## 10.168 Colimits and maps of finite presentation, II

This section is a continuation of Section 10.127.

We start with an application of the openness of flatness. It says that we can approximate flat modules by flat modules which is useful.

Lemma 10.168.1. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume that

1. $R \to S$ is of finite presentation,

2. $M$ is a finitely presented $S$-module, and

3. $M$ is flat over $R$.

In this case we have the following:

1. There exists a finite type $\mathbf{Z}$-algebra $R_0$ and a finite type ring map $R_0 \to S_0$ and a finite $S_0$-module $M_0$ such that $M_0$ is flat over $R_0$, together with a ring maps $R_0 \to R$ and $S_0 \to S$ and an $S_0$-module map $M_0 \to M$ such that $S \cong R \otimes _{R_0} S_0$ and $M = S \otimes _{S_0} M_0$.

2. If $R = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } R_\lambda$ is written as a directed colimit, then there exists a $\lambda$ and a ring map $R_\lambda \to S_\lambda$ of finite presentation, and an $S_\lambda$-module $M_\lambda$ of finite presentation such that $M_\lambda$ is flat over $R_\lambda$ and such that $S = R \otimes _{R_\lambda } S_\lambda$ and $M = S \otimes _{S_{\lambda }} M_\lambda$.

3. If

$(R \to S, M) = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } (R_\lambda \to S_\lambda , M_\lambda )$

is written as a directed colimit such that

1. $R_\mu \otimes _{R_\lambda } S_\lambda \to S_\mu$ and $S_\mu \otimes _{S_\lambda } M_\lambda \to M_\mu$ are isomorphisms for $\mu \geq \lambda$,

2. $R_\lambda \to S_\lambda$ is of finite presentation,

3. $M_\lambda$ is a finitely presented $S_\lambda$-module,

then for all sufficiently large $\lambda$ the module $M_\lambda$ is flat over $R_\lambda$.

Proof. We first write $(R \to S, M)$ as the directed colimit of a system $(R_\lambda \to S_\lambda , M_\lambda )$ as in as in Lemma 10.127.18. Let $\mathfrak q \subset S$ be a prime. Let $\mathfrak p \subset R$, $\mathfrak q_\lambda \subset S_\lambda$, and $\mathfrak p_\lambda \subset R_\lambda$ the corresponding primes. As seen in the proof of Theorem 10.129.4

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

is a system as in Lemma 10.127.13, and hence by Lemma 10.128.3 we see that for some $\lambda _{\mathfrak q} \in \Lambda$ for all $\lambda \geq \lambda _{\mathfrak q}$ the module $M_\lambda$ is flat over $R_\lambda$ at the prime $\mathfrak q_{\lambda }$.

By Theorem 10.129.4 we get an open subset $U_\lambda \subset \mathop{\mathrm{Spec}}(S_\lambda )$ such that $M_\lambda$ flat over $R_\lambda$ at all the primes of $U_\lambda$. Denote $V_\lambda \subset \mathop{\mathrm{Spec}}(S)$ the inverse image of $U_\lambda$ under the map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(S_\lambda )$. The argument above shows that for every $\mathfrak q \in \mathop{\mathrm{Spec}}(S)$ there exists a $\lambda _{\mathfrak q}$ such that $\mathfrak q \in V_\lambda$ for all $\lambda \geq \lambda _{\mathfrak q}$. Since $\mathop{\mathrm{Spec}}(S)$ is quasi-compact we see this implies there exists a single $\lambda _0 \in \Lambda$ such that $V_{\lambda _0} = \mathop{\mathrm{Spec}}(S)$.

The complement $\mathop{\mathrm{Spec}}(S_{\lambda _0}) \setminus U_{\lambda _0}$ is $V(I)$ for some ideal $I \subset S_{\lambda _0}$. As $V_{\lambda _0} = \mathop{\mathrm{Spec}}(S)$ we see that $IS = S$. Choose $f_1, \ldots , f_ r \in I$ and $s_1, \ldots , s_ n \in S$ such that $\sum f_ i s_ i = 1$. Since $\mathop{\mathrm{colim}}\nolimits S_\lambda = S$, after increasing $\lambda _0$ we may assume there exist $s_{i, \lambda _0} \in S_{\lambda _0}$ such that $\sum f_ i s_{i, \lambda _0} = 1$. Hence for this $\lambda _0$ we have $U_{\lambda _0} = \mathop{\mathrm{Spec}}(S_{\lambda _0})$. This proves (1).

