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.

  • $\begingroup$ This doesn't jive with the usual definition of regular, right? $\endgroup$ – Dylan Moreland Jul 8 '12 at 2:52
  • 5
    $\begingroup$ @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... $\endgroup$ – Pete L. Clark Jul 8 '12 at 3:31
  • $\begingroup$ @PeteL.Clark Wonderful, thanks! $\endgroup$ – Dylan Moreland Jul 8 '12 at 3:32
  • $\begingroup$ This is proved without using quotient rings in this answer $\endgroup$ – Xam Sep 10 '17 at 16:45

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.

| cite | improve this answer | |
  • $\begingroup$ Nice, smooth, beautiful...+1 $\endgroup$ – DonAntonio Jul 8 '12 at 1:48
  • $\begingroup$ Thanks for the kind words :) $\endgroup$ – Zev Chonoles Jul 8 '12 at 1:49
  • $\begingroup$ @ZevChonoles Thanks! That seems simple after I look at the solution:) $\endgroup$ – Lyapunov Jul 8 '12 at 2:16
  • $\begingroup$ That mouse-over thing is cool! I had no idea you could do that. $\endgroup$ – Keenan Kidwell Jul 8 '12 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). $\endgroup$ – Zev Chonoles Jul 8 '12 at 6:06

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ Damn, this is also nice, smooth and beautiful! Twice such answers to the same question...lucky OP! +1 , of course. $\endgroup$ – DonAntonio Jul 8 '12 at 10:28

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 $R$.

| cite | improve this answer | |

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.