# show if $P$ is minimal prime ideal of $R$ then every element of $PR_P$ is nilpotent.

Show if $$P$$ is minimal prime ideal of $$R$$ then every element of $$PR_P$$ is nilpotent.

The only idea that I come to mind is, we know $$PR_P$$ is the maximal ideal of $$R_P$$. Since $$P$$ is a prime ideal of $$R$$ then $$PR_P$$ also is a prime ideal of $$R_P$$. hence $$PR_P$$ also is the only prime ideal of $$R_P$$. Since the radical ideal is the intersection of all the prime ideals, $$PR_P$$ also is a radical ideal of $$R_P$$.

I don't know if that would help for the proof, and I am not sure how to carry on. Please help. Thank you.

• Dear user138017, you're almost done! You just realized that you may restrict to the case of a local ring with a unique prime ideal, and now you only need to recall that, in any commutative ring, the intersection of all prime ideals is precisely the set of nilpotents. Feb 16, 2015 at 8:11
• @Hanno, I don't remember I have learnt that the intersection of all prime ideals is precisely the set of nilpotents, I will go to look it up, thanks a lot. Feb 16, 2015 at 8:14
• You need it to know that ${\mathfrak p}R_{\mathfrak p}$ is the only prime of $R_{\mathfrak p}$; in general, the primes of $R_{\mathfrak p}$ are those primes ${\mathfrak q}$ in $R$ satisfying ${\mathfrak q}\subset{\mathfrak p}$. Feb 16, 2015 at 8:27
• Feb 16, 2015 at 19:23
• @Hanno: do you have a reference for this fact? $\tag*{}$ EDIT: I've just found it sorry… It is for instance "The prime ideals of $S^{-1}A$ are in one-to-one correspondence with the prime ideals of $A$ which don't meet $S$", proposition 3.11, Atiyah–MacDonald, Introduction to Commutative algebra. Sep 25, 2016 at 20:20

Maybe no one would care about this, but I just want to write a proof about this statement. Let $R$ be a commutative ring (with $1$) and $S$ be a multiplicatively subset of $R$ (we can assume that $0\notin S$).

If we consider the natural homomorphism $\phi \colon R\rightarrow S^{-1}R$ given by $a\mapsto \frac{a}{1}$, then for ideals $I$ of $R$ and $\mathscr{I}$ of $S^{-1}R$, $I^e=\langle \phi(I)\rangle$ is an ideal of $S^{-1}R$ called the extension of $I$, and $\mathscr{I}^c=\phi^{-1}(\mathscr{I})$ is an ideal of $R$ called the contraction of $\mathscr{I}$.

Proposition.- Let the situation be as in the previous paragraph. Then there exits a bijective map $\Phi$ between the sets $A=\{P\in \text{Spec}(R): P\cap S=\emptyset\}$ and $B=\text{Spec}(S^{-1}R)$ given by $$\Phi\colon A\rightarrow B$$ $$\;\;\;\;\;P\mapsto P^e.$$

Whose inverse is $\Psi\colon B\rightarrow A$ given by $\mathscr{P}\mapsto \mathscr{P}^c$. Moreover, both $\Phi$ and $\Psi$ preserve inclusions.

Proof: This is the corollary of theorem 5.32 given in Sharp's "Steps in Commutative Algebra".

Now, in the particular case $S=R\setminus P$, where $P$ is minimal prime ideal of $R$, we have for $P'\in \text{Spec}(R)$, $P'\cap (R\setminus P)=\emptyset$ if and only if $P'\subseteq P$, which implies that $P'=P$. So it follows that $P^e=PR_P$ is the only prime ideal of $R_P$. But it's a standar result that $\text{Nil}(R_P)$ is the intersection of the elements of $\text{Spec}(R_P)$ and therefore $\text{Nil}(R_P)=PR_P$. Hence, every element of $PR_P$ is nilpotent.

• Thanks a lot for your answering! Dec 18, 2016 at 21:31
• @user138017 you're welcome.
– Xam
Dec 18, 2016 at 21:46