bijection between prime ideals of $R_p$ and prime ideals of $R$ contained in $P$

Question

Given a ring $R$, I want to show that the localization of $R$ at the prime ideal $P$ of $R$ (denoted as $R_P$) is isomorphic to the set of prime ideals of $R$ contained in $P$. That is:

$$ \text{Spectrum}(R_P)\cong \{I\subseteq P \mid \text{$I$ is an ideal of $R$}\} $$

From the statment, I can see that $Q\subseteq R_P$ is a prime ideal, then any $x\in Q$ is of the form $x=\frac{a}{b}$, where $a\in R$, but $b\notin P$, from the definition of $R_P$. But how can I show that each such $Q$ relates to an ideal of $R$ contained in $P$.

It's hard to believe that a ring is isomorphic to a set of prime ideals. You must be talking about the spectrum of $R_P$ which is, btw, a set of prime ideals.. — user26857, May 13 '15 at 20:13
yes, so I have a set of prime ideals in $R_P$ mapping to a set of ideals in $R$ contained in $P$ — Andrew Brick, May 13 '15 at 20:17
Is there any textbook in commutative algebra not proving this? — user26857, May 13 '15 at 20:18
I'm not sure, I've checked the recommended textbook for my module, but can't seem to find the proof. It is mentioned in my lecture notes, but the proof is left as an excercise — Andrew Brick, May 13 '15 at 20:19
Can you please edit your question to address user26857's comment: your question confuses the localisation of a ring with a set of ideals of a ring. — Rob Arthan, May 13 '15 at 21:04
Andrew Brick i think the true statement is this: $Spec(R_P)\cong {I\in Spec(R)|I\subseteq P }$; if so please edit. — user 1, May 14 '15 at 06:21

Bernard · Accepted Answer · 2016-03-01T01:26:33.687

4

Sketch:

Let $\,i\colon R\to R_\mathfrak p$, $\,x\mapsto \dfrac x1$ be the canonical morphism. If $\mathfrak q\in\operatorname{Spec}R_{\mathfrak p}$, $\,\mathfrak p'=i^{-1}(\mathfrak q)$ is a prime ideal of $R$ contained in $\mathfrak p$.

It is straightforward to check the inverse of this mapping is $\,.\mathfrak p'\mapsto \mathfrak p'R_\mathfrak p$

edited Mar 01 '16 at 01:26

answered May 13 '15 at 20:47

Bernard

179,256

The O.P. doesn't seem to have thought of this. I only made the relation explicit. – Bernard May 13 '15 at 20:52
For the first direction in the bijection, I think you meant $\mathfrak q\in\operatorname{Spec}R_{\mathfrak p}$. Right? – Mar 01 '16 at 01:23
Absolutely. Thanks for pointing it. – Bernard Mar 01 '16 at 01:27

user 1 · Answer 2 · 2015-05-14T06:23:23.670

0

Proof for $Spec(R_P)\cong \{I\in Spec(R)|I\subseteq P \}$:
More generally, The book, "Steps in Commutative Algebra" by Sharp has:

enter image description here

Consider $S=R-P$.

Theorem 5.37 of the book has similar result for even primary ideals.

edited May 14 '15 at 06:23

answered May 14 '15 at 06:15

user 1

7,652

5.23 NOTATION. let S be a multiplicatively closed subset of the commutative ringR. let $f : R \to S^{-1}R$ denote the natural ring homomorphism. Use the extension and contraction notation for $f$ . – user 1 May 14 '15 at 06:31

bijection between prime ideals of $R_p$ and prime ideals of $R$ contained in $P$

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