5
$\begingroup$

I'm working on the following problem and would like some guidance

$R$ is a finite ring with $x^5 = x$ for every $x$ in $R$. Determine the structure of $R$.

My first thoughts were to factor and look at cases whether $R$ has zero divisors.
If $R$ has no zero divisors, then $x^4 = 1$ for nonzero $x$, and so $R$ is a field, namely $\mathbb{Z}/5$

Then you could look at rings of the form $\mathbb{Z}/n$ and focus on the multiplicative group. But this does not guarantee you'll find all commutative rings and it does not include non-commutative rings. So I think you have to use some ideas with the jacobson radical and semisimplicity.

Source: Fall 1996

$\endgroup$
9
  • 1
    $\begingroup$ Careful: do rings necessarily have an identity for you? $\endgroup$
    – Randall
    Aug 9 '18 at 0:43
  • 1
    $\begingroup$ good point! I guess since they didn't specify, the rings don't have to have an identity. Would that significantly change the techniques used to attack this kind of problem? $\endgroup$
    – iYOA
    Aug 9 '18 at 1:09
  • 3
    $\begingroup$ I think the most widely accepted modern convention is to assume that "rings" have an identity unless otherwise specified, in which case they are often called "rngs", I would assume an identity if I attacked this problem. And yes, I think things might be done somewhat differently in the identity-less case. $\endgroup$
    – Robert Lewis
    Aug 9 '18 at 1:16
  • $\begingroup$ What exactly do you mean by the "structure of $R$?" $\endgroup$
    – Robert Lewis
    Aug 9 '18 at 15:53
  • $\begingroup$ @Randall It turns out that the conditions in the problem imply there is an identity, although that is not completely obvious at the outset, and it would have been nice if the user specified. $\endgroup$
    – rschwieb
    Aug 10 '18 at 2:31
3
$\begingroup$

Because $xx^3x=x$ for all $x$, the ring is von Neumann regular.

Exercise: $R$ also has no nonzero nilpotent elements.

Exercise: If there are no nonzero nilpotents, that implies all idempotents are central.

Exercise: the idempotents are partially ordered by the relation $e\leq f$ iff $ef=e$.

Exercise: every two idempotents have a common upper bound . Hint: $e+f-ef$

Exercise: By finiteness, there is a unique maximal idempotent with respect to that order, and it is the identity of the ring (Hint: Let $e$ be a maximal idempotent. For arbitrary $a$, there exists $x$ such that $axa=a$. Notice that $ax$ and $xa$ are idempotents and look first at $ea=eaxa$.)

Exercise: a finite von Neumann regular ring is semisimple (having an identity speeds this along, somewhat.)

Since the ring has no nonzero nilpotents, it is a finite product of division rings. (Using Artin-Wedderburn.)

Since finite division rings are fields, it is a finite product of finite fields. (Wedderburn's little theorem.)

In each such field, since every element is a root of $x^5-x$, each of these fields can have no more than $5$ elements.

You can verify that $F_5, F_3, F_2$ all work, but $F_4$ does not.

So, now you should be able to see completely what such a ring looks like. I leave it to you to put into words.

Alternatively, you can apply a sledgehammer due to Jacobson to say very early on that $R$ is commutative, and find an identity the same way. I tried to avoid that.

$\endgroup$
6
  • 1
    $\begingroup$ You seem to suggest replacing Jacoson's "sledgehammer" with a large-ish collection of small-ish hammers! Once again you educate. Cheers! $\endgroup$
    – Robert Lewis
    Aug 10 '18 at 2:54
  • 2
    $\begingroup$ @RobertLewis I don’t think any of the lemmas or fundamental theorems I cited really count as sledgehammers. Jacobson’s theorem is quite a bit harder and not so well known. Jacobson’s theorem only substitutes for a small piece, too. It’s not like it simplifies things much. $\endgroup$
    – rschwieb
    Aug 10 '18 at 3:09
  • $\begingroup$ Regarding Jacobson's sledgehammer, can I say that because it's commutative and finite, its also finitely generated as an abelian group, so by the theorem of finitely generated abelian groups it is some direct sum of stuff of the form $\mathbb{Z}/n_i$ $\endgroup$
    – iYOA
    Aug 10 '18 at 7:13
  • $\begingroup$ @iYOA That is definitely information about the group structure, but it doesn't tell you much about the ring structure. Every finite field has an additive group structure that is a direct sum of $\mathbb Z/(p)$'s $\endgroup$
    – rschwieb
    Aug 10 '18 at 10:53
  • $\begingroup$ ah good point! So where exactly does jacobson's theorem simplify things? I'm still currently working through the steps $\endgroup$
    – iYOA
    Aug 11 '18 at 8:11

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.