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The Stacks project

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


Comments (1)

Comment #983 by on

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


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