basis $\mathcal{B}=\{b_1,b_2,...,b_9\}$ and we can probably create a decreasing series of groups each of which is generated by some subset of the basis elements, a la

$$ R \supset <b_1,...,b_8> \supset <b_1,...,b_7> \supset ... \supset <b_1> \supset \{0\}. $$

I think the following is true:

Each quotient is isomorphic to $<b_1>$ which is isomorphic to the base field $\mathbb{R}$. There isn't any proper additive subgroup $S$ of $\mathbb{R}$ which is closed under multiplication by $\mathbb{R}$, because, given a nonzero element $s \in S$ and an arbitrary element $x \in \mathbb{R}$, $\exists$ an element $y \in \mathbb{R}$ such that $ys=x$. Thus, each quotient is simple and this is a composition series.

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Subrings and ideals are not the same! – Dylan Moreland Dec 15 '11 at 2:45
Yes. Where I went wrong was considering subrings at all, right? Because ideals in $R$ are $R$-modules and it makes sense to consider their factors. This is all we need. – user18297 Dec 15 '11 at 3:14
up vote 2 down vote accepted

as a vector space over $\mathbb{R}$, $M_3$ is just $\mathbb{R}^9$ and the composition series is $0,\mathbb{R},\mathbb{R}^2,...,\mathbb{R}^9$.

as a (left) module over itself $M_3$ is the direct sum of the column spaces, and a composition series is $$ 0, \left( \begin{array}{ccc} 0&0&*\\ 0&0&*\\ 0&0&*\\ \end{array} \right), \left( \begin{array}{ccc} 0&*&*\\ 0&*&*\\ 0&*&*\\ \end{array} \right), M_3 $$

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