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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?

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1 Answer 1

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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 $\mathbf{p}$.

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  • $\begingroup$ 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. $\endgroup$ Commented Aug 9, 2022 at 13:54
  • $\begingroup$ I think my first line is an answer to his first lines "I want to show that..." or am I mistaken? $\endgroup$
    – Simonsays
    Commented Aug 9, 2022 at 13:58
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    $\begingroup$ Ooh, I missed that when I wrote my answer. You're right, $f(p)$ isn't even necessarily an ideal! $\endgroup$ Commented Aug 9, 2022 at 14:58
  • $\begingroup$ exactly, it is slightly confusing and the OP is asking two slightly different questions at the same time $\endgroup$
    – Simonsays
    Commented Aug 9, 2022 at 15:05
  • $\begingroup$ 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? $\endgroup$
    – Plantation
    Commented Aug 10, 2022 at 2:06

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