Lemma 37.61.3. Let $X \to Y$ be a morphism of schemes such that $X \to X \times _ Y X$ is flat. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. If $\mathcal{F}$ is flat over $Y$, then $\mathcal{F}$ is flat over $X$.

Proof. Let $x \in X$ with image $y = f(x)$ in $Y$. Since $X \to X \times _ Y X$ is flat, we see that $\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{X, x} \to \mathcal{O}_{X, x}$ is flat. Hence the result follows from More on Algebra, Lemma 15.104.2 and the definitions. $\square$

Comment #6299 by on

It seems to me that we can reproduce the argument in Tag 092C instead of referring to it, since it is even clearer in this context, without passing to stalks:

We need to check that the functor $\mathcal G\mapsto\mathcal G\otimes\mathcal F$ is exact. The point is that we can take the "external tensor product" (this is essentially the argument in Tag 092C, but it seems even clearer when we write it in the language of algebraic geometry): let $p_1,p_2\colon X\times_YX\to X$ be two projections of which a common section is the diagonal $j\colon X\to X\times_YX$. Then we have $\mathcal G\otimes\mathcal F\cong j^*(p_1^*\mathcal G\otimes p_2^*\mathcal F)$. Note that the pushforward of $p_1^*\mathcal G\otimes p_2^*\mathcal F$ to $Y$ is the tensor product of $\mathcal G$ and $\mathcal F$ over $\mathcal O_Y$, the result follows from the assumed flatness.

Comment #6411 by on

OK, yes, but I am going to leave as is for now.

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