Lemma 15.85.2. Let $R \to S$ and $R \to R'$ be ring maps. Let $\alpha : P \to S$ be a presentation of $S$ over $R$. Then $\alpha ' : P \otimes _ R R' \to S \otimes _ R R'$ is a presentation of $S' = S \otimes _ R R'$ over $R'$. The canonical map
\[ NL(\alpha ) \otimes _ S S' \to \mathop{N\! L}\nolimits (\alpha ') \]
is an isomorphism on $H^0$ and surjective on $H^{-1}$. In particular, the canonical map
\[ \mathop{N\! L}\nolimits _{S/R} \otimes _ S S' \to \mathop{N\! L}\nolimits _{S'/R'} \]
is an isomorphism on $H^0$ and surjective on $H^{-1}$.
Proof.
Denote $I = \mathop{\mathrm{Ker}}(P \to S)$. Denote $P' = P \otimes _ R R'$ and $I' = \mathop{\mathrm{Ker}}(P' \to S')$. Suppose $P$ is a polynomial algebra on $x_ j$ for $j \in J$. The map displayed in the lemma becomes
\[ \xymatrix{ \bigoplus _{j \in J} S' \text{d}x_ j \ar[r] & \bigoplus _{j \in J} S' \text{d}x_ j \\ I/I^2 \otimes _ S S' \ar[r] \ar[u] & I'/(I')^2 \ar[u] } \]
where the left column is $\mathop{N\! L}\nolimits (\alpha ) \otimes _ S S'$ and the right column is $\mathop{N\! L}\nolimits (\alpha ')$. By right exactness of tensor product we see that $I \otimes _ R R' \to I'$ is surjective. Hence the bottom arrow is a surjection. This proves the first statement of the lemma. The statement for $\mathop{N\! L}\nolimits _{S/R} \otimes _ S S' \to \mathop{N\! L}\nolimits _{S'/R'}$ follows as these complexes are homotopic to $\mathop{N\! L}\nolimits (\alpha ) \otimes _ S S'$ and $\mathop{N\! L}\nolimits (\alpha ')$.
$\square$
Comments (0)