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Let $R$ be a commutative ring with $1\neq 0$. $R$ is said to be von Neumann regular if for all $a\in R$, there is some $x\in R$ such that $a^2x=a.$ Prove that if $R$ is von Neumann regular and $P$ a prime ideal, then $P$ is maximal.

My idea: We know that $P$ is a prime ideal and $R$ is a commutative ring, so $R/P$ is an integral domain. If we can show that $R/P$ is a field, then $P$ is maximal. Further, every finite integral domain is a field, although I am not sure it will be helpful here.

Any suggestions/comments/answers are welcome. Thanks.

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  • $\begingroup$ This doesn't jive with the usual definition of regular, right? $\endgroup$ Jul 8, 2012 at 2:52
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    $\begingroup$ @Dylan: these are the von Neumann regular rings: see e.g. en.wikipedia.org/wiki/Von_Neumann_regular. As the article says, they are precisely the absolutely flat rings, i.e., the rings for which every module is flat. Maybe this terminology is safer... $\endgroup$ Jul 8, 2012 at 3:31
  • $\begingroup$ @PeteL.Clark Wonderful, thanks! $\endgroup$ Jul 8, 2012 at 3:32
  • $\begingroup$ This is proved without using quotient rings in this answer $\endgroup$
    – Xam
    Sep 10, 2017 at 16:45

3 Answers 3

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This answer is the same as Zev's, but perhaps stated more "conceptually". For what it's worth:

A commutative ring is von Neumann regular if for all $a \in R$, there is $x \in R$ such that $a^2 x = a$.

Here are two straightforward facts:

Fact 1: Every quotient of a von Neumann regular ring is von Neumann regular.

[The defining condition is an identity, and if an identity holds in a ring it holds in any quotient.]

Fact 2: An integral domain which is von Neumann regular is a field.

[Fact 2 is literally the first thing that springs to mind when I see the somewhat strange defining condition. What does $a^2 x = a$ mean? Well, if we're allowed to cancel the $a$'s, it means $ax = 1$!]

Thus if $\mathfrak{p}$ is a prime ideal in a von Neumann regular ring, $R/\mathfrak{p}$ is a von Neumann regular domain, hence a field, so $\mathfrak{p}$ is maximal.

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    $\begingroup$ Damn, this is also nice, smooth and beautiful! Twice such answers to the same question...lucky OP! +1 , of course. $\endgroup$
    – DonAntonio
    Jul 8, 2012 at 10:28
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You're on the right track! For any prime ideal $P\in R$ and $a\notin P$, we have $a^2x=a$ in $R$ for some $x\in R$, so that in $R/P$, we have $\bar{a}^2\bar{x}=\bar{a}$ (where $\bar{s}$ means the equivalence class $s+P$). Do you see how to proceed?

Rest of solution (mouse over to reveal):

Rewriting, we have$$\bar{a}^2\bar{x}-\bar{a}=\bar{a}(\bar{a}\bar{x}-\bar{1})=\bar{0}.$$Because $a\notin P$, we have $\bar{a}\neq\bar{0}$, so that because $R/P$ is an integral domain, we can conclude $\bar{a}\bar{x}-\bar{1}=0$. Thus any $\bar{a}\neq\bar{0}$ in $R/P$ has an inverse, so $R/P$ is a field.

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  • $\begingroup$ @ZevChonoles Thanks! That seems simple after I look at the solution:) $\endgroup$
    – Lyapunov
    Jul 8, 2012 at 2:16
  • $\begingroup$ That mouse-over thing is cool! I had no idea you could do that. $\endgroup$ Jul 8, 2012 at 4:01
  • $\begingroup$ @Keenan: It's quite useful for making complete answers that don't give everything away all at once :) The Markdown help page describes the code for it (towards the bottom). $\endgroup$ Jul 8, 2012 at 6:06
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Let's show that every $a + P\in \dfrac{R}{P}$ such that $a + P \neq 0 + P$ is an invertible element. Since $R$ is a von Neumann regular ring, for such an $a \in R$ there must exist some $b \in R$ such that $aba = a$, so $a+P = aba + P$, so $a+P = (ab + P)\cdot (a + P)$, so $ab + P = 1 + P$, hence $(a + P)^{-1} = b+P$. Since every nonzero element of $\dfrac{R}{P}$ is invertible, it follows that $\dfrac{R}{P}$ is a field, so $P$is a maximal ideal of $R$.

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