6

Let $R$ be a finite ring such that for any $x \in R$ there exists $y \in R$ with $xyx = x$. Show that $R$ is unital and that if $ab = 1$, then $ba = 1$.

Thoughts so far: If I can show that the Jacobson radical is trivial, then I see how to get the result (since then it would be semi-simple, therefore equal to a direct product of matrix rings over division rings, hence equal to a direct product of matrix rings over fields (since the division rings would be finite), and this satisfies the property. So I'm wondering about how to prove that $J(R)$ is trivial. Any hints?

darij grinberg
  • 17,777
user19817
  • 455
  • Start from here: If $a \in J(R)$, then $1-ra$ is a unit for all $r \in R$. – moonlight Sep 02 '15 at 08:10
  • @user26857: Sorry, I misread the question. Please disregard my earlier comment. I am not too familiar with non-unital rings and don't know how to proceed along OPs original approach, but it seems to me that the usual proof that a finitely generated right ideal of a von Neumann regular ring is generated by an idempotent can be adapted (with a minor modification) to show that $R=eR$ for some idempotent $e=1$ (and, probably, along the way, to show that $R$ is semisimple). – moonlight Sep 02 '15 at 12:38
  • 4
    One more note: In fact $ab=1$ implies $ba=1$ in all unital finite rings (see here), even if the ring is not semisimple. So it really only is the existence of the unit for which one needs the vNr property. – moonlight Sep 02 '15 at 12:53
  • It holds for rings (with identity) satisfying much looser chain conditions. It holds for right Noetherian rings, and even rings satisfying the ACC on annihilators. A ring with the property that $xy=1$ implies $yx=1$ is called a Dedekind finite ring (or directly finite ring.) (Thought you might find this of interest @moonlight ) – rschwieb Sep 03 '15 at 14:39
  • @rschwieb: Thanks! I also noticed that when I went through the proof sketched above. But I wonder if there is an easier proof in the finite case? Otherwise the assumption for $R$ to be finite is a bit puzzling to me. – moonlight Sep 03 '15 at 14:41
  • @moonlight Yeah, finiteness is certainly a lot easier to use than the chain conditions. My guess is that it is simply for simplicity. Perhaps there is a clever trick in conjunction with VNR too. – rschwieb Sep 03 '15 at 15:27
  • @moonlight Let me know if you see any big simplification (or hole) in the answer I attempted. I could see no way around dealing directly with the idempotents. – rschwieb Sep 10 '15 at 19:01

1 Answers1

5

Lemma 1

Each principal left ideal is generated by an idempotent. Proof: Consider $Ra$ for some $a\in R$. Then $a=axa$, and $xa$ is idempotent. Finally, $Ra=Raxa\subseteq Rxa\subseteq Ra$. Thus $Ra=Rxa$.

Lemma 2

The sum of two principal left ideals is again principal. Proof: Consider $Re+Rf$ where $e,f$ are idempotents for the two principal left ideals found via lemma 1. Since $R(f-fe)$ is principal, it is equal to $Rg$ for some idempotent $g$. Furthermore, $ge=gge\in Rge=R(f-fe)e=0$.

Claim 1: $Re+Rg=R(e+g)$. Both $e=ee$ and $g=gg$ are in the left hand side, so $e+g$ is in the left hand side, proving $\supseteq$. On the other hand, $g=g(e+g)$ and $e=e(e+g)-eg(e+g)$, proving that both $Re$ and $Rg$ and $Re+Rg\subseteq R(e+g)$.

Claim 2: $Re+Rg=Re+Rf$. Recall that we set $Rg=R(f-fe)$. Now $f-fe=ff-fe\in Rf+Re$ proves $Rg\subseteq Re+Rf$, and of course $Re\subseteq Re+Rf$, so we have achieved $Re+Rg\subseteq Re+Rf$. In the other direction, $Rf \subseteq Re+Rg$ since $f = \underbrace{f-fe}_{\in R(f-fe) = Rg} + \underbrace{f}_{\in R}e \in Rg+Re = Re+Rg$, and thus $Re+Rf \subseteq Re+Rg$.

The two claims combine to prove that $Re+Rf$ is principal. By induction we conclude that in fact any finitely generated left ideal is principal.

Lemma 3:

By symmetry, the sum of two principal right ideals is principal, and finitely generated right ideals are principal.

Proof that $R$ has identity

Since $R$ is finite, it is finitely generated as a left $R$-module (because the assumption $x = xyx$ yields $x \in Rx$ for each $x \in R$). There is therefore some idempotent $e$ such that $R=Re$. Then for every $r\in R$, $r=se$ is unchanged by multiplication on the right by $e$. This shows $e$ is a right identity. By Lemma 3 and symmetry, there is a left identity $e'$. But $e'=e'e=e$, so it turns out that $e$ is the identity of $R$.

Finite rings are Dedekind finite

Now that the problem about existence of identity has been solved, the Dedekind finiteness follows as in other posts. If $ab=1$, $b$ creates an injective map of $R$ into itself by left multiplication, but this must also be surjective, etc.

darij grinberg
  • 17,777
rschwieb
  • 153,510