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I was recently reading a post on MSE which had an argument like:

If $P$ is a prime ideal of a ring $R$ all of whose elements satisfy $x^n = x$, then $R/P$ is an integral domain with the same property. This implies that $R/P$ is a finite integral domain, hence a finite field $\mathbb{F}_q$. Hence all prime ideals of $R$ are maximal (so $R$ is zero-dimensional).

Incidentally, I was trying to prove that : If $R$ is a ring such that $\forall x\in R, x^n=x$ for some $n>1$ then, every prime ideal is a maximal ideal.

However, I just don't get how can $R/P$ is finite?

rschwieb
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Thomas Finley
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  • The "incidentally" paragraph is addressed well on the site. I'm going to adjust the title so it reflects your actual question, instead of the one that fools people into thinking this is a duplicate. – rschwieb Dec 18 '23 at 19:21
  • @rschwieb Thanks! I searched for this question previously on this site, but was unable to find it. Are you using any "search software"? – Thomas Finley Dec 19 '23 at 03:15
  • I usually use a tag and very relevant keywords, and that works 90% of the time. Failing that, I try google with site:math.stackexchange.com specified – rschwieb Dec 19 '23 at 03:24
  • @rschwieb Ok, say, what did you try for this question? I have a request: Can you please take a look at this: https://math.stackexchange.com/questions/4829900/let-r-be-a-ring-and-p-be-an-ideal-containing-an-ideal-i-of-r-show-that ? It's also about ideals. I am having a hard time solving it. – Thomas Finley Dec 19 '23 at 03:55
  • It would have been something like [ring-theory] "prime ideal" maximal, and then I recall I found something linked as a duplicate to the one I linked here. – rschwieb Dec 19 '23 at 14:33
  • As for the question you linked, I think you should follow the first comment and use the third isomorphism theorem which makes it very straightforward. It doesn't have anything to do with commutativity: the ideals between $R$ and $I$ and those of $R/I$ correspond in a way that respects order, so if $I\subseteq M\subseteq R$, then $M$ is maximal in $R$ iff it is maximal in $R/I$. – rschwieb Dec 19 '23 at 14:36
  • I suppose the first commenter erred in saying "field" where they should have said "simple ring." – rschwieb Dec 19 '23 at 14:44
  • @rschwieb Can you say what a simple ring is? Never have I heard about it. – Thomas Finley Dec 19 '23 at 14:46
  • LMWTFY In general, a "simple object" is something that doesn't have nontrivial kernels of morphisms. It's the same idea as "simple group" or "simple module." – rschwieb Dec 19 '23 at 14:49

1 Answers1

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Every element of $R/P$ is a root of $x^n-x$. But over an integral domain, nonzero polynomials have only finitely many roots.

In the comments, a variant of the question was raised. Suppose that we have the weaker condition that for all $x \in R$, there is some $n \in \Bbb N,n \geq 2$ (depending on $x$) such that $x^n=x$. Then we can still show that all prime ideals in $R$ are maximal, but $R/P$ might not be finite for all prime ideals $P$.

To see that $R/P$ is a field assuming that condition, we can choose $x \in R/P$ non-zero. Then we get $x^n=x$ for some $n\geq 2$. Because $R/P$ is a domain, we may cancel $x$ and get $x^{n-1}=1$. This implies that $x$ is a unit, because $1=xx^{n-2}$.

To see that $R/P$ might not be finite, consider $R=\overline{\Bbb F_p}$ (the algebraic closure of $\Bbb F_p$) and $P=0$. Here every element satisfies $x^{p^k}=x$ for some $k\in \Bbb N$.

Lukas Heger
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