This result takes some work to prove, and (perhaps) deserves its own section. Here it is.
Theorem 37.15.1. Let $S$ be a scheme. Let $f : X \to S$ be a morphism which is locally of finite presentation. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module which is locally of finite presentation. Then is open in $X$.
\[ U = \{ x \in X \mid \mathcal{F}\text{ is flat over }S\text{ at }x\} \]
Proof. We may test for openness locally on $X$ hence we may assume that $f$ is a morphism of affine schemes. In this case the theorem is exactly Algebra, Theorem 10.129.4. $\square$
Lemma 37.15.2. Let $S$ be a scheme. Let be a cartesian diagram of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $x' \in X'$ with images $x = g'(x')$ and $s' = f'(x')$.
If $\mathcal{F}$ is flat over $S$ at $x$, then $(g')^*\mathcal{F}$ is flat over $S'$ at $x'$.
If $g$ is flat at $s'$ and $(g')^*\mathcal{F}$ is flat over $S'$ at $x'$, then $\mathcal{F}$ is flat over $S$ at $x$.
\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S } \]
In particular, if $g$ is flat, $f$ is locally of finite presentation, and $\mathcal{F}$ is locally of finite presentation, then formation of the open subset of Theorem 37.15.1 commutes with base change.
Proof. Consider the commutative diagram of local rings
Note that $\mathcal{O}_{X', x'}$ is a localization of $\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{S', s'}$, and that $((g')^*\mathcal{F})_{x'}$ is equal to $\mathcal{F}_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{X', x'}$. Hence the lemma follows from Algebra, Lemma 10.100.1. $\square$
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