Lemma 38.12.1. Let $R$ be a ring. Let $R \to S$ be a finitely presented flat ring map with geometrically integral fibres. Let $\mathfrak q \subset S$ be a prime ideal lying over the prime $\mathfrak r \subset R$. Set $\mathfrak p = \mathfrak r S$. Let $N$ be a finitely presented $S$-module. There exists $r \geq 0$ and an $S$-module map

$\alpha : S^{\oplus r} \longrightarrow N$

such that $\alpha : \kappa (\mathfrak p)^{\oplus r} \to N \otimes _ S \kappa (\mathfrak p)$ is an isomorphism. For any such $\alpha$ the following are equivalent:

1. $N_{\mathfrak q}$ is $R$-flat,

2. there exists an $f \in R$, $f \not\in \mathfrak r$ such that $\alpha _ f : S_ f^{\oplus r} \to N_ f$ is $R_ f$-universally injective and a $g \in S$, $g \not\in \mathfrak q$ such that $\mathop{\mathrm{Coker}}(\alpha )_ g$ is $R$-flat,

3. $\alpha _{\mathfrak r}$ is $R_{\mathfrak r}$-universally injective and $\mathop{\mathrm{Coker}}(\alpha )_{\mathfrak q}$ is $R$-flat

4. $\alpha _{\mathfrak r}$ is injective and $\mathop{\mathrm{Coker}}(\alpha )_{\mathfrak q}$ is $R$-flat,

5. $\alpha _{\mathfrak p}$ is an isomorphism and $\mathop{\mathrm{Coker}}(\alpha )_{\mathfrak q}$ is $R$-flat, and

6. $\alpha _{\mathfrak q}$ is injective and $\mathop{\mathrm{Coker}}(\alpha )_{\mathfrak q}$ is $R$-flat.

Proof. To obtain $\alpha$ set $r = \dim _{\kappa (\mathfrak p)} N \otimes _ S \kappa (\mathfrak p)$ and pick $x_1, \ldots , x_ r \in N$ which form a basis of $N \otimes _ S \kappa (\mathfrak p)$. Define $\alpha (s_1, \ldots , s_ r) = \sum s_ i x_ i$. This proves the existence.

Fix a choice of $\alpha$. We may apply Lemma 38.10.1 to the map $\alpha _{\mathfrak r} : S_{\mathfrak r}^{\oplus r} \to N_{\mathfrak r}$. Hence we see that (1), (3), (4), (5), and (6) are all equivalent. Since it is also clear that (2) implies (3) we see that all we have to do is show that (1) implies (2).

Assume (1). By openness of flatness, see Algebra, Theorem 10.129.4, the set

$U_1 = \{ \mathfrak q' \subset S \mid N_{\mathfrak q'}\text{ is flat over }R\}$

is open in $\mathop{\mathrm{Spec}}(S)$. It contains $\mathfrak q$ by assumption and hence $\mathfrak p$. Because $S^{\oplus r}$ and $N$ are finitely presented $S$-modules the set

$U_2 = \{ \mathfrak q' \subset S \mid \alpha _{\mathfrak q'}\text{ is an isomorphism}\}$

is open in $\mathop{\mathrm{Spec}}(S)$, see Algebra, Lemma 10.79.2. It contains $\mathfrak p$ by (5). As $R \to S$ is finitely presented and flat the map $\Phi : \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is open, see Algebra, Proposition 10.41.8. For any prime $\mathfrak r' \in \Phi (U_1 \cap U_2)$ we see that there exists a prime $\mathfrak q'$ lying over $\mathfrak r'$ such that $N_{\mathfrak q'}$ is flat and such that $\alpha _{\mathfrak q'}$ is an isomorphism, which implies that $\alpha \otimes \kappa (\mathfrak p')$ is an isomorphism where $\mathfrak p' = \mathfrak r' S$. Thus $\alpha _{\mathfrak r'}$ is $R_{\mathfrak r'}$-universally injective by the implication (1) $\Rightarrow$ (3). Hence if we pick $f \in R$, $f \not\in \mathfrak r$ such that $D(f) \subset \Phi (U_1 \cap U_2)$ then we conclude that $\alpha _ f$ is $R_ f$-universally injective, see Algebra, Lemma 10.82.12. The same reasoning also shows that for any $\mathfrak q' \in U_1 \cap \Phi ^{-1}(\Phi (U_1 \cap U_2))$ the module $\mathop{\mathrm{Coker}}(\alpha )_{\mathfrak q'}$ is $R$-flat. Note that $\mathfrak q \in U_1 \cap \Phi ^{-1}(\Phi (U_1 \cap U_2))$. Hence we can find a $g \in S$, $g \not\in \mathfrak q$ such that $D(g) \subset U_1 \cap \Phi ^{-1}(\Phi (U_1 \cap U_2))$ and we win. $\square$

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