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?
Best Answer
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 $\mathbf{p}$.