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This is a question I posted in this topic: Pathologies in "rng". However, I decided that the question deserves its own thread, and I want to know if anybody can answer it. For terminologies, please read my reply in that link.

Let $R$ be a (not necessarily unital) ring and $n\in\mathbb{N}$. What are all left ideals and two-sided ideals of the ring of matrices $S:=\mathrm{Mat}_{n\times n}(R)$?

Assume that $R$ is unital. Then, there is a one-to-one correspondence between the set $\mathcal{L}\left(R^n\right)$ of all left $R$-submodules of the unitary $R$-module $R^n$ and the set $\mathcal{L}(S)$ of all left ideals of $S$ which associates each $V\in\mathcal{L}\left(R^n\right)$ with the left ideal of $S$ consisting of matrices $\left[v_{i,j}\right]_{i,j\in[n]}$, where $[n]:=\{1,2,\ldots,n\}$, and $\left(v_{i,1},v_{i,2},\ldots,v_{i,n}\right) \in V$ for every $i\in[n]$. There is also a one-to-one correspondence between the set $\mathcal{T}(R)$ of two-sided ideal of $R$ and the set $\mathcal{T}(S)$ of two-sided ideals of $S$. This correspondence associates $I\in\mathcal{T}(R)$ with $\text{Mat}_{n\times n}(I)$.

Now, what happens if $R$ is nonunital? I don't know the answer, and I suspect that it is an open question.

EDIT: When $R$ is a trivial ring (i.e., an additive abelian group $R$ with the trivial multiplication: $r\cdot s:=0_R$ for all $r,s\in R$), we have a somewhat good characterization of left ideals and two-sided ideals of $\text{Mat}_{n\times n}(R)$.

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  • Well, in the case of a trivial ring, also the matrix ring has trivial multiplication. – egreg Jun 23 '15 at 19:51
  • Yes, I know that it means that left ideals and two-sided ideals of $S$ would be all additive abelian subgroups of $S$. I was seeking a description through $R$, if you understand what I mean. – Batominovski Jun 23 '15 at 19:53
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    Since the ring structure on $R$ bears no information, … What you can say is that in the case $R$ is trivial, then $M_n(R)=R^{n^2}$ as abelian groups (which has a rather complicated structure, if $R$ is not an “easy” abelian group). – egreg Jun 23 '15 at 19:54
  • You are absolutely right. I guess I was trying to make some analogy to the unital case, and see if the unital closure $\hat{R}=\mathbb{Z}\oplus R$ has anything to do with the answer (which it sort of does, as abelian subgroups of $R^{n^2}$ are left $\hat{R}$-submodules of $R^{n^2}$). Unfortunately, I don't think knowing the answer for trivial rings help crack the general case. Thanks. – Batominovski Jun 23 '15 at 20:01

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