Proof of (2). Let $(R_0 \to S_0, M_0)$ be as in (1) and suppose that $R = \mathop{\mathrm{colim}}\nolimits R_\lambda$. Since $R_0$ is a finite type $\mathbf{Z}$ algebra, there exists a $\lambda$ and a map $R_0 \to R_\lambda$ such that $R_0 \to R_\lambda \to R$ is the given map $R_0 \to R$ (see Lemma 10.127.3). Then, part (2) follows by taking $S_\lambda = R_\lambda \otimes _{R_0} S_0$ and $M_\lambda = S_\lambda \otimes _{S_0} M_0$.

Finally, we come to the proof of (3). Let $(R_\lambda \to S_\lambda , M_\lambda )$ be as in (3). Choose $(R_0 \to S_0, M_0)$ and $R_0 \to R$ as in (1). As in the proof of (2), there exists a $\lambda _0$ and a ring map $R_0 \to R_{\lambda _0}$ such that $R_0 \to R_{\lambda _0} \to R$ is the given map $R_0 \to R$. Since $S_0$ is of finite presentation over $R_0$ and since $S = \mathop{\mathrm{colim}}\nolimits S_\lambda$ we see that for some $\lambda _1 \geq \lambda _0$ we get an $R_0$-algebra map $S_0 \to S_{\lambda _1}$ such that the composition $S_0 \to S_{\lambda _1} \to S$ is the given map $S_0 \to S$ (see Lemma 10.127.3). For all $\lambda \geq \lambda _1$ this gives maps

$\Psi _{\lambda } : R_\lambda \otimes _{R_0} S_0 \longrightarrow R_\lambda \otimes _{R_{\lambda _1}} S_{\lambda _1} \cong S_\lambda$

the last isomorphism by assumption. By construction $\mathop{\mathrm{colim}}\nolimits _\lambda \Psi _\lambda$ is an isomorphism. Hence $\Psi _\lambda$ is an isomorphism for all $\lambda$ large enough by Lemma 10.127.8. In the same vein, there exists a $\lambda _2 \geq \lambda _1$ and an $S_0$-module map $M_0 \to M_{\lambda _2}$ such that $M_0 \to M_{\lambda _2} \to M$ is the given map $M_0 \to M$ (see Lemma 10.127.5). For $\lambda \geq \lambda _2$ there is an induced map

$S_\lambda \otimes _{S_0} M_0 \longrightarrow S_\lambda \otimes _{S_{\lambda _2}} M_{\lambda _2} \cong M_\lambda$

and for $\lambda$ large enough this map is an isomorphism by Lemma 10.127.6. This implies (3) because $M_0$ is flat over $R_0$. $\square$

Lemma 10.168.2. Let $R \to A \to B$ be ring maps. Assume $A \to B$ faithfully flat of finite presentation. Then there exists a commutative diagram

$\xymatrix{ R \ar[r] \ar@{=}[d] & A_0 \ar[d] \ar[r] & B_0 \ar[d] \\ R \ar[r] & A \ar[r] & B }$

with $R \to A_0$ of finite presentation, $A_0 \to B_0$ faithfully flat of finite presentation and $B = A \otimes _{A_0} B_0$.

