3
$\begingroup$

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.

$\endgroup$
7
  • 1
    $\begingroup$ 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. $\endgroup$
    – Hanno
    Commented Feb 16, 2015 at 8:11
  • $\begingroup$ @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. $\endgroup$
    – user138017
    Commented Feb 16, 2015 at 8:14
  • 2
    $\begingroup$ 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 ${\mathfrak q}\subset{\mathfrak p}$. $\endgroup$
    – Hanno
    Commented Feb 16, 2015 at 8:27
  • 1
    $\begingroup$ See this question. $\endgroup$ Commented Feb 16, 2015 at 19:23
  • 1
    $\begingroup$ @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. $\endgroup$
    – Watson
    Commented Sep 25, 2016 at 20:20

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thanks a lot for your answering! $\endgroup$
    – user138017
    Commented Dec 18, 2016 at 21:31
  • $\begingroup$ @user138017 you're welcome. $\endgroup$
    – Xam
    Commented Dec 18, 2016 at 21:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .