Theorem 10.129.4. Let $R$ be a ring. Let $R \to S$ be a ring map of finite presentation. Let $M$ be a finitely presented $S$-module. The set

is open in $\mathop{\mathrm{Spec}}(S)$.

Theorem 10.129.4. Let $R$ be a ring. Let $R \to S$ be a ring map of finite presentation. Let $M$ be a finitely presented $S$-module. The set

\[ \{ \mathfrak q \in \mathop{\mathrm{Spec}}(S) \mid M_{\mathfrak q}\text{ is flat over }R\} \]

is open in $\mathop{\mathrm{Spec}}(S)$.

**Proof.**
Let $\mathfrak q \in \mathop{\mathrm{Spec}}(S)$ be a prime. Let $\mathfrak p \subset R$ be the inverse image of $\mathfrak q$ in $R$. Note that $M_{\mathfrak q}$ is flat over $R$ if and only if it is flat over $R_{\mathfrak p}$. Let us assume that $M_{\mathfrak q}$ is flat over $R$. We claim that there exists a $g \in S$, $g \not\in \mathfrak q$ such that $M_ g$ is flat over $R$.

We first reduce to the case where $R$ and $S$ are of finite type over $\mathbf{Z}$. Choose a directed set $\Lambda $ and a system $(R_\lambda \to S_\lambda , M_\lambda )$ as in Lemma 10.127.18. Set $\mathfrak p_\lambda $ equal to the inverse image of $\mathfrak p$ in $R_\lambda $. Set $\mathfrak q_\lambda $ equal to the inverse image of $\mathfrak q$ in $S_\lambda $. Then the system

\[ ((R_\lambda )_{\mathfrak p_\lambda }, (S_\lambda )_{\mathfrak q_\lambda }, (M_\lambda )_{\mathfrak q_{\lambda }}) \]

is a system as in Lemma 10.127.13. Hence by Lemma 10.128.3 we see that for some $\lambda $ the module $M_\lambda $ is flat over $R_\lambda $ at the prime $\mathfrak q_{\lambda }$. Suppose we can prove our claim for the system $(R_\lambda \to S_\lambda , M_\lambda , \mathfrak q_{\lambda })$. In other words, suppose that we can find a $g \in S_\lambda $, $g \not\in \mathfrak q_\lambda $ such that $(M_\lambda )_ g$ is flat over $R_\lambda $. By Lemma 10.127.18 we have $M = M_\lambda \otimes _{R_\lambda } R$ and hence also $M_ g = (M_\lambda )_ g \otimes _{R_\lambda } R$. Thus by Lemma 10.39.7 we deduce the claim for the system $(R \to S, M, \mathfrak q)$.

At this point we may assume that $R$ and $S$ are of finite type over $\mathbf{Z}$. We may write $S$ as a quotient of a polynomial ring $R[x_1, \ldots , x_ n]$. Of course, we may replace $S$ by $R[x_1, \ldots , x_ n]$ and assume that $S$ is a polynomial ring over $R$. In particular we see that $R \to S$ is flat and all fibres rings $S \otimes _ R \kappa (\mathfrak p)$ have global dimension $n$.

Choose a resolution $F_\bullet $ of $M$ over $S$ with each $F_ i$ finite free, see Lemma 10.71.1. Let $K_ n = \mathop{\mathrm{Ker}}(F_{n-1} \to F_{n-2})$. Note that $(K_ n)_{\mathfrak q}$ is flat over $R$, since each $F_ i$ is flat over $R$ and by assumption on $M$, see Lemma 10.39.13. In addition, the sequence

\[ 0 \to K_ n/\mathfrak p K_ n \to F_{n-1}/ \mathfrak p F_{n-1} \to \ldots \to F_0 / \mathfrak p F_0 \to M/\mathfrak p M \to 0 \]

is exact upon localizing at $\mathfrak q$, because of vanishing of $\text{Tor}_ i^{R_\mathfrak p}(\kappa (\mathfrak p), M_{\mathfrak q})$. Since the global dimension of $S_\mathfrak q/\mathfrak p S_{\mathfrak q}$ is $n$ we conclude that $K_ n / \mathfrak p K_ n$ localized at $\mathfrak q$ is a finite free module over $S_\mathfrak q/\mathfrak p S_{\mathfrak q}$. By Lemma 10.99.4 $(K_ n)_{\mathfrak q}$ is free over $S_{\mathfrak q}$. In particular, there exists a $g \in S$, $g \not\in \mathfrak q$ such that $(K_ n)_ g$ is finite free over $S_ g$.

By Lemma 10.129.3 there exists a further localization $S_ g$ such that the complex

\[ 0 \to K_ n \to F_{n-1} \to \ldots \to F_0 \]

is exact on *all fibres* of $R \to S$. By Lemma 10.99.5 this implies that the cokernel of $F_1 \to F_0$ is flat. This proves the theorem in the Noetherian case.
$\square$

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