Proof. We first prove the lemma with $R$ replaced $\mathbf{Z}$. By Lemma 10.168.1 there exists a diagram

$\xymatrix{ A_0 \ar[r] & A \\ B_0 \ar[u] \ar[r] & B \ar[u] }$

where $A_0$ is of finite type over $\mathbf{Z}$, $B_0$ is flat of finite presentation over $A_0$ such that $B = A \otimes _{A_0} B_0$. As $A_0 \to B_0$ is flat of finite presentation we see that the image of $\mathop{\mathrm{Spec}}(B_0) \to \mathop{\mathrm{Spec}}(A_0)$ is open, see Proposition 10.41.8. Hence the complement of the image is $V(I_0)$ for some ideal $I_0 \subset A_0$. As $A \to B$ is faithfully flat the map $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is surjective, see Lemma 10.39.16. Now we use that the base change of the image is the image of the base change. Hence $I_0A = A$. Pick a relation $\sum f_ i r_ i = 1$, with $r_ i \in A$, $f_ i \in I_0$. Then after enlarging $A_0$ to contain the elements $r_ i$ (and correspondingly enlarging $B_0$) we see that $A_0 \to B_0$ is surjective on spectra also, i.e., faithfully flat.

Thus the lemma holds in case $R = \mathbf{Z}$. In the general case, take the solution $A_0' \to B_0'$ just obtained and set $A_0 = A_0' \otimes _{\mathbf{Z}} R$, $B_0 = B_0' \otimes _{\mathbf{Z}} R$. $\square$

Lemma 10.168.3. Let $A = \mathop{\mathrm{colim}}\nolimits _{i \in I} A_ i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi _0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume

1. $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is finite,

2. $C_0$ is of finite type over $B_0$.

Then there exists an $i \geq 0$ such that the map $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is finite.

Proof. Let $x_1, \ldots , x_ m$ be generators for $C_0$ over $B_0$. Pick monic polynomials $P_ j \in A \otimes _{A_0} B_0[T]$ such that $P_ j(1 \otimes x_ j) = 0$ in $A \otimes _{A_0} C_0$. For some $i \geq 0$ we can find $P_{j, i} \in A_ i \otimes _{A_0} B_0[T]$ mapping to $P_ j$. Since $\otimes$ commutes with colimits we see that $P_{j, i}(1 \otimes x_ j)$ is zero in $A_ i \otimes _{A_0} C_0$ after possibly increasing $i$. Then this $i$ works. $\square$

Lemma 10.168.4. Let $A = \mathop{\mathrm{colim}}\nolimits _{i \in I} A_ i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi _0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume

1. $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is surjective,

2. $C_0$ is of finite type over $B_0$.

Then for some $i \geq 0$ the map $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is surjective.

Proof. Let $x_1, \ldots , x_ m$ be generators for $C_0$ over $B_0$. Pick $b_ j \in A \otimes _{A_0} B_0$ mapping to $1 \otimes x_ j$ in $A \otimes _{A_0} C_0$. For some $i \geq 0$ we can find $b_{j, i} \in A_ i \otimes _{A_0} B_0$ mapping to $b_ j$. Then this $i$ works. $\square$

Lemma 10.168.5. Let $A = \mathop{\mathrm{colim}}\nolimits _{i \in I} A_ i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi _0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume

1. $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is unramified,

2. $C_0$ is of finite type over $B_0$.

Then for some $i \geq 0$ the map $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is unramified.

Proof. Set $B_ i = A_ i \otimes _{A_0} B_0$, $C_ i = A_ i \otimes _{A_0} C_0$, $B = A \otimes _{A_0} B_0$, and $C = A \otimes _{A_0} C_0$. Let $x_1, \ldots , x_ m$ be generators for $C_0$ over $B_0$. Then $\text{d}x_1, \ldots , \text{d}x_ m$ generate $\Omega _{C_0/B_0}$ over $C_0$ and their images generate $\Omega _{C_ i/B_ i}$ over $C_ i$ (Lemmas 10.131.14 and 10.131.9). Observe that $0 = \Omega _{C/B} = \mathop{\mathrm{colim}}\nolimits \Omega _{C_ i/B_ i}$ (Lemma 10.131.5). Thus there is an $i$ such that $\text{d}x_1, \ldots , \text{d}x_ m$ map to zero and hence $\Omega _{C_ i/B_ i} = 0$ as desired. $\square$

Lemma 10.168.6. Let $A = \mathop{\mathrm{colim}}\nolimits _{i \in I} A_ i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi _0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume

1. $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is an isomorphism,

2. $B_0 \to C_0$ is of finite presentation.

