Generic freeness.

Lemma 10.118.2. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume

1. $R$ is a domain,

2. $R \to S$ is of finite presentation, and

3. $M$ is an $S$-module of finite presentation.

Then there exists a nonzero $f \in R$ such that $M_ f$ is a free $R_ f$-module.

Proof. Write $S = R[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$. For $g \in R[x_1, \ldots , x_ n]$ denote $\overline{g}$ its image in $S$. We may write $M = S^{\oplus t}/\sum Sn_ i$ for some $n_ i \in S^{\oplus t}$. Write $n_ i = (\overline{g}_{i1}, \ldots , \overline{g}_{it})$ for some $g_{ij} \in R[x_1, \ldots , x_ n]$. Let $R_0 \subset R$ be the subring generated by all the coefficients of all the elements $g_ i, g_{ij} \in R[x_1, \ldots , x_ n]$. Define $S_0 = R_0[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$. Define $M_0 = S_0^{\oplus t}/\sum S_0n_ i$. Then $R_0$ is a domain of finite type over $\mathbf{Z}$ and hence Noetherian (see Lemma 10.31.1). Moreover via the injection $R_0 \to R$ we have $S \cong R \otimes _{R_0} S_0$ and $M \cong R \otimes _{R_0} M_0$. Applying Lemma 10.118.1 we obtain a nonzero $f \in R_0$ such that $(M_0)_ f$ is a free $(R_0)_ f$-module. Hence $M_ f = R_ f \otimes _{(R_0)_ f} (M_0)_ f$ is a free $R_ f$-module. $\square$

Comment #983 by on

Suggested slogan: Finitely presented modules over domains are generically free.

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