I am trying to show that a flat morphism satisfies the going down property.
For this, the reference I am following reduces the problem to a certain claim which uses the following result of section 10.16 in the Stacks project, regarding the spectrum of a ring and related properties:
Lemma 10.16.9. Let $\varphi : R \to S$ be a ring map. Let $\mathfrak{p}$ be a prime of $R$. The following are equivalent:
- (1) $S \otimes_R \kappa(\mathfrak{p}) \neq 0$,
- (2) $S_{\mathfrak{p}}/\mathfrak{p}S_\mathfrak{p} \neq 0$,
- (3) $(S/\mathfrak{p}S)_{\mathfrak{p}} \neq 0$,
- (4) $\mathfrak{p} = \varphi^{-1}(\mathfrak{p}S)$.
If I understand correctly, here $\kappa(\mathfrak{p}) = R_{\mathfrak{p}}/\mathfrak{p}R_{\mathfrak{p}} \simeq R/\mathfrak{p} \otimes R_\mathfrak{p}$, and so
$$ S_{\mathfrak{p}}/\mathfrak{p}S_\mathfrak{p} \simeq R/\mathfrak{p} \otimes_R R_\mathfrak{p} \otimes_R S \simeq (S/\mathfrak{p}S)_\mathfrak{p} $$
shows that $(1) \iff (2) \iff (3)$.
Why are $(1) - (3)$ equivalent to $(4)$?
Edit: Mindlack has answered the former, I am replacing what was after it by what I actually meant.
The reference I am following shows that a flat map satisfies going down in the following way: given $\mathfrak{p} \subset \mathfrak{p}'$ and a lift $\mathfrak{q}'$ of the latter, via localizing we may assume that $\varphi : R \to S$ is a local flat map, and we only have to prove that a prime $\mathfrak{p}$ in $R$ lifts to a prime in $S$.
With these hypothesis, we can see that $C = \kappa(\mathfrak{p}) \otimes_R S$ is non zero, and this should mean that $S$ has a prime lift of $\mathfrak{p}$. That last part I dont see.
Edit 2: I think I got the last part. Since $R \setminus \mathfrak{p}$ is multiplicatively closed, so is its image via $\varphi$, and so the localization $S_\mathfrak{p}$ coincides with $T^{-1}S$ where $T=\varphi(R \setminus \mathfrak{p})$.
We can also consider the ideal generated by $\varphi(\mathfrak{p})$, and the quotient gives $(2)$. Hence, primes in this ring are in correspondence with primes in $S$ such that:
- $\mathfrak{q} \cap \varphi(R \setminus \mathfrak{p}) = \emptyset$
- $\mathfrak{q} \supset \mathfrak{p}R = \varphi(\mathfrak{p}), i.e. \varphi^{-1}(\mathfrak{q}) \supset \mathfrak{p}$.
These are exactly the lifts of $\mathfrak{p}$ in $S$, and so $(1)$ implies that $\mathfrak{p}$ has a lift.
(1) is equivalent to $S \otimes_R R/\mathfrak{p} \otimes_R R_{\mathfrak{p}}$ being nonzero, ie $(R \backslash \mathfrak{p})^{-1}\,(S/\mathfrak{p}S)$ being nonzero.
This is equivalent to the following sentence: there is no $x \in R \backslash \mathfrak{p}$ such that $x \cdot 1_{S/\mathfrak{p}S} =0$.
Said sentence is equivalent to: “there is no $x \in R \backslash \mathfrak{p}$ such that $\varphi(x) \in \mathfrak{p}S$”, ie $\varphi^{-1}(\mathfrak{p}S) \subset \mathfrak{p}$.
This last sentence is equivalent to (4).
I’m afraid I don’t really understand your second question. Which quotients should be equal? Note that (unless $\mathfrak{p}S$ is prime) I don’t think you can localize $S$ at $\mathfrak{p}S$. You can localize $S$ at $\mathfrak{p}$ (as a $R$-module) or at a multiplicative part of $S$.