Then for some $i \geq 0$ the map $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is an isomorphism.

Proof. By Lemma 10.168.4 there exists an $i$ such that $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is surjective. Since the map is of finite presentation the kernel is a finitely generated ideal. Let $g_1, \ldots , g_ r \in A_ i \otimes _{A_0} B_0$ generate the kernel. Then we may pick $i' \geq i$ such that $g_ j$ map to zero in $A_{i'} \otimes _{A_0} B_0$. Then $A_{i'} \otimes _{A_0} B_0 \to A_{i'} \otimes _{A_0} C_0$ is an isomorphism. $\square$

Lemma 10.168.7. Let $A = \mathop{\mathrm{colim}}\nolimits _{i \in I} A_ i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi _0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume

1. $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is étale,

2. $B_0 \to C_0$ is of finite presentation.

Then for some $i \geq 0$ the map $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is étale.

Proof. Write $C_0 = B_0[x_1, \ldots , x_ n]/(f_{1, 0}, \ldots , f_{m, 0})$. Write $B_ i = A_ i \otimes _{A_0} B_0$ and $C_ i = A_ i \otimes _{A_0} C_0$. Note that $C_ i = B_ i[x_1, \ldots , x_ n]/(f_{1, i}, \ldots , f_{m, i})$ where $f_{j, i}$ is the image of $f_{j, 0}$ in the polynomial ring over $B_ i$. Write $B = A \otimes _{A_0} B_0$ and $C = A \otimes _{A_0} C_0$. Note that $C = B[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ where $f_ j$ is the image of $f_{j, 0}$ in the polynomial ring over $B$. The assumption is that the map

$\text{d} : (f_1, \ldots , f_ m)/(f_1, \ldots , f_ m)^2 \longrightarrow \bigoplus C \text{d}x_ k$

is an isomorphism. Thus for sufficiently large $i$ we can find elements

$\xi _{k, i} \in (f_{1, i}, \ldots , f_{m, i})/(f_{1, i}, \ldots , f_{m, i})^2$

with $\text{d}\xi _{k, i} = \text{d}x_ k$ in $\bigoplus C_ i \text{d}x_ k$. Moreover, on increasing $i$ if necessary, we see that $\sum (\partial f_{j, i}/\partial x_ k) \xi _{k, i} = f_{j, i} \bmod (f_{1, i}, \ldots , f_{m, i})^2$ since this is true in the limit. Then this $i$ works. $\square$

Lemma 10.168.8. Let $A = \mathop{\mathrm{colim}}\nolimits _{i \in I} A_ i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi _0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume

1. $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is smooth,

2. $B_0 \to C_0$ is of finite presentation.

Then for some $i \geq 0$ the map $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is smooth.

Proof. Write $C_0 = B_0[x_1, \ldots , x_ n]/(f_{1, 0}, \ldots , f_{m, 0})$. Write $B_ i = A_ i \otimes _{A_0} B_0$ and $C_ i = A_ i \otimes _{A_0} C_0$. Note that $C_ i = B_ i[x_1, \ldots , x_ n]/(f_{1, i}, \ldots , f_{m, i})$ where $f_{j, i}$ is the image of $f_{j, 0}$ in the polynomial ring over $B_ i$. Write $B = A \otimes _{A_0} B_0$ and $C = A \otimes _{A_0} C_0$. Note that $C = B[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ where $f_ j$ is the image of $f_{j, 0}$ in the polynomial ring over $B$. The assumption is that the map

$\text{d} : (f_1, \ldots , f_ m)/(f_1, \ldots , f_ m)^2 \longrightarrow \bigoplus C \text{d}x_ k$

is a split injection. Let $\xi _ k \in (f_1, \ldots , f_ m)/(f_1, \ldots , f_ m)^2$ be elements such that $\sum (\partial f_ j/\partial x_ k) \xi _ k = f_ j \bmod (f_1, \ldots , f_ m)^2$. Then for sufficiently large $i$ we can find elements

