Lemma 15.31.3 (063Q)—The Stacks project

Lemma 15.31.3. Let $A \to B$ be a ring map. Let $f_1, \ldots , f_ r$ be a sequence in $B$ such that $B/(f_1, \ldots , f_ r)$ is $A$-flat. Let $A \to A'$ be a ring map. Then the canonical map

\[ H_1(K_\bullet (B, f_1, \ldots , f_ r)) \otimes _ A A' \longrightarrow H_1(K_\bullet (B', f'_1, \ldots , f'_ r)) \]

is surjective. Here $B' = B \otimes _ A A'$ and $f_ i' \in B'$ is the image of $f_ i$.

Proof. The sequence

\[ \wedge ^2(B^{\oplus r}) \to B^{\oplus r} \to B \to B/J \to 0 \]

is a complex of $A$-modules with $B/J$ flat over $A$ and cohomology group $H_1 = H_1(K_\bullet (B, f_1, \ldots , f_ r))$ in the spot $B^{\oplus r}$. If we tensor this with $A'$ we obtain a complex

\[ \wedge ^2((B')^{\oplus r}) \to (B')^{\oplus r} \to B' \to B'/J' \to 0 \]

which is exact at $B'$ and $B'/J'$. In order to compute its cohomology group $H'_1 = H_1(K_\bullet (B', f'_1, \ldots , f'_ r))$ at $(B')^{\oplus r}$ we split the first sequence above into the exact sequences $0 \to J \to B \to B/J \to 0$, $0 \to K \to B^{\oplus r} \to J \to 0$, and $\wedge ^2(B^{\oplus r}) \to K \to H_1 \to 0$. Tensoring over $A$ with $A'$ we obtain the exact sequences

\[ \begin{matrix} 0 \to J \otimes _ A A' \to B \otimes _ A A' \to (B/J) \otimes _ A A' \to 0 \\ K \otimes _ A A' \to B^{\oplus r} \otimes _ A A' \to J \otimes _ A A' \to 0 \\ \wedge ^2(B^{\oplus r}) \otimes _ A A' \to K \otimes _ A A' \to H_1 \otimes _ A A' \to 0 \end{matrix} \]

where the first one is exact as $B/J$ is flat over $A$, see Algebra, Lemma 10.39.12. We conclude that $J' = J \otimes _ A A' \subset B'$ and that $K \otimes _ A A' \to \mathop{\mathrm{Ker}}((B')^{\oplus r} \to B')$ is surjective. Thus

\begin{align*} H_1 \otimes _ A A' & = \mathop{\mathrm{Coker}}\left(\wedge ^2(B^{\oplus r}) \otimes _ A A' \to K \otimes _ A A'\right) \\ & \to \mathop{\mathrm{Coker}}\left( \wedge ^2((B')^{\oplus r}) \to \mathop{\mathrm{Ker}}((B')^{\oplus r} \to B') \right) = H'_1 \end{align*}

is surjective too. $\square$

Comments (3)

Comment #7125 by Xuande Liu on March 21, 2022 at 16:49

I cannot figure out why $Ker((B')^{\oplus r}\to B')=K\otimes_A A'$ . Could someone give me a hint?

Comment #7126 by Johan on March 21, 2022 at 19:16

Oops! The lemma should say that the map is just surjective and this follows because $K \otimes_A A'$ surjects onto the kernel of $(B')^{\oplus r} \to B'$ . I just checked and the lemma originally used to just say this but in an edit in 2016 changed the statement to "isomorphism" without justifying why! Luckily, the only lemmas using this lemma rely on the original statement of surjectivity! I'm going to fix this right now. Thanks very much!

Comment #7127 by Johan on March 21, 2022 at 19:52

Thanks again for pointing this out. Fixed here. The webpage should be updated in a few minutes. Furthermore, in case you are interested, I observe that if $B$ is flat over $A$ , then $J$ is a kernel of a surjection of flat $A$ -modules, hence $J$ is itself flat over $A$ and we do get $K \otimes_A A' \to (B')^{\oplus r}$ injective and we do get an isomorphism in the statement of the lemma.

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