# If every element in a prime ideal $I$ is nilpotent, then $I=\sqrt{0}$?

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?

• Where do you get $\bigcap_{P\in Spec{R}}P \subset I$ from? – Arthur Sep 12 '16 at 13:18
• Because $I$ is a prime ideal, $I\in Spec{R}$ then like what I write? Is it wrong? – Soulostar Sep 12 '16 at 13:22
• 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. – egreg Sep 12 '16 at 13:37
• @egreg We can consider that it is also nilpotent – Soulostar Sep 12 '16 at 13:47
• @LêThếLong Of course not. – 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 $x^{n-1}/1\ne0/1$.

• I read the Rotman 's book "An introduction to homological algebra". Proposition 4.76 say that. Could you check it to me? – Soulostar Sep 13 '16 at 0:20
• 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". – Soulostar Sep 13 '16 at 0:42
• @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. – tracing Sep 13 '16 at 1:46
• Thank you so much. Now I am satisfied. – Soulostar Sep 13 '16 at 1:55
• @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. – egreg Sep 13 '16 at 7:07