# On the full rings of matrices

Let $R$ be a ring with elements $a,b$ such that $a^2 = 0 = b^2$ and $a+b$ is a unit. How to show that $R$ is a ring of $2\times2$ matrices?

COMMENT: The converse is clearly true, just take $a = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$ and $b = \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}$. So it would be interesting if some one could generalize this to dimension $n$.

• @StevenLandsburg I don't understand why you put this on hold !? – user40768 Oct 5 '13 at 6:06
• @andreyRekalo I don't understand why you put this on hold !? – user40768 Oct 5 '13 at 6:08
• @StevenLandsburg Was that really unclear to you that I meant non-trivial matrices? Even in that case you could simply leave a note or even edit my post. – user40768 Oct 5 '13 at 19:03
• @StevenLandsburg Dear Steven, I totally agree with user40768. I think closing a post needs more responsibility. – user39321 Oct 5 '13 at 19:12
• The original wording was "Let $R$ be a ring with elements $a,b$ such that $a^2 = 0 = b^2$ and $a+b$ is a unit. How to show that $R$ is a ring of matrices?" The missing assumption that the dimension is 2x2 and the phrasing "how to show" with no motivation probably gave the impression that this is a HW problem. I do think there is some onus on the OP to formulate a question well and that it was ok to close on these grounds. If there is a subsequent edit improving the question, then it is fine to reopen (as it seems will happen here). This is how it is supposed to work, IMHO. – Benjamin Steinberg Oct 7 '13 at 18:15

## 1 Answer

Let $c$ denote the inverse of $a+b$, i.e., $ac+bc=ca+cb=1$. Multiplying this by $a$ and $b$ from the left and from the right and using $a^2=b^2=0$, we obtain $abc=a$, $cba=a$, and $cab=b$.

Let us show that $R=aR\oplus bR$. It follows from $ac+bc=1$ that $R=aR+bR$. If $ax=by$, then $bax=0$, implying $0=cbax=ax$ in view of $cba=a$.

The $R$-modules $aR$ and $bR$ are isomorphic. Indeed, the rule $bcx\mapsto ax$, $x\in R$, defines an $R$-module homomorphism $bR=bcR\to aR$ because $bcx=0$ implies $ax=0$ due to $abc=a$. Its kernel vanishes: $ax=0$ implies $abcx=0$ because $abc=a$ and, hence, $0=cabcx=bcx$.

It remains to apply Propositions 5 and 6 on page 52 of "Structure of Rings" by N.Jacobson (1964) and to conclude that $R$ is a ring of $2\times2$ matrices.

Edit. By suggestion of Martin Brandenburg, I reproduce the Propositions quite literally:

Proposition 5 $\dots$ Conversely, if $\frak A$ is a ring with an identity $1$ and ${\frak A}={\frak I}_1\oplus\dots\oplus{\frak I}_n$ is a direct decomposition of $\frak A$ into right ideals which are isomorphic $\frak A$-modules, then there exists a set of matrix units $\{e_{ij}\mid i,j=1,\dots,n\}$ such that ${\frak I}_j=e_{jj}{\frak A}$.

Proposition 6. Let $\{e_{ij}\mid i,j=1,\dots,n\}$ be a set of matrix units in a ring $\frak A$ with identity $1$, and $\frak B$ the subring consisting of the elements which commute with the $e_{ij}$, $i,j=1,\dots,n$. Then every element of $\frak A$ can be written in one and only one way as $\sum b_{ij}e_{ij}$ where $b_{ij}\in{\frak B}$ for all $i$ and $j$. Hence ${\frak A}\cong{\frak B}_n$. The ring $\frak B$ is isomorphic to $e_{11}{\frak A}e_{11}$.

Second thought Edit. Those Propositions are an old stuff. It is probably better to prove them here in two lines.

For any $R$-module $M$, we have $\text{End}_R(\underbrace{M\oplus\dots\oplus M}_n)\cong\text{Matr}_n(\text{End}_RM)$. On the other hand, $\text{End}_RR\cong R$.

Question. The problem looks a bit artificial. Is it a known exercise?

• Could you add the statements of these Propositions? Thanks. – Martin Brandenburg Oct 3 '13 at 19:36
• Thanks. Would you generalize to $n\times n$ matrices? – user40768 Oct 5 '13 at 6:14
• Yes, it does not look very difficult to formulate and then to prove (just following the above exposition) a similar statement concerning $n\times n$ matrices. (You are welcome to try it yourself.) – Sasha Anan'in Oct 5 '13 at 6:33
• @SashaAnan'in Please be explicit. If you mean $a^n = 0 = b^n$ and $a+b$ unit, its not true. – user39121 Oct 5 '13 at 7:16
• @SashaAnan'in I am not sure about your suggestion. – user39321 Oct 5 '13 at 19:13