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 $n$.

  • 11
    $\begingroup$ @StevenLandsburg I don't understand why you put this on hold !? $\endgroup$ – user40768 Oct 5 '13 at 6:06
  • 6
    $\begingroup$ @andreyRekalo I don't understand why you put this on hold !? $\endgroup$ – user40768 Oct 5 '13 at 6:08
  • 8
    $\begingroup$ @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. $\endgroup$ – user40768 Oct 5 '13 at 19:03
  • 6
    $\begingroup$ @StevenLandsburg Dear Steven, I totally agree with user40768. I think closing a post needs more responsibility. $\endgroup$ – user39321 Oct 5 '13 at 19:12
  • 7
    $\begingroup$ 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. $\endgroup$ – Benjamin Steinberg Oct 7 '13 at 18:15

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?

  • 2
    $\begingroup$ Could you add the statements of these Propositions? Thanks. $\endgroup$ – Martin Brandenburg Oct 3 '13 at 19:36
  • 1
    $\begingroup$ Thanks. Would you generalize to $n\times n$ matrices? $\endgroup$ – user40768 Oct 5 '13 at 6:14
  • $\begingroup$ 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.) $\endgroup$ – Sasha Anan'in Oct 5 '13 at 6:33
  • 1
    $\begingroup$ @SashaAnan'in Please be explicit. If you mean $a^n = 0 = b^n$ and $a+b$ unit, its not true. $\endgroup$ – user39121 Oct 5 '13 at 7:16
  • 2
    $\begingroup$ @SashaAnan'in I am not sure about your suggestion. $\endgroup$ – user39321 Oct 5 '13 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.