Theorem 37.16.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $x \in X$. Set $y = f(x)$ and $s \in S$ the image of $x$ in $S$. Assume $S$, $X$, $Y$ locally Noetherian, $\mathcal{F}$ coherent, and $\mathcal{F}_ x \not= 0$. Then the following are equivalent:

1. $\mathcal{F}$ is flat over $S$ at $x$, and $\mathcal{F}_ s$ is flat over $Y_ s$ at $x$, and

2. $Y$ is flat over $S$ at $y$ and $\mathcal{F}$ is flat over $Y$ at $x$.

Proof. Consider the ring maps

$\mathcal{O}_{S, s} \longrightarrow \mathcal{O}_{Y, y} \longrightarrow \mathcal{O}_{X, x}$

and the module $\mathcal{F}_ x$. The stalk of $\mathcal{F}_ s$ at $x$ is the module $\mathcal{F}_ x/\mathfrak m_ s \mathcal{F}_ x$ and the local ring of $Y_ s$ at $y$ is $\mathcal{O}_{Y, y}/\mathfrak m_ s \mathcal{O}_{Y, y}$. Thus the implication (1) $\Rightarrow$ (2) is Algebra, Lemma 10.99.15. If (2) holds, then the first ring map is faithfully flat and $\mathcal{F}_ x$ is flat over $\mathcal{O}_{Y, y}$ so by Algebra, Lemma 10.39.4 we see that $\mathcal{F}_ x$ is flat over $\mathcal{O}_{S, s}$. Moreover, $\mathcal{F}_ x/\mathfrak m_ s \mathcal{F}_ x$ is the base change of the flat module $\mathcal{F}_ x$ by $\mathcal{O}_{Y, y} \to \mathcal{O}_{Y, y}/\mathfrak m_ s \mathcal{O}_{Y, y}$, hence flat by Algebra, Lemma 10.39.7. $\square$

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).