The Stacks project

Lemma 90.5.2. Let $(f, f')$ be a morphism of first order thickenings of ringed spaces as in Situation 90.3.1. Let $\mathcal{F}'$ be an $\mathcal{O}_{X'}$-module and set $\mathcal{F} = i^*\mathcal{F}'$. Assume that $\mathcal{F}$ is flat over $S$ and that $(f, f')$ is a strict morphism of thickenings (Definition 90.3.2). Then the following are equivalent

  1. $\mathcal{F}'$ is flat over $S'$, and

  2. the canonical map $f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I}\mathcal{F}'$ is an isomorphism.

Moreover, in this case the maps

\[ f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I}\mathcal{F}' \]

are isomorphisms.

Proof. The map $f^*\mathcal{J} \to \mathcal{I}$ is surjective as $(f, f')$ is a strict morphism of thickenings. Hence the final statement is a consequence of (2).

Proof of the equivalence of (1) and (2). We may check these conditions at stalks. Let $x \in X \subset X'$ be a point with image $s = f(x) \in S \subset S'$. Set $A' = \mathcal{O}_{S', s}$, $B' = \mathcal{O}_{X', x}$, $A = \mathcal{O}_{S, s}$, and $B = \mathcal{O}_{X, x}$. Then $A = A'/J$ and $B = B'/I$ for some square zero ideals. Since $(f, f')$ is a strict morphism of thickenings we have $I = JB'$. Let $M' = \mathcal{F}'_ x$ and $M = \mathcal{F}_ x$. Then $M'$ is a $B'$-module and $M$ is a $B$-module. Since $\mathcal{F} = i^*\mathcal{F}'$ we see that the kernel of the surjection $M' \to M$ is $IM' = JM'$. Thus we have a short exact sequence

\[ 0 \to JM' \to M' \to M \to 0 \]

Using Sheaves, Lemma 6.26.4 and Modules, Lemma 17.16.1 to identify stalks of pullbacks and tensor products we see that the stalk at $x$ of the canonical map of the lemma is the map

\[ (J \otimes _ A B) \otimes _ B M = J \otimes _ A M = J \otimes _{A'} M' \longrightarrow JM' \]

The assumption that $\mathcal{F}$ is flat over $S$ signifies that $M$ is a flat $A$-module.

Assume (1). Flatness implies $\text{Tor}_1^{A'}(M', A) = 0$ by Algebra, Lemma 10.75.8. This means $J \otimes _{A'} M' \to M'$ is injective by Algebra, Remark 10.75.9. Hence $J \otimes _ A M \to JM'$ is an isomorphism.

Assume (2). Then $J \otimes _{A'} M' \to M'$ is injective. Hence $\text{Tor}_1^{A'}(M', A) = 0$ by Algebra, Remark 10.75.9. Hence $M'$ is flat over $A'$ by Algebra, Lemma 10.99.8. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08LI. Beware of the difference between the letter 'O' and the digit '0'.