The Stacks project

Proof. As a first step we reduce this lemma to the case where $R$ is of finite type over $\mathbf{Z}$. By Lemma 10.168.2 there exists a diagram

where $R_0$ is of finite type over $\mathbf{Z}$, and $S_0$ is faithfully flat of finite presentation over $R_0$ such that $S = R \otimes _{R_0} S_0$. If we prove the lemma for the ring map $R_0 \to S_0$, then the lemma follows for $R \to S$ by base change, as the base change of a quasi-finite ring map is quasi-finite, see Lemma 10.122.8. (Of course we also use that base changes of flat maps are flat and base changes of maps of finite presentation are of finite presentation.)

Assume $R \to S$ is a faithfully flat ring map of finite presentation and that $R$ is Noetherian (which we may assume by the preceding paragraph). Let $W \subset \mathop{\mathrm{Spec}}(S)$ be the open set of Lemma 10.130.4. As $R \to S$ is faithfully flat the map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective, see Lemma 10.39.16. By Lemma 10.130.5 the map $W \to \mathop{\mathrm{Spec}}(R)$ is also surjective. Hence by replacing $S$ with a product $S_{g_1} \times \ldots \times S_{g_ m}$ we may assume $W = \mathop{\mathrm{Spec}}(S)$; here we use that $\mathop{\mathrm{Spec}}(R)$ is quasi-compact (Lemma 10.17.10), and that the map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is open (Proposition 10.41.8). Suppose that $\mathfrak p \subset R$ is a prime. Choose a prime $\mathfrak q \subset S$ lying over $\mathfrak p$ which corresponds to a maximal ideal of the fibre ring $S \otimes _ R \kappa (\mathfrak p)$. The Noetherian local ring $\overline{S}_{\mathfrak q} = S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is Cohen-Macaulay, say of dimension $d$. We may choose $f_1, \ldots , f_ d$ in the maximal ideal of $S_{\mathfrak q}$ which map to a regular sequence in $\overline{S}_{\mathfrak q}$. Choose a common denominator $g \in S$, $g \not\in \mathfrak q$ of $f_1, \ldots , f_ d$, and consider the $R$-algebra

By construction there is a prime ideal $\mathfrak q' \subset S'$ lying over $\mathfrak p$ and corresponding to $\mathfrak q$ (via $S_ g \to S'_ g$). Also by construction the ring map $R \to S'$ is quasi-finite at $\mathfrak q$ as the local ring

has dimension zero, see Lemma 10.122.2. Also by construction $R \to S'$ is of finite presentation. Finally, by Lemma 10.99.3 the local ring map $R_{\mathfrak p} \to S'_{\mathfrak q'}$ is flat (this is where we use that $R$ is Noetherian). Hence, by openness of flatness (Theorem 10.129.4), and openness of quasi-finiteness (Lemma 10.123.13) we may after replacing $g$ by $gg'$ for a suitable $g' \in S$, $g' \not\in \mathfrak q$ assume that $R \to S'$ is flat and quasi-finite. The image $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(R)$ is open and contains $\mathfrak p$. In other words we have shown a ring $S'$ as in the statement of the lemma exists (except possibly the faithfulness part) whose image contains any given prime. Using one more time the quasi-compactness of $\mathop{\mathrm{Spec}}(R)$ we see that a finite product of such rings does the job. $\square$