Here is the “Critère de platitude par fibres” in the Noetherian case.
Theorem 76.24.1. Let $S$ be a scheme. Let $f : X \to Y$ and $Y \to Z$ be morphisms of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume
$X$, $Y$, $Z$ locally Noetherian, and
$\mathcal{F}$ a coherent $\mathcal{O}_ X$-module.
Let $x \in |X|$ and let $y \in |Y|$ and $z \in |Z|$ be the images of $x$. If $\mathcal{F}_{\overline{x}} \not= 0$, then the following are equivalent:
$\mathcal{F}$ is flat over $Z$ at $x$ and the restriction of $\mathcal{F}$ to its fibre over $z$ is flat at $x$ over the fibre of $Y$ over $z$, and
$Y$ is flat over $Z$ at $y$ and $\mathcal{F}$ is flat over $Y$ at $x$.
Proof. Choose a diagram as in Lemma 76.23.1 part (3). It follows from the definitions that this reduces to the corresponding theorem for the morphisms of schemes $U \to V \to W$, the quasi-coherent sheaf $a^*\mathcal{F}$, and the point $u \in U$. Thus the theorem follows from the corresponding result for schemes which is More on Morphisms, Theorem 37.16.1. $\square$
Lemma 76.24.2. Let $S$ be a scheme. Let $f : X \to Y$ and $Y \to Z$ be a morphism of algebraic spaces over $S$. Assume
$X$, $Y$, $Z$ locally Noetherian,
$X$ is flat over $Z$,
for every $z \in |Z|$ the fibre of $X$ over $z$ is flat over the fibre of $Y$ over $z$.
Then $f$ is flat. If $f$ is also surjective, then $Y$ is flat over $Z$.
Proof. This is a special case of Theorem 76.24.1. $\square$
Just like for checking smoothness, if the base is Noetherian it suffices to check flatness over Artinian rings. Here is a sample statement.
Lemma 76.24.3. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $X$ be an algebraic space locally of finite presentation over $S = \mathop{\mathrm{Spec}}(A)$. For $n \geq 1$ set $S_ n = \mathop{\mathrm{Spec}}(A/I^ n)$ and $X_ n = S_ n \times _ S X$. Let $\mathcal{F}$ be coherent $\mathcal{O}_ X$-module. If for every $n \geq 1$ the pullback $\mathcal{F}_ n$ of $\mathcal{F}$ to $X$ is flat over $S_ n$, then the (open) locus where $\mathcal{F}$ is flat over $X$ contains the inverse image of $V(I)$ under $X \to S$.
Proof. The locus where $\mathcal{F}$ is flat over $S$ is open in $|X|$ by Theorem 76.22.1. The statement is insensitive to replacing $X$ by the members of an étale covering, hence we may assume $X$ is an affine scheme. In this case the result follows immediately from Algebra, Lemma 10.99.11. Some details omitted. $\square$
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