Lemma 15.19.3. In Situation 15.19.1 let $R' \to R''$ be an $R$-algebra map. Let $I' \subset R'$ and $I'R'' \subset I'' \subset R''$ be ideals. Assume

the map $V(I'') \to V(I')$ induced by $\mathop{\mathrm{Spec}}(R'') \to \mathop{\mathrm{Spec}}(R')$ is surjective, and

$R''_{\mathfrak p''}$ is flat over $R'$ for all primes $\mathfrak p'' \in V(I'')$.

If (15.19.1.1) holds for $(R'', I'')$, then (15.19.1.1) holds for $(R', I')$.

**Proof.**
Assume (15.19.1.1) holds for $(R'', I'')$. Pick a prime $I'S' + JS' \subset \mathfrak q' \subset S'$. Let $I' \subset \mathfrak p' \subset R'$ be the corresponding prime of $R'$. By assumption there exists a prime $\mathfrak p'' \in V(I'')$ of $R''$ lying over $\mathfrak p'$ and $R'_{\mathfrak p'} \to R''_{\mathfrak p''}$ is flat. Choose a prime $\overline{\mathfrak q}'' \subset \kappa (\mathfrak q') \otimes _{\kappa (\mathfrak p')} \kappa (\mathfrak p'')$. This corresponds to a prime $\mathfrak q'' \subset S'' = S' \otimes _{R'} R''$ which lies over $\mathfrak q'$ and over $\mathfrak p''$. In particular we see that $I''S'' \subset \mathfrak q''$ and that $JS'' \subset \mathfrak q''$. Note that $(S' \otimes _{R'} R'')_{\mathfrak q''}$ is a localization of $S'_{\mathfrak q'} \otimes _{R'_{\mathfrak p'}} R''_{\mathfrak p''}$. By assumption the module $(M' \otimes _{R'} R'')_{\mathfrak q''}$ is flat over $R''_{\mathfrak p''}$. Hence Algebra, Lemma 10.100.1 implies that $M'_{\mathfrak q'}$ is flat over $R'_{\mathfrak p'}$ which is what we wanted to prove.
$\square$

## Comments (0)