I see a problem when I read this topic.

As we know that $\sqrt{0}=\bigcap_{P\in \mathrm{Spec}{R}}P$, if every element in a prime ideal $I$ is nilpotent, then $I\subset \sqrt{0}$. Then $I \subset \sqrt{0}= \bigcap_{P\in\mathrm{Spec}{R}}P \subset I$, so $I=\sqrt{0}$. I think something goes wrong but I do not know. Could you point out it to me?

  • 1
    $\begingroup$ Where do you get $ \bigcap_{P\in Spec{R}}P \subset I$ from? $\endgroup$ – Arthur Sep 12 '16 at 13:18
  • $\begingroup$ Because $I$ is a prime ideal, $I\in Spec{R}$ then like what I write? Is it wrong? $\endgroup$ – Soulostar Sep 12 '16 at 13:22
  • 1
    $\begingroup$ If a prime ideal consists of nilpotent elements, then it coincides with $\sqrt{0}$, that's right. But the topic in the linked question is about every element of a minimal prime being a zero divisor, which is not the same as being nilpotent. $\endgroup$ – egreg Sep 12 '16 at 13:37
  • $\begingroup$ @egreg We can consider that it is also nilpotent $\endgroup$ – Soulostar Sep 12 '16 at 13:47
  • $\begingroup$ @LêThếLong Of course not. $\endgroup$ – egreg Sep 12 '16 at 13:50

What goes wrong is that generally a minimal prime ideal does not consist of nilpotent elements.

Example. Consider a field $F$ and the ring $R=F\times F$. This ring only has four ideals, $\{0\}\times\{0\}$, $F\times\{0\}$, $\{0\}\times F$ and $R$. The two middle ones are minimal (and maximal) prime ideals. On the other hand, the ring has no nilpotent element.

What's true is that if a prime ideal consists of nilpotent elements then it equals $\sqrt{0}$. However, what's generally true is that a minimal prime ideal consists of zero divisors, which is a very different property than nilpotency.

Your doubts seem to come from a wrong statement in Rotman's book. Let's see what happens.

If $P$ is a minimal prime ideal of $R$, then the localization $R_P$ has exactly one prime ideal and therefore any element in it is nilpotent. Therefore, if $x\in P$, then $x/1$ is nilpotent in $R_P$, so $x^n/1=0$ for some $n>0$, which means

there is $s\in R\setminus P$ with $x^ns=0$.

Hence $x$ is a zero divisor. Indeed, if we choose $n$ minimal, we have $x^{n-1}s\ne 0$, because $x^{n-1}/1\ne0/1$.

  • $\begingroup$ I read the Rotman 's book "An introduction to homological algebra". Proposition 4.76 say that. Could you check it to me? $\endgroup$ – Soulostar Sep 13 '16 at 0:20
  • $\begingroup$ Is it the solution is wrong at "x is nilpotent if and only if x/1 is nilpotent". I think it is just true in the part "if". $\endgroup$ – Soulostar Sep 13 '16 at 0:42
  • $\begingroup$ @LêThếLong I just looked at the statement in Rotman you quote in your comment, and it is wrong. I am surprised that there is such a serious and basic error in this text, but there it is. As you say, the if part is true, but in the other direction, all you get is that $x^n s = 0$ for some element $s \not\in \mathfrak p$. This gives that $x$ is a zero divisor, but you can't do better than that, as the most basic examples show. $\endgroup$ – tracing Sep 13 '16 at 1:46
  • $\begingroup$ Thank you so much. Now I am satisfied. $\endgroup$ – Soulostar Sep 13 '16 at 1:55
  • $\begingroup$ @LêThếLong Sorry, I don't have the book. Anyway, the example should be clear enough to remove your doubts. When you localize at a minimal prime, you indeed get that all elements in the (unique) prime ideal are nilpotent, but it's clearly false that if $x/1$ is nilpotent then $x$ is nilpotent. $\endgroup$ – egreg Sep 13 '16 at 7:07

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.