The Stacks project

Proof. We first reduce to the case where $X \to S$ has a section. Say $S = \mathop{\mathrm{Spec}}(A)$. Choose a closed point $\xi $ of $X_\eta $. Choose an extension of discrete valuation rings $A \subset B$ such that the fraction field of $B$ is $\kappa (\xi )$. This is possible by Krull-Akizuki (Algebra, Lemma 10.120.18) and the fact that $\kappa (\xi )$ is a finite extension of the fraction field of $A$. By the valuative criterion of properness (Morphisms, Lemma 29.42.1) we get a $B$-valued point $\tau : \mathop{\mathrm{Spec}}(B) \to X$ which induces a section $\sigma : \mathop{\mathrm{Spec}}(B) \to X_ B$. For a finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$ let $\mathcal{E}_ B$ be the pullback to the base change $X_ B$. By flat base change (Cohomology of Schemes, Lemma 30.5.2) we see that $H^0(X_ B, \mathcal{E}_ B) = H^0(X, \mathcal{E}) \otimes _ A B$. Thus if $\mathcal{E}_ B$ is free of rank $r$, then the sections in $H^0(X, \mathcal{E})$ generate the free $B$-module $\tau ^*\mathcal{E} = \sigma ^*\mathcal{E}_ B$. In particular, we can find $r$ global sections $s_1, \ldots , s_ r$ of $\mathcal{E}$ which generate $\tau ^*\mathcal{E}$. Then

is a map of finite locally free $\mathcal{O}_ X$-modules of rank $r$ and the pullback to $X_ B$ is a map of free $\mathcal{O}_{X_ B}$-modules which restricts to an isomorphism in one point of each fibre. Taking the determinant we get a function $g \in \Gamma (X_\eta , \mathcal{O}_{X_ B})$ which is invertible in one point of each fibre. As the fibres are proper and connected, we see that $g$ must be invertible (details omitted; hint: use Varieties, Lemma 33.9.3). Thus it suffices to prove the lemma for the base change $X_ B \to \mathop{\mathrm{Spec}}(B)$.

Assume we have a section $\sigma : S \to X$. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ X$-module which is assumed free on the generic fibre and on $X_ n$ (we will choose $n$ later). Choose an isomorphism $\sigma ^*\mathcal{E} = \mathcal{O}_ S^{\oplus r}$. Consider the map

in $D(A)$. Arguing as above, we see $\mathcal{E}$ is free if (and only if) the induced map $H^0(K) = H^0(X, \mathcal{E}) \to A^{\oplus r}$ is surjective.

Set $L = R\Gamma (X, \mathcal{O}_ X^{\oplus r})$ and observe that the corresponding map $L \to A^{\oplus r}$ has the desired property. Observe that $K \otimes _ A Q(A) \cong L \otimes _ A Q(A)$ by flat base change and the assumption that $\mathcal{E}$ is free on the generic fibre. Let $\pi \in A$ be a uniformizer. Observe that

and similarly for $L$. Denote $\mathcal{E}_{tors} \subset \mathcal{E}$ the coherent subsheaf of sections supported on the special fibre and similarly for other $\mathcal{O}_ X$-modules. Choose $k > 0$ such that $(\mathcal{O}_ X)_{tors} \to \mathcal{O}_ X/\pi ^ k \mathcal{O}_ X$ is injective (Cohomology of Schemes, Lemma 30.10.3). Since $\mathcal{E}$ is locally free, we see that $\mathcal{E}_{tors} \subset \mathcal{E}/\pi ^ k\mathcal{E}$. Then for $n \geq m + k$ we have isomorphisms

in $D(\mathcal{O}_ X)$. This determines an isomorphism

in $D(A)$ (holds when $n \geq m + k$). Observe that these isomorphisms are compatible with pulling back by $\sigma $ hence in particular we conclude that $K \otimes _ A^\mathbf {L} A/\pi ^ m A \to (A/\pi ^ m A)^{\oplus r}$ defines an surjection on degree $0$ cohomology modules (as this is true for $L$). Since $A$ is a discrete valuation ring, we have

in $D(A)$. See More on Algebra, Example 15.69.3. The cohomology groups $H^ i(K) = H^ i(X, \mathcal{E})$ and $H^ i(L) = H^ i(X, \mathcal{O}_ X)^{\oplus r}$ are finite $A$-modules by Cohomology of Schemes, Lemma 30.19.2. By More on Algebra, Lemma 15.124.3 these modules are direct sums of cyclic modules. We have seen above that the rank $\beta _ i$ of the free part of $H^ i(K)$ and $H^ i(L)$ are the same. Next, observe that

and similarly for $K$. Let $e$ be the largest integer such that $A/\pi ^ eA$ occurs as a summand of $H^ i(X, \mathcal{O}_ X)$, or equivalently $H^ i(L)$, for some $i$. Then taking $m = e + 1$ we see that $H^ i(L \otimes _ A^\mathbf {L} A/\pi ^ m A)$ is a direct sum of $\beta _ i$ copies of $A/\pi ^ m A$ and some other cyclic modules each annihilated by $\pi ^ e$. By the same reasoning for $K$ and the isomorphism $K \otimes _ A^\mathbf {L} A/\pi ^ m A \cong L \otimes _ A^\mathbf {L} A/\pi ^ m A$ it follows that $H^ i(K)$ cannot have any cyclic summands of the form $A/\pi ^ l A$ with $l > e$. (It also follows that $K$ is isomorphic to $L$ as an object of $D(A)$, but we won't need this.) Then the only way the map

is surjective, is if it is surjective on the first summand. This is what we wanted to show. (To be precise, the integer $n$ in the statement of the lemma, if there is a section $\sigma $, should be equal to $k + e + 1$ where $k$ and $e$ are as above and depend only on $X$.) $\square$