The Stacks project

Proof. Since $B' = B \times _ A A'$ we obtain a commutative square of spectra, which induces a continuous map

as the source is a pushout in the category of topological spaces (which exists by Topology, Section 5.29).

To show the map $can$ is surjective, let $\mathfrak q' \subset B'$ be a prime ideal. If $I \subset \mathfrak q'$ (here and below we take the liberty of considering $I$ as an ideal of $B'$ as well as an ideal of $A'$), then $\mathfrak q'$ corresponds to a prime ideal of $B$ and is in the image. If not, then pick $h \in I$, $h \not\in \mathfrak q'$. In this case $B_ h = A_ h = 0$ and the ring map $B'_ h \to A'_ h$ is an isomorphism, see Lemma 15.5.3. Thus we see that $\mathfrak q'$ corresponds to a unique prime ideal $\mathfrak p' \subset A'$ which does not contain $I$.

Since $B' \to B$ is surjective, we see that $can$ is injective on the summand $\mathop{\mathrm{Spec}}(B)$. We have seen above that $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(B')$ is injective on the complement of $V(I) \subset \mathop{\mathrm{Spec}}(A')$. Since $V(I) \subset \mathop{\mathrm{Spec}}(A')$ is exactly the image of $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A')$ a trivial set theoretic argument shows that $can$ is injective.

To finish the proof we have to show that $can$ is open. To do this, observe that an open of the pushout is of the form $V \amalg U'$ where $V \subset \mathop{\mathrm{Spec}}(B)$ and $U' \subset \mathop{\mathrm{Spec}}(A')$ are opens whose inverse images in $\mathop{\mathrm{Spec}}(A)$ agree. Let $v \in V$. We can find a $g \in B$ such that $v \in D(g) \subset V$. Let $f \in A$ be the image. Pick $f' \in A'$ mapping to $f$. Then $D(f') \cap U' \cap V(I) = D(f') \cap V(I)$. Hence $V(I) \cap D(f')$ and $D(f') \cap (U')^ c$ are disjoint closed subsets of $D(f') = \mathop{\mathrm{Spec}}(A'_{f'})$. Write $(U')^ c = V(J)$ for some ideal $J \subset A'$. Since $A'_{f'} \to A'_{f'}/IA'_{f'} \times A'_{f'}/JA'_{f'}$ is surjective by the disjointness just shown, we can find an $a'' \in A'_{f'}$ mapping to $1$ in $A'_{f'}/IA'_{f'}$ and mapping to zero in $A'_{f'}/JA'_{f'}$. Clearing denominators, we find an element $a' \in J$ mapping to $f^ n$ in $A$. Then $D(a'f') \subset U'$. Let $h' = (g^{n + 1}, a'f') \in B'$. Since $B'_{h'} = B_{g^{n + 1}} \times _{A_{f^{n + 1}}} A'_{a'f'}$ by a previously cited lemma, we see that $D(h')$ pulls back to an open neighbourhood of $v$ in the pushout, i.e., the image of $V \amalg U'$ contains an open neighbourhood of the image of $v$. We omit the (easier) proof that the same thing is true for $u' \in U'$ with $u' \not\in V(I)$. $\square$

Proof. Let $a' \in A'$ with image $a \in A$. Let $x^ d + b_1 x^{d - 1} + \ldots + b_ d$ be a monical polynomial with coefficients in $B$ satisfied by $a$. Choose $b'_ i \in B'$ mapping to $b_ i \in B$ (possible). Then $(a')^ d + b'_1 (a')^{d - 1} + \ldots + b'_ d$ is in the kernel of $A' \to A$. Since $\mathop{\mathrm{Ker}}(B' \to B) = \mathop{\mathrm{Ker}}(A' \to A)$ we can modify our choice of $b'_ d$ to get $(a')^ d + b'_1 (a')^{d - 1} + \ldots + b'_ d = 0$ as desired. $\square$

Proof. The adjointness statement follows from the more general Lemma 15.5.4. To prove the final assertion, recall that $B' = B \times _ A A'$ and $N' = N \times _{\varphi , M} M'$ and extend these equalities to

where $I, J, K, L$ are the kernels of the horizontal maps of the original diagrams. We present the proof as a sequence of observations:

1. $K = IM'$ (see statement lemma),

2. $B' \to B$ is surjective with kernel $J$ and $J \to I$ is bijective,

3. $N' \to N$ is surjective with kernel $L$ and $L \to K$ is bijective,

4. $JN' \subset L$,

5. $\mathop{\mathrm{Im}}(N \to M)$ generates $M$ as an $A$-module (because $N \otimes _ B A = M$),

