1
$\begingroup$

Suppose $ R $ is a ring, and $ \mathfrak{p} \in \text{Spec}(R). $

I have been told that $ \text{Spec}(R_{\mathfrak{p}}) \cong \lbrace \mathfrak{q} \in \text{Spec}(R)\;| \mathfrak{q} \subset \mathfrak{p} \rbrace, $ and I am trying to figure out whether this is true or not, but I am stuck trying to show that there is an injective map $ \alpha : \text{Spec}(R_{\mathfrak{p}}) \longrightarrow \lbrace \mathfrak{q} \in \text{Spec}(R)\;| \mathfrak{q} \subset \mathfrak{p} \rbrace.$

I suspect that there is a way to show this by exploiting the fact that any (prime) ideal of $ \text{Spec}(R_{p}) $ cannot contain units of $ R_{\mathfrak{p}} $, but I can't find a way through.

Any assistance would be much appreciated.

$\endgroup$
2
$\begingroup$

In a word, yes.

In general, for a multiplicatively closed subset $S$ of $R$, there is a natural correspondence between the prime ideals of $S^{-1}R$ and the prime ideals of $R$ disjoint from $S$. Your example is the case where $S=R-\newcommand{\fp}{\mathfrak{p}} \newcommand{\fq}{\mathfrak{q}}\fp$.

The prime ideals of $S^{-1}R$ all have the form $S^{-1}\fq$ where $\fq\subseteq \fp$ is a prime ideal of $S$. It is straightforward to check that these $S^{-1}\fq$ are prime in $S^{-1}R$. Conversely, let $Q$ be a prime ideal of $S^{-1}R$. Then $\fq=\{a\in R:a/1\in Q\}$ is a prime ideal of $R$, being the inverse image of $Q$ under the map $\phi:a\mapsto a/1$ from $R$ to $S^{-1}R$. If $b/s\in Q$, then $b/1=(b/s)(s/1)\in Q$ so $b\in\fq$ and so $Q=S^{-1}\fq$ etc.

Texts on commutative algebra will have more details.

Your $\alpha$ is $Q\mapsto\phi^{-1}(Q)$.

$\endgroup$
  • $\begingroup$ Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end. $\endgroup$ – Confused Student Jan 20 at 4:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.