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

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

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.
$\square$

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