Openness of the finite free locus

Setup at a point of $Z_r$: let $x\in X$ with ${\mathcal{F}}_x\cong {\mathcal{O}}_{X,x}^r$; by [F1] choose an affine open $U=\mathrm{Spec}A$ containing $x$ with $\mathcal{F}{\mathrm{\mid }}_U\cong \tilde{M}$ for a finitely presented $A$-module $M$, and let $\mathfrak{p}\subseteq A$ be the prime with $x=\mathfrak{p}$. Then $M_{\mathfrak{p}}\cong A_{\mathfrak{p}}^r$, and by [F4] the fibre $\mathcal{F}(\mathfrak{p})=M_{\mathfrak{p}}{\otimes }_{A_{\mathfrak{p}}}\kappa (\mathfrak{p})=M{\otimes }_A\kappa (\mathfrak{p})$ is $\kappa (\mathfrak{p}{)}^r$, of dimension $r$.

A surjection on an affine neighbourhood: choose a finite generating list of $M$. Its images span $M{\otimes }_A\kappa (\mathfrak{p})$ over $\kappa (\mathfrak{p})$, since every tensor is a finite sum of scalar multiples of those images. Extract a basis from this finite spanning list and denote the corresponding elements of $M$ by $m_1,\dots ,m_r$. The $m_i$ are sections of $\mathcal{F}$ over $U$, so [F3(ii)] provides an affine open $V\subseteq U$ containing $x$ such that the induced morphism $\phi :{\mathcal{O}}_V^r\to \mathcal{F}{\mathrm{\mid }}_V$ is an epimorphism; after further shrinking $V$ to a finite-presentation chart supplied by [F1], it remains an epimorphism. For $r=0$ it says that $\mathcal{F}(\mathfrak{p})=0$.

The kernel of $\phi$ is finitely generated: write $V=\mathrm{Spec}B$ and $\mathcal{F}{\mathrm{\mid }}_V\cong \tilde{N}$ with $N$ a finitely presented $B$-module; by [F1] the epimorphism $\phi$ corresponds to a $B$-linear surjection $\psi :B^r\to N$, and [F6(a)] shows that $K=\ker \psi$ is finitely generated.

The map on stalks is an isomorphism: let $\mathfrak{q}\subseteq B$ be the prime with $x=\mathfrak{q}$. Both $B_{\mathfrak{q}}^r$ and $N_{\mathfrak{q}}$ are free of rank $r$ over the local ring $B_{\mathfrak{q}}$: the first is clear, and for the second ${\mathcal{F}}_x\cong {\mathcal{O}}_{X,x}^r=B_{\mathfrak{q}}^r$ while ${\mathcal{F}}_x=N_{\mathfrak{q}}$ because $\mathcal{F}{\mathrm{\mid }}_V\cong \tilde{N}$. The localised map ${\psi }_{\mathfrak{q}}:B_{\mathfrak{q}}^r\to N_{\mathfrak{q}}$ is surjective, so by [F6(c)] it is an isomorphism; in particular $K_{\mathfrak{q}}=\ker ({\psi }_{\mathfrak{q}})=0$ by [F5]. If $r=0$ this says that the stalk vanishes, and $\mathcal{F}(\mathfrak{p})=0$, so by [F3(i)] $\mathcal{F}$ vanishes on a neighbourhood of $x$ and $\mathcal{F}$ is free of rank $0$ there; assume $r\ge 1$ from now on.

Killing the kernel: since $K$ is finitely generated and $K_{\mathfrak{q}}=0$, [F6(b)] gives $g\in B\setminus \mathfrak{q}$ with $K_g=0$; then ${\psi }_g:B_g^r\to N_g$ is surjective with zero kernel, hence an isomorphism. Restricting the isomorphism $\mathcal{F}{\mathrm{\mid }}_V\cong \tilde{N}$ to $D(g)$ and using [F7], $\mathcal{F}{\mathrm{\mid }}_{D(g)}\cong \widetilde{(N_g)}\cong \widetilde{(B_g^r)}={\mathcal{O}}_{D(g)}^r$, and $D(g)$ is an open neighbourhood of $x$ in $X$.

Claims 1 and 2: the argument of steps 1.1, 2.1, 3.1, 4.1 and 5.1 applies to every $x\in Z_r(\mathcal{F})$ and produces an open neighbourhood $D(g)$ of $x$ with $\mathcal{F}{\mathrm{\mid }}_{D(g)}\cong {\mathcal{O}}_{D(g)}^r$; conversely, if $\mathcal{F}{\mathrm{\mid }}_W\cong {\mathcal{O}}_W^r$ then $F_y\cong {\mathcal{O}}_{X,y}^r$ for every $y\in W$, so such a $W$ is contained in $Z_r(\mathcal{F})$. Hence $Z_r(\mathcal{F})$ is open and is covered by the opens $D(g)$ on which $\mathcal{F}$ is free of rank $r$, proving claims 1 and 2.

Claim 3, sharpness: let $k$ be a field, $A=k[t]\mathrm{/}(t^2)$ with class $\epsilon$ of $t$, so that ${\epsilon }^2=0$ and $\epsilon \mathrm{\ne }0$, and $M=A\mathrm{/}(\epsilon )=k$; let $x$ be the unique point of $X=\mathrm{Spec}A$, so that ${\mathcal{O}}_{X,x}=A$ and $\kappa (x)=k$. Then $\mathcal{F}(x)=M{\otimes }_{A}k=M\mathrm{/}\epsilon M=k$, of dimension $1$, while ${\mathcal{F}}_x=M=k$ is not free of rank $1$ over $A$: a free module of rank $1$ is isomorphic to $A$, whose dimension over $k$ is $2$, and moreover $\epsilon$ annihilates every element of $M$ while $\epsilon \mathrm{\ne }0$ in $A$. The kernel of the surjection ${\mathcal{O}}_X\to \mathcal{F}$ that sends $1$ to the class of $1\in M$ is the subsheaf $\widetilde{(\epsilon )}$, nonzero at $x$; thus the fibre dimension $1$ and the vanishing of this kernel are genuinely different conditions, and $x\mathrm{\notin }{Z}_1(\mathcal{F})$ even though ${\dim }_{\kappa (x)}\mathcal{F}(x)=1$ and the set $\{x:\dim \mathcal{F}(x)=1\}$ happens to be all of $X$. This proves claim 3.

Choice accounting: all selections are finite: the $r$ sections $m_1,\dots ,m_r$ in step 2.1, the finitely many generators of $K$ in step 3.1 and the finitely many denominators in step 5.1. The assumed Axiom of Choice [F9] is consumed through the associated-sheaf machinery of [F1] and [F7] and through [F3(ii)] at step 2.1 (or [F3(i)] for $r=0$ at step 4.1); the local argument itself makes no new arbitrary-index selection. No Noetherian hypothesis is used.