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

Comment #7126 by Johan on

Comment #7127 by Johan on