Lemma 10.118.1. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume
$R$ is Noetherian,
$R$ is a domain,
$R \to S$ is of finite type, and
$M$ is a finite type $S$-module.
Then there exists a nonzero $f \in R$ such that $M_ f$ is a free $R_ f$-module.
Proof. Let $K$ be the fraction field of $R$. Set $S_ K = K \otimes _ R S$. This is an algebra of finite type over $K$. We will argue by induction on $d = \dim (S_ K)$ (which is finite for example by Noether normalization, see Section 10.115). Fix $d \geq 0$. Assume we know that the lemma holds in all cases where $\dim (S_ K) < d$.
Suppose given $R \to S$ and $M$ as in the lemma with $\dim (S_ K) = d$. By Lemma 10.62.1 there exists a filtration $0 \subset M_1 \subset M_2 \subset \ldots \subset M_ n = M$ so that $M_ i/M_{i - 1}$ is isomorphic to $S/\mathfrak q$ for some prime $\mathfrak q$ of $S$. Note that $\dim ((S/\mathfrak q)_ K) \leq \dim (S_ K)$. Also, note that an extension of free modules is free (see basic notion 50). Thus we may assume $M = S$ and that $S$ is a domain of finite type over $R$.
If $R \to S$ has a nontrivial kernel, then take a nonzero $f \in R$ in this kernel. In this case $S_ f = 0$ and the lemma holds. (This is really the case $d = -1$ and the start of the induction.) Hence we may assume that $R \to S$ is a finite type extension of Noetherian domains.
Apply Lemma 10.115.7 and replace $R$ by $R_ f$ (with $f$ as in the lemma) to get a factorization
where the second extension is finite. Choose $z_1, \ldots , z_ r \in S$ which form a basis for the fraction field of $S$ over the fraction field of $R[y_1, \ldots , y_ d]$. This gives a short exact sequence
By construction $N$ is a finite $R[y_1, \ldots , y_ d]$-module whose support does not contain the generic point $(0)$ of $\mathop{\mathrm{Spec}}(R[y_1, \ldots , y_ d])$. By Lemma 10.40.5 there exists a nonzero $g \in R[y_1, \ldots , y_ d]$ such that $g$ annihilates $N$, so we may view $N$ as a finite module over $S' = R[y_1, \ldots , y_ d]/(g)$. Since $\dim (S'_ K) < d$ by induction there exists a nonzero $f \in R$ such that $N_ f$ is a free $R_ f$-module. Since $(R[y_1, \ldots , y_ d])_ f \cong R_ f[y_1, \ldots , y_ d]$ is free also we conclude by the already mentioned fact that an extension of free modules is free. $\square$
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