6. $\mathop{\mathrm{Im}}(N' \to M')$ generates $M'$ as an $A'$-module (because it holds modulo $K$ and $L$ maps isomorphically to $K$),

7. $JN' = L$ (because $L \cong K = I M'$ is generated by images of elements $x n'$ with $x \in I$ and $n' \in N'$ by the previous statement),

8. $N' \otimes _{B'} B = N$ (because $N = N'/L$, $B = B'/J$, and the previous statement),

9. there is a map $\gamma : N' \otimes _{B'} A' \to M'$,

10. $\gamma$ is surjective (see above),

11. the kernel of the composition $N' \otimes _{B'} A' \to M' \to M$ is generated by elements $l \otimes 1$ and $n' \otimes x$ with $l \in K$, $n' \in N'$, $x \in I$ (because $M = N \otimes _ B A$ by assumption and because $N' \to N$ and $A' \to A$ are surjective with kernels $L$ and $I$),

12. any element of $N' \otimes _{B'} A'$ in the submodule generated by the elements $l \otimes 1$ and $n' \otimes x$ with $l \in L$, $n' \in N'$, $x \in I$ can be written as $l \otimes 1$ for some $l \in L$ (because $J$ maps isomorphically to $I$ we see that $n' \otimes x = n'x \otimes 1$ in $N' \otimes _{B'} A'$; similarly $x n' \otimes a' = n' \otimes xa' = n'(xa') \otimes 1$ in $N' \otimes _{B'} A'$ when $n' \in N'$, $x \in J$ and $a' \in A'$; since we have seen that $JN' = L$ this proves the assertion),

13. the kernel of $\gamma$ is zero (because by (10) and (11) any element of the kernel is of the form $l \otimes 1$ with $l \in L$ which is mapped to $l \in K \subset M'$ by $\gamma$).

This finishes the proof. $\square$

Proof. As in the proof of Lemma 15.6.4 let $J \subset B'$ be the kernel of the map $B' \to B$. Then $L' \otimes _{B'} B = L'/JL'$. Hence to prove surjectivity it suffices to show that elements of the form $(0, z)$ of the fibre product are in the image of the map of the lemma. The kernel of the map $L' \otimes _{B'} A' \to L' \otimes _{B'} A$ is the image of $L' \otimes _{B'} I \to L' \otimes _{B'} A'$. Since the map $J \to I$ induced by $B' \to A'$ is an isomorphism the composition

induces a surjection of $L' \otimes _{B'} J$ onto the set of elements of the form $(0, z)$. To see the map is not injective in general we present a simple example. Namely, take a field $k$, set $B' = k[x, y]/(xy)$, $A' = B'/(x)$, $B = B'/(y)$, $A = B'/(x, y)$ and $L' = B'/(x - y)$. In that case the class of $x$ in $L'$ is nonzero but is mapped to zero under the displayed arrow. $\square$

Proof. Pick $(x_2, y_2) \in N_2 \times _{\varphi _2, M_2} M'_2$. Choose $x_1 \in N_1$ mapping to $x_2$. Since $M'_1 \to M_1$ is surjective we can find $y_1 \in M'_1$ mapping to $\varphi _1(x_1)$. Then $(x_1, y_1)$ maps to $(x_2, y'_2)$ in $N_2 \times _{\varphi _2, M_2} M'_2$. Thus it suffices to show that elements of the form $(0, y_2)$ are in the image of the map. Here we see that $y_2 \in IM'_2$. Write $y_2 = \sum t_ i y_{2, i}$ with $t_ i \in I$. Choose $y_{1, i} \in M'_1$ mapping to $y_{2, i}$. Then $y_1 = \sum t_ iy_{1, i} \in IM'_1$ and the element $(0, y_1)$ does the job. $\square$

Proof. We will use the results of Lemma 15.6.4 without further mention. Choose generators $y_1, \ldots , y_ r$ of $N$ over $B$ and generators $x_1, \ldots , x_ s$ of $M'$ over $A'$. Using that $N = N' \otimes _{B'} B$ and $B' \to B$ is surjective we can find $u_1, \ldots , u_ r \in N'$ mapping to $y_1, \ldots , y_ r$ in $N$. Using that $M' = N' \otimes _{B'} A'$ we can find $v_1, \ldots , v_ t \in N'$ such that $x_ i = \sum v_ j \otimes a'_{ij}$ for some $a'_{ij} \in A'$. In particular we see that the images $\overline{v}_ j \in M'$ of the $v_ j$ generate $M'$ over $A'$. We claim that $u_1, \ldots , u_ r, v_1, \ldots , v_ t$ generate $N'$ as a $B'$-module. Namely, pick $\xi \in N'$. We first choose $b'_1, \ldots , b'_ r \in B'$ such that $\xi$ and $\sum b'_ i u_ i$ map to the same element of $N$. This is possible because $B' \to B$ is surjective and $y_1, \ldots , y_ r$ generate $N$ over $B$. The difference $\xi - \sum b'_ i u_ i$ is of the form $(0, \theta )$ for some $\theta$ in $IM'$. Say $\theta$ is $\sum t_ j\overline{v}_ j$ with $t_ j \in I$. As $J = \mathop{\mathrm{Ker}}(B' \to B)$ maps isomorphically to $I$ we can choose $s_ j \in J \subset B'$ mapping to $t_ j$. Because $N' = N \times _{\varphi , M} M'$ it follows that $\xi = \sum b'_ i u_ i + \sum s_ j v_ j$ as desired. $\square$

