8
$\begingroup$

On my algebra final exam, there was a problem that essentially asked the following:

Let $R$ be a commutative ring such that for all $r\in R$, there exists $n_r\in\Bbb{Z}^{>1}$ with $r^{n_r} = r$. Prove that all prime ideals are maximal.

The solution which I believe was desired goes like this:

Let $\mathfrak{p}\subseteq R$ be prime. Take $a\in R\setminus\mathfrak{p}$, and consider $a + \mathfrak{p} = a^{n_a} + \mathfrak{p}\in R/\mathfrak{p}$. As a ring mod a prime ideal is an integral domain has no zero divisors, we can cancel. Assuming that $R$ has unity, we find $1 + \mathfrak{p} = a^{n_a - 1} + \mathfrak{p}$, and since $n_a > 1$, we can write $$ a\cdot a^{n_a - 2} + \mathfrak{p} = \left(a + \mathfrak{p}\right)\left(a^{n_a - 2} + \mathfrak{p}\right) = 1 + \mathfrak{p}, $$ so we have found an inverse for any nonzero element in $R/\mathfrak{p}$. Since $R/\mathfrak{p}$ is commutative, $R/\mathfrak{p}$ is a field, and hence $\mathfrak{p}$ is maximal.

I had a problem with the cancellation step. It seems to require $R$ that have unity, whereas the problem statement does not require $R$ to have unity. I think this was a misstatement on my professor's part, but I cannot seem to find a counterexample. It isn't too much trouble to find an $R$ (without unity) with the property that for all $ r\in R$, there exists $ n_r\in\Bbb{Z}^{> 1}$ such that $r^{n_r} = r$: take for example, the subring of $\left(\Bbb{Z}/p\Bbb{Z}\right)^{\Bbb{N}}$ (countably infinite product of $\Bbb{Z}/p\Bbb{Z}$'s) where all but finitely many entries in a "vector" are nonzero. It's easy to see that this ring does not have unity; however, it still satisfies the property that every prime ideal is maximal. I tried to come up with a genuine counterexample, but I couldn't find one. My idea was to modify the example above by considering an infinite product of some integral domain $\mathcal{O}$ (not a field) where $a^{n_a} = a$ for some $n_a > 1$ for each $a\in\mathcal{O}$, but I couldn't find such an $\mathcal{O}$. So long story short, my question is:

Is there a counterexample to the original claim when $R$ does not have $1$?

$\endgroup$
  • $\begingroup$ Dear Stahl, A small remark: I don't think you can find an integral domain that is not a field with the property you want, since by the problem you solved, any such integral domain is in fact a field. Regards, $\endgroup$ – Matt E May 9 '13 at 18:46
  • $\begingroup$ @MattE: excellent point, I missed that when trying to create a counterexample. That shows that if a counterexample exists at all, it won't be of the form I was thinking about before. $\endgroup$ – Stahl May 9 '13 at 18:50
  • $\begingroup$ @Stahl: Rings are almost assumed to be unital, unless otherwise stated. Especially when dealing with prime ideals, maximal ideals (which behave quite badly in the non-unital case, in fact the usual definitions have to replaced by better ones). $\endgroup$ – Martin Brandenburg May 9 '13 at 19:15
  • 1
    $\begingroup$ @MartinBrandenburg I realize this; however, the professor made it semi-explicit that we were not adhering to the convention of rings being assumed to be unital throughout the course. $\endgroup$ – Stahl May 9 '13 at 19:20
  • $\begingroup$ How did you define prime and maximal ideals then? $\endgroup$ – Martin Brandenburg May 9 '13 at 20:22
7
$\begingroup$

Quotienting out by a prime ideal in the candidate ring $R$, we reduce ourselves to the following question:

If $R$ is a commutative ring with no zero divisors for which $r^{n_r} = r$ for some $n_r > 1$ and all $r \in R$, then is $R$ simple?

Choose $r \neq 0$ in $R$, and let $s$ be any other element of $R$. Then $r^{n_r} s = r s,$ and so cancelling $r$ from both sides (possible since $r$ is non-zero), we have $r^{n_r -1} s = s$. Here $s$ is arbitrary, and so in fact we find that $R$ admits a unit, namely $r^{n_r - 1}$. Thus $R$ is actually a field (by the argument in the OP) and so is simple.

So there is no counterexample.

$\endgroup$
  • 1
    $\begingroup$ I cannot remember having ever seen any result concerning prime ideals in commutative rngs that was this interesting. Nice! $\endgroup$ – rschwieb May 9 '13 at 19:44

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.