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?
 A: 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 $x^{n-1}/1\ne0/1$.
