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.


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

  • $\begingroup$ 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

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