Show that $\mathrm{Spec}\, R_{\mathfrak p}=\{\mathfrak qR_{\mathfrak p}\mid \mathfrak q\subset \mathfrak p, \mathfrak q\in\mathrm{Spec}(R)\}$

Question

Let $R$ a commutative ring. I'm trying to show that $$\mathrm{Spec}( R_{\mathfrak p})=\{\mathfrak qR_{\mathfrak p}\mid \mathfrak q\subset \mathfrak p, \mathfrak q\in Spec(R)\}.$$ I recall that $$R_{\mathfrak p}=\left\{\frac{x}{s}\mid s\notin \mathfrak p\right\}.$$

For the inclusion $\supset$ I did as follow. Suppose by contradiction that it's not prime, i.e., there is $\frac{x}{s},\frac{x}{s'}\in \mathfrak qR_{\mathfrak p}$ s.t. neither $\frac{x}{s}$ nor $\frac{x'}{s'}$ are in $\mathfrak q R_{\mathfrak p}$.

I can't arrive to a contradiction. For the other inclusion, I have no idea.

Any help is welcome.

Answer 1

There is a canonical homomorphism $\varphi:R\rightarrow R_{\mathfrak{p}}$ by sending $x\mapsto x/1$, prime ideals $\mathfrak{q}$ in $R_{\mathfrak{p}}$ go to prime ideals in $R$ under the preimage of $\varphi$. In symbols $\varphi^{-1}(\mathfrak{q})$ is a prime ideal in $R$. If $\mathfrak{q}':=\varphi(\mathfrak{q})$ is not contained in $\mathfrak{p}$ then there are elements in $\mathfrak{q}'$ inside $R-\mathfrak{p}$ these elements are units in $\mathfrak{q}$ so this is a contradiction. We must have $\mathfrak{q'}\subset \mathfrak{p}$.

Answer 2

Ok so I'm going to sketch the main steps of the proof. First let's set ourselves straight:

$$\mathfrak{q}R_{\mathfrak p}=\left\{\dfrac{x}{s}\bigg| x\in \mathfrak{q},\ s\notin \mathfrak{p}\right\}$$

is the ideal generated in $R_{\mathfrak p}$ by the image of $\mathfrak q \subset R$ by $\phi: R\rightarrow R_{\mathfrak p},\ x\mapsto \dfrac{x}{1}$. You actually need to prove that it is an ideal but it should be easy.

$\supseteq$: First let's prove it is prime. So you fix $\dfrac{x}{s},\ \dfrac{x'}{s'}\in R_{\mathfrak{p}}$ such that $\dfrac{xx'}{ss'}=\dfrac{y}{t}\in \mathfrak{q}R_{\mathfrak{p}}$, where $y\in \mathfrak{q}$ and $t\notin\mathfrak{p}$. Thus by the définition of localisation you get $\sigma\notin\mathfrak{p}$ such that: $$\sigma(yss' -txx')= 0$$ Thus you get $(\sigma t)xx'=\sigma yss'\in\mathfrak{q}$ but $\sigma t\notin \mathfrak{p}$ so $\notin\mathfrak{q}$ thus $xx'\in\mathfrak{q}$ thus $x\in \mathfrak{q}$ or $x'\in \mathfrak{q}$.

$\subseteq$: You take $\mathfrak Q$ prime in $R_{\mathfrak p}$ and consider $$\mathfrak{q}=\phi^{-1}(\mathfrak{Q})$$ By construction you know that it is prime.

In fact what we need to do, is prove that these constructions are reciprocal one to another.

Thus you need to prove $(1)$: $$\phi^{-1}(\mathfrak{q} R_{\mathfrak p})=\mathfrak q$$

and $(2)$: $$\phi^{-1}(\mathfrak{Q})R_{\mathfrak p}=\mathfrak Q$$

You should be able to handle this easily, much like what I have done in $\supseteq$. Tell me if you need more help.