Affine-local flatness

Fix affine opens $U=\mathrm{Spec}B\subseteq X$ and $V=\mathrm{Spec}A\subseteq S$ with $f(U)\subseteq V$; by [F6] the restriction corresponds to a ring map $\phi :A\to B$. Let $x\in U$ correspond to $\mathfrak{q}\in \mathrm{Spec}B$ and put $\mathfrak{p}=\mathfrak{q}\cap A$, so $s=f(x)$ corresponds to $\mathfrak{p}$. By [F1], ${\mathcal{O}}_{X,x}={\mathcal{O}}_{U,x}\cong B_{\mathfrak{q}}$ and ${\mathcal{O}}_{S,s}={\mathcal{O}}_{V,s}\cong A_{\mathfrak{p}}$, and the local ring map is the localisation of $\phi$ at $\mathfrak{q}$. Hence by [F2], $f$ is flat at $x$ if and only if $B_{\mathfrak{q}}$ is flat over $A_{\mathfrak{p}}$.

If $B$ is flat over $A$, then $B_{\mathfrak{q}}$ is flat over $A_{\mathfrak{p}}$ for every $\mathfrak{q}\in \mathrm{Spec}B$: localising $B$ at the multiplicative set $A\setminus \mathfrak{p}$ gives the flat $A_{\mathfrak{p}}$-module $B{\otimes }_A{A}_{\mathfrak{p}}$, and localising that further at the image of $B\setminus \mathfrak{q}$ gives $B_{\mathfrak{q}}$, flat over $A_{\mathfrak{p}}$ by [F4] applied twice. Thus $B$ flat over $A$ implies $f$ is flat at every point of $U$.

Conversely assume $B_{\mathfrak{q}}$ flat over $A_{\mathfrak{p}}$ for every $\mathfrak{q}\in \mathrm{Spec}B$. To prove $B$ flat over $A$ it suffices by [F3] to show that for every finitely generated ideal $I\subseteq A$ the kernel $K$ of $I{\otimes }_{A}B\to B$ is zero. By exactness of localisation [F5], for $\mathfrak{q}\in \mathrm{Spec}B$ we have $K{\otimes }_B{B}_{\mathfrak{q}}=\ker (I{\otimes }_A{B}_{\mathfrak{q}}\to B_{\mathfrak{q}})$, and this vanishes: $B_{\mathfrak{q}}$ is flat over $A_{\mathfrak{p}}$, so the multiplication map $I_{\mathfrak{p}}{\otimes }_{A_{\mathfrak{p}}}{B}_{\mathfrak{q}}\to B_{\mathfrak{q}}$ is injective by [F3], and $I{\otimes }_A{B}_{\mathfrak{q}}=I_{\mathfrak{p}}{\otimes }_{A_{\mathfrak{p}}}{B}_{\mathfrak{q}}$. If $K\mathrm{\ne }0$, then choosing a maximal ideal of $B$ containing the annihilator of a nonzero element of $K$ — here the Axiom of Choice is used, [F7] — produces $\mathfrak{q}\in \mathrm{Spec}B$ with $K_{\mathfrak{q}}\mathrm{\ne }0$. Hence $K=0$ for every finitely generated ideal $I$, and $B$ is flat over $A$ by [F3].

The last sentence of the statement: if $U^{\mathrm{\prime }}=\mathrm{Spec}{B}^{\mathrm{\prime }}\subseteq U$, $V^{\mathrm{\prime }}=\mathrm{Spec}{A}^{\mathrm{\prime }}\subseteq V$ is a refinement of affine charts with $f(U^{\mathrm{\prime }})\subseteq V^{\mathrm{\prime }}$, then for $x\in U^{\mathrm{\prime }}$ the local rings computed in the small and large charts are canonically the same local rings by [F1] applied to the two affine descriptions of the same open neighbourhoods, so the test of step 1.1 gives the same answer for the two charts. Hence flatness of $f$ at $x$ is independent of the chosen affine charts, flatness on $U$ is exactly flatness of $B$ over $A$ by steps 1.2 and 2.1, and the affine tests are compatible on refinements. [F1, step 1.1, step 1.2, step 2.1] $\mathrm{\square }$