Proof. In the proof we will use Lemma 15.6.4 without further mention.

Proof of (1). Set $J = \mathop{\mathrm{Ker}}(B' \to B)$. This is an ideal of $B'$ mapping isomorphically to $I = \mathop{\mathrm{Ker}}(A' \to A)$. Let $\mathfrak b' \subset B'$ be an ideal. We have to show that $\mathfrak b' \otimes _{B'} N' \to N'$ is injective, see Algebra, Lemma 10.39.5. We know that

is injective as $N$ is flat over $B$. As $\mathfrak b' \cap J \to \mathfrak b' \to \mathfrak b'/(\mathfrak b' \cap J) \to 0$ is exact, we conclude that it suffices to show that $(\mathfrak b' \cap J) \otimes _{B'} N' \to N'$ is injective. Thus we may assume that $\mathfrak b' \subset J$. Next, since $J \to I$ is an isomorphism we have

which maps injectively into $M'$ as $M'$ is a flat $A'$-module. Hence $J \otimes _{B'} N' \to N'$ is injective and we conclude that $\text{Tor}_1^{B'}(B'/J, N') = 0$, see Algebra, Remark 10.75.9. Thus we may apply Algebra, Lemma 10.99.8 to $N'$ over $B'$ and the ideal $J$. Going back to our ideal $\mathfrak b' \subset J$, let $\mathfrak b' \subset \mathfrak b'' \subset J$ be the smallest ideal whose image in $I$ is an $A'$-submodule of $I$. In other words, we have $\mathfrak b'' = A' \mathfrak b'$ if we view $J = I$ as $A'$-module. Then $\mathfrak b''/\mathfrak b'$ is killed by $J$ and we get a short exact sequence

by the vanishing of $\text{Tor}_1^{B'}(\mathfrak b''/\mathfrak b', N')$ we get from the application of the lemma. Thus we may replace $\mathfrak b'$ by $\mathfrak b''$. In particular we may assume $\mathfrak b'$ is an $A'$-module and maps to an ideal of $A'$. Then

This tensor product maps injectively into $M'$ by our assumption that $M'$ is flat over $A'$. We conclude that $\mathfrak b' \otimes _{B'} N' \to N' \to M'$ is injective and hence the first map is injective as desired.

Proof of (2). This follows by tensoring the short exact sequence $0 \to B' \to B \oplus A' \to A \to 0$ with $L'$ over $B'$.

Proof of (3). Immediate consequence of (1) and (2). $\square$

Proof. Recall that a module is finite projective if and only if it is finitely presented and flat, see Algebra, Lemma 10.78.2. Using Lemmas 15.6.8 and 15.6.7 we reduce to showing that $N' = N \times _{\varphi , M} M'$ is a $B'$-module of finite presentation if $N$ finite projective over $B$ and $M'$ finite projective over $A'$.

By Lemma 15.6.7 the module $N'$ is finite over $B'$. Choose a surjection $(B')^{\oplus n} \to N'$ with kernel $K'$. By base change we obtain maps $B^{\oplus n} \to N$, $(A')^{\oplus n} \to M'$, and $A^{\oplus n} \to M$ with kernels $K_ B$, $K_{A'}$, and $K_ A$. There is a canonical map

On the other hand, since $N' = N \times _{\varphi , M} M'$ and $B' = B \times _ A A'$ there is also a canonical map $K_ B \times _{K_ A} K_{A'} \to K'$ inverse to the displayed arrow. Hence the displayed map is an isomorphism. By Algebra, Lemma 10.5.3 the modules $K_ B$ and $K_{A'}$ are finite. We conclude from Lemma 15.6.7 that $K'$ is a finite $B'$-module provided that $K_ B \to K_ A$ and $K_{A'} \to K_ A$ induce isomorphisms $K_ B \otimes _ B A = K_ A = K_{A'} \otimes _{A'} A$. This is true because the flatness assumptions implies the sequences

stay exact upon tensoring, see Algebra, Lemma 10.39.12. $\square$