The Stacks project

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

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

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

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

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

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 $A_0 = A_0' \otimes _{\mathbf{Z}} R$, $B_0 = B_0' \otimes _{\mathbf{Z}} R$. $\square$

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. $\square$

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 $b_ j$. After increasing $i$ we may assume that $b_{j, i}$ maps to $1 \otimes x_ j$ in $A_ i \otimes _{A_0} C_0$ for all $j = 1, \ldots , m$. Then this $i$ works. $\square$

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. $\square$

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. $\square$

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

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

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. $\square$

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

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

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. $\square$

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

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

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). $\square$

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 10.17.8), 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. $\square$