3

Let $R$ be an integral domain which is Noetherian, let $P$ be an invertible maximal ideal, and let $Q<P$ be a prime ideal. How to show that $Q=0$?

I have proved that $Q=QP$, and still haven't used the condition that $R$ is Noetherian and $P$ is maximal.

Can anyone help?

rschwieb
  • 153,510
hxhxhx88
  • 5,257
  • 1
    Look at $R_P$ and use Nakayama. – Cantlog Nov 07 '13 at 16:51
  • @Cantlog, sorry, what is $R_P$? – hxhxhx88 Nov 07 '13 at 21:36
  • This notation is pretty standard in commutative algebra, it denotes the localization of $R$ at $P$. – Cantlog Nov 07 '13 at 23:09
  • @Cantlog, in the wiki I found one version of the Nakayama Lemma, stating that: "Let I be an ideal in R, and M a finitely-generated module over R. If IM = M, then there exists an r ∈ R with r ≡ 1 (mod I), such that rM = 0." I think by this, we do not need to consider $R_P$. In fact, there exists $p\in P$ such that $(1+p)Q=0$, if $Q\neq$, then there exists $0\neq q\in Q$ such that $(1+p)q=0$, following that $-p=1\in P$, contradicting $P$ is maximal. Is my argument right? If so, I think we do not need $P$ to be maximal, just $P\neq R$ is enough.. – hxhxhx88 Nov 08 '13 at 00:40
  • Yes, in the initial statement you don't need $P$ maximal, just prime. In the localization $R_P$, the corresponding ideal $PR_P$ is maximal. – Cantlog Nov 08 '13 at 18:44

1 Answers1

1

From $Q=QP$ we get $QR_P=(QR_P)(PR_P)$ and by Nakayama lemma $QR_P=0$, so $Q=0$.

user26857
  • 52,094