Neither commutative rings nor division rings are viable approaches to studying rings of matrices. However, there is a very cool notion of a reversible ring, which looks like it can fill this void. I have a few basic questions, but first, here's a little information for the interested reader.

Prelude. Call a ring reversible iff $$xy = 0 \implies yx = 0.$$

Therefore: in any reversible ring, left zero divisors and right zero divisors coincide. We may therefore simply speak of "zero divisors," without qualification.

Furthermore, it is proved here that in any reversible ring with unity, we have $$xy = 1 \implies yx = 1.$$

Thus: in any reversible ring with unity, left units and and right units coincide. We may therefore speak of "units," without qualification.

Now it is easy to see the following.

  1. If $x$ is a zero divisor, then so too are $xy$ and $yx$.
  2. If $x$ and $y$ are units, then so too is $xy$.
  3. No element of a reversible ring with unity is both a zero divisor and a unit.

Questions. It seems reasonable that for all fields $F$ and all $n \in \mathbb{N},$ the following hold.

  1. The ring of $n \times n$ matrices over $F$ is reversible.

  2. Furthermore, every element of this ring is either a zero divisor, or a unit.

Any ideas how to prove these statements?

I also suspect that there exists a reversible ring with an element that is neither a zero divisor, nor a unit. Is this true? Assuming it is, is there a special name for rings (like those of matrices over a field) that are partitioned by the zero divisors and the units? I think this is a very cool variation on the notion of a division ring, since it includes matrix rings.

  • $\begingroup$ You have no quantifiers. Is it, for some $x$ and for all $y$? or both for all? where is the $x$ and $y$ anyways?? $\endgroup$ – Don Larynx Nov 8 '13 at 12:34
  • $\begingroup$ @DonLarynx its for all $x$ and all $y$ in the ring, $xy = 0 \implies yx = 0$. $\endgroup$ – goblin Nov 8 '13 at 12:36

Matrix rings $M_n(\mathbb{F})$ over any field $\mathbb{F}$ (for $n \geq 2$) are not reversible. To see this, note that $$ \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}, $$ but $$ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, $$

  • $\begingroup$ Oh wow. You're totally right. $\endgroup$ – goblin Nov 8 '13 at 12:38
  • $\begingroup$ Tyler, do you know whether its still true that for all $x$ in a matrix ring, if $xy=0$ for some non-zero $y$, then $yx = 0$ for some non-zero $y$? $\endgroup$ – goblin Nov 8 '13 at 12:40
  • $\begingroup$ I'm not sure if this is what you are asking, but if $xy = 0$ for all $x$, then in particular it is true for $x = I_n$ (the identity matrix). That is, $y = 0$. $\endgroup$ – tylerc0816 Nov 8 '13 at 12:41
  • $\begingroup$ No I mean is it true that for all $x$ in a matrix ring, we have that (if $xy = 0$ for some non-zero $y$, then $yx = 0$ for some non-zero $y$). $\endgroup$ – goblin Nov 8 '13 at 12:43
  • 1
    $\begingroup$ @user18921 It's true that in any right Artinian ring (in particular matrix rings over right Artinian rings), all elements are units or zero divisors. If you are interested in reversible rings, you should really take a look at Greg Marks's paper: Reversible and symmetric rings Journal of Pure and Applied Algebra Volume 174, Issue 3, 24 October 2002, Pages 311–318 $\endgroup$ – rschwieb Nov 8 '13 at 14:53

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.