$\xi _{k, i} \in (f_{1, i}, \ldots , f_{m, i})/(f_{1, i}, \ldots , f_{m, i})^2$

with $\sum (\partial f_{j, i}/\partial x_ k) \xi _{k, i} = f_{j, i} \bmod (f_{1, i}, \ldots , f_{m, i})^2$ since this is true in the limit. Then this $i$ works. $\square$

Lemma 10.168.9. Let $A = \mathop{\mathrm{colim}}\nolimits _{i \in I} A_ i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi _0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume

1. $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is syntomic (resp. a relative global complete intersection),

2. $C_0$ is of finite presentation over $B_0$.

Then there exists an $i \geq 0$ such that the map $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is syntomic (resp. a relative global complete intersection).

Proof. Assume $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is a relative global complete intersection. By Lemma 10.136.11 there exists a finite type $\mathbf{Z}$-algebra $R$, a ring map $R \to A \otimes _{A_0} B_0$, a relative global complete intersection $R \to S$, and an isomorphism

$(A \otimes _{A_0} B_0) \otimes _ R S \longrightarrow A \otimes _{A_0} C_0$

Because $R$ is of finite type (and hence finite presentation) over $\mathbf{Z}$, there exists an $i$ and a map $R \to A_ i \otimes _{A_0} B_0$ lifting the map $R \to A \otimes _{A_0} B_0$, see Lemma 10.127.3. Using the same lemma, there exists an $i' \geq i$ such that $(A_ i \otimes _{A_0} B_0) \otimes _ R S \to A \otimes _{A_0} C_0$ comes from a map $(A_ i \otimes _{A_0} B_0) \otimes _ R S \to A_{i'} \otimes _{A_0} C_0$. Thus we may assume, after replacing $i$ by $i'$, that the displayed map comes from an $A_ i \otimes _{A_0} B_0$-algebra map

$(A_ i \otimes _{A_0} B_0) \otimes _ R S \longrightarrow A_ i \otimes _{A_0} C_0$

By Lemma 10.168.6 after increasing $i$ this map is an isomorphism. This finishes the proof in this case because the base change of a relative global complete intersection is a relative global complete intersection by Lemma 10.136.9.

Assume $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is syntomic. Then there exist elements $g_1, \ldots , g_ m$ in $A \otimes _{A_0} C_0$ generating the unit ideal such that $A \otimes _{A_0} B_0 \to (A \otimes _{A_0} C_0)_{g_ j}$ is a relative global complete intersection, see Lemma 10.136.15. We can find an $i$ and elements $g_{i, j} \in A_ i \otimes _{A_0} C_0$ mapping to $g_ j$. After increasing $i$ we may assume $g_{i, 1}, \ldots , g_{i, m}$ generate the unit ideal of $A_ i \otimes _{A_0} C_0$. The result of the previous paragraph implies that, after increasing $i$, we may assume the maps $A_ i \otimes _{A_0} B_0 \to (A_ i \otimes _{A_0} C_0)_{g_{i, j}}$ are relative global complete intersections. Then $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is syntomic by Lemma 10.136.4 (and the already used Lemma 10.136.15). $\square$

The following lemma is an application of the results above which doesn't seem to fit well anywhere else.

Lemma 10.168.10. Let $R \to S$ be a faithfully flat ring map of finite presentation. Then there exists a commutative diagram

$\xymatrix{ S \ar[rr] & & S' \\ & R \ar[lu] \ar[ru] }$

where $R \to S'$ is quasi-finite, faithfully flat and of finite presentation.

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

$\xymatrix{ S_0 \ar[r] & S \\ R_0 \ar[u] \ar[r] & R \ar[u] }$

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

$S' = S_ g/(f_1, \ldots , f_ d).$

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

$S'_{\mathfrak q'}/\mathfrak pS'_{\mathfrak q'} = S_{\mathfrak q}/(f_1, \ldots , f_ d) + \mathfrak pS_{\mathfrak q} = \overline{S}_{\mathfrak q}/(\overline{f}_1, \ldots , \overline{f}_ d)$

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. $\square$

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