Lemma 10.134.16 (00S6)—The Stacks project

Lemma 10.134.16. Let $R \to S$ be a ring map of finite type. Let $g \in S$. For any presentations $\alpha : R[x_1, \ldots , x_ n] \to S$, and $\beta : R[y_1, \ldots , y_ m] \to S_ g$ we have

\[ (I/I^2)_ g \oplus S^{\oplus m}_ g \cong J/J^2 \oplus S_ g^{\oplus n} \]

as $S_ g$-modules where $I = \mathop{\mathrm{Ker}}(\alpha )$ and $J = \mathop{\mathrm{Ker}}(\beta )$.

Proof. Let $\beta ' : R[x_1, \ldots , x_ n, x] \to S_ g$ be the presentation of Lemma 10.134.12 constructed starting with $\alpha $. Then we know that $\mathop{N\! L}\nolimits (\alpha ) \otimes _ S S_ g$ is homotopy equivalent to $\mathop{N\! L}\nolimits (\beta ')$. We know that $\mathop{N\! L}\nolimits (\beta )$ and $\mathop{N\! L}\nolimits (\beta ')$ are homotopy equivalent by Lemma 10.134.2. We conclude that $\mathop{N\! L}\nolimits (\alpha ) \otimes _ S S_ g$ is homotopy equivalent to $\mathop{N\! L}\nolimits (\beta )$. Finally, we apply Lemma 10.134.15. $\square$

Comments (3)

Comment #7436 by nkym on June 20, 2022 at 23:36

The result for a single choice of $\alpha$ and $\beta$ seem to yield not the result for the general case, but a similar isomorphism with more $S_g$ added than $m$ or $n$ .

Comment #7441 by Johan on June 21, 2022 at 11:34

Good catch! The fix is as follows. Let $\beta' : R[x_1, \ldots, x_n, x] \to S_g$ be the representation given in Lemma 10.134.12 using $\alpha$ . Then we know that $NL(\alpha) \otimes_B B_g$ is homotopy equivalent to $NL(\beta')$ . We know that $NL(\beta)$ and $NL(\beta)$ are homotopy equivalent by Lemma 10.134.2. Hence we see that $NL(\alpha) \otimes_B B_g$ and $NL(\beta)$ are homotopy equivalent. Finally, we apply Lemma 10.134.14.

Comment #7598 by Stacks Project on August 13, 2022 at 18:29

Thanks. Fixed here.

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