# Injective ring map sends minimal prime ideals to prime ideals?

I want to show that

"If $$\varphi : R \hookrightarrow S$$ is an injective ring map and $$\mathfrak{p}$$ is an minimal prime of $$R$$, then $$\varphi(\mathfrak{p})$$ is a prime ideal of $$S$$."

Is this true?

I tried to use https://stacks.math.columbia.edu/tag/00FK and there is a point that I can't understand.

In the link, they state that

Let $$R⊂S$$ be an injective ring map. Then $$\operatorname{Spec}(S)→\operatorname{Spec(R)}$$ hits all the minimal primes.

And they argue as

" Let $$\mathfrak{p}⊂R$$ be a minimal prime. In this case $$R_{\mathfrak{p}}$$ has a unique prime ideal. Hence it suffices to show that $$S_{\mathfrak{p}}$$ is not zero. And this follows from the fact that localization is exact, see Proposition 10.9.12."

My question is,

Q.1) Here, $$S_\mathfrak{p}$$ is a module localized at $$\mathfrak{p}$$? Can it be a ring too?

Q.2) Why it suffices to show that $$S_{\mathfrak{p}}$$ is nonzero? In this case, why $$\mathfrak{p}$$ is a prime ideal of $$S$$ or there is a prime ideal $$\mathfrak{q}$$ of $$S$$ such that $$\mathfrak{p}= R \cap \mathfrak{q}$$?

Can anyone help?

First of all, I don't think its true what you want to prove: Take a field $$k$$ and $$k \hookrightarrow k[x]/x^2$$. Then $$(0) \subset k$$ is prime but its extension $$(0) \subset k[x]/x^2$$ is not prime because $$k[x]/x^2$$ is not reduced.
A few remarks to you question: note first that for a ring $$S$$ with prime ideal $$\mathbf{p}$$, you have $$\mathbf{Spec}(S_{\mathbf{p}})\cong\{\mathbf{q} \in \mathbf{Spec}(S): \mathbf{q} \subset \mathbf{p} \}$$.
Here, $$S_\mathbf{p}=S \otimes_R R_\mathbf{p}$$ is an identification of $$R$$-modules. You're right that here $$\mathbf{p}$$ is not a prime of $$S$$, but of $$R$$ and $$S_\mathbf{p}$$ is actually the local ring of a prime $$\mathbf{q}$$ lying over $$\mathbf{p}$$ in $$S$$ (which is still a ring). Localising at $$\mathbf{p}$$ gives a map $$\phi: R_\mathbf{p} \hookrightarrow S_\mathbf{p}$$ (is exact).
If $$S_\mathbf{p} \neq 0$$ then it contains a prime $$\mathbf{q}$$ and the map $$\mathbf{Spec}(S_\mathbf{p}) \rightarrow \mathbf{Spec}(R_\mathbf{p})$$ is just $$\mathbf{q} \mapsto \phi^{-1}(\mathbf{q})=\mathbf{p}$$. Not that the last equality holds because $$R_\mathbf{p}$$ has only one prime so it must be $$\mathbf{p}$$.
• Your first line is completely unrelated to the problem at hand: what the OP wants to show is that for all minimal primes $p\subset R$ there is a prime $q\subset S$ so that $f^{-1}(q)=p$ where $f:R\to S$ is the ring map. Commented Aug 9, 2022 at 13:54
• Ooh, I missed that when I wrote my answer. You're right, $f(p)$ isn't even necessarily an ideal! Commented Aug 9, 2022 at 14:58
• You wrote, $\mathbf{Spec}(S_{\mathbf{p}})\cong\{\mathbf{q} \in \mathbf{Spec}(S): \mathbf{q} \subset \mathbf{p} \}$. Let $\varphi : R \to S$ be the injective map. If $\operatorname{Spec}(S_p) \cong \{q \in \operatorname{Spec}(S) : \varphi^{-1}(q) \subset p \}$ is also true, then if $S_p$ is nonzero, then there exists $q \in \operatorname{Spec}(S)$ such that $\varphi^{-1}(q) \subset p$. Since $p$ is minimal prime, $\varphi^{-1}(q) = p$ and we are done (correct argument?). Additional argument is redundant? Commented Aug 10, 2022 at 2:06