Lemma 38.12.9. Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. Let $x \in X$ with image $s \in S$. Assume that
$f$ is locally of finite type,
$\mathcal{F}$ is of finite type, and
$\mathcal{F}$ is flat at $x$ over $S$.
Then there exists an elementary étale neighbourhood $(S', s') \to (S, s)$ and a commutative diagram of pointed schemes
\[ \xymatrix{ (X, x) \ar[d] & (X', x') \ar[l]^ g \ar[d] \\ (S, s) & (\mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'}), s') \ar[l] } \]
such that $X' \to X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'})$ is étale, $\kappa (x) = \kappa (x')$, the scheme $X'$ is affine, and such that $\Gamma (X', g^*\mathcal{F})$ is a free $\mathcal{O}_{S', s'}$-module.
Proof.
(The only difference with Lemma 38.12.8 is that we do not assume $f$ is of finite presentation.) The problem is local on $X$ and $S$. Hence we may assume $X$ and $S$ are affine, say $X = \mathop{\mathrm{Spec}}(B)$ and $S = \mathop{\mathrm{Spec}}(A)$. Since $B$ is a finite type $A$-algebra we can find a surjection $A[x_1, \ldots , x_ n] \to B$. In other words, we can choose a closed immersion $i : X \to \mathbf{A}^ n_ S$. Set $t = i(x)$ and $\mathcal{G} = i_*\mathcal{F}$. Note that $\mathcal{G}_ t \cong \mathcal{F}_ x$ are $\mathcal{O}_{S, s}$-modules. Hence $\mathcal{G}$ is flat over $S$ at $t$. We apply Lemma 38.12.8 to the morphism $\mathbf{A}^ n_ S \to S$, the point $t$, and the sheaf $\mathcal{G}$. Thus we can find an elementary étale neighbourhood $(S', s') \to (S, s)$ and a commutative diagram of pointed schemes
\[ \xymatrix{ (\mathbf{A}^ n_ S, t) \ar[d] & (Y, y) \ar[l]^ h \ar[d] \\ (S, s) & (\mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'}), s') \ar[l] } \]
such that $Y \to \mathbf{A}^ n_{\mathcal{O}_{S', s'}}$ is étale, $\kappa (t) = \kappa (y)$, the scheme $Y$ is affine, and such that $\Gamma (Y, h^*\mathcal{G})$ is a projective $\mathcal{O}_{S', s'}$-module. Then a solution to the original problem is given by the closed subscheme $X' = Y \times _{\mathbf{A}^ n_ S} X$ of $Y$.
$\square$
Comments (0)