The Stacks project

Lemma 37.30.3. Let $f : X \to S$ be a proper flat morphism of finite presentation. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Assume

  1. $S$ is the spectrum of a valuation ring,

  2. $\mathcal{L}$ is trivial on the generic fibre $X_\eta $ of $f$,

  3. the closed fibre $X_0$ of $f$ is integral,

  4. $H^0(X_\eta , \mathcal{O}_{X_\eta })$ is equal to the function field of $S$.

Then $\mathcal{L}$ is trivial.

Proof. Write $S = \mathop{\mathrm{Spec}}(A)$. We will first prove the lemma when $A$ is a discrete valuation ring (as this is the case most often used in practice). Let $\pi \in A$ be a uniformizer. Take a trivializing section $s \in \Gamma (X_\eta , \mathcal{L}_\eta )$. After replacing $s$ by $\pi ^ n s$ if necessary we can assume that $s \in \Gamma (X, \mathcal{L})$. If $s|_{X_0} = 0$, then we see that $s$ is divisible by $\pi $ (because $X_0$ is the scheme theoretic fibre and $X$ is flat over $A$). Thus we may assume that $s|_{X_0}$ is nonzero. Then the zero locus $Z(s)$ of $s$ is contained in $X_0$ but does not contain the generic point of $X_0$ (because $X_0$ is integral). This means that the $Z(s)$ has codimension $\geq 2$ in $X$ which contradicts Divisors, Lemma 31.15.3 unless $Z(s) = \emptyset $ as desired.

Proof in the general case. Since the valuation ring $A$ is coherent (Algebra, Example 10.90.2) we see that $H^0(X, \mathcal{L})$ is a coherent $A$-module, see Derived Categories of Schemes, Lemma 36.33.1. Equivalently, $H^0(X, \mathcal{L})$ is a finitely presented $A$-module (Algebra, Lemma 10.90.4). Since $H^0(X, \mathcal{L})$ is torsion free (by flatness of $X$ over $A$), we see from More on Algebra, Lemma 15.121.3 that $H^0(X, \mathcal{L}) = A^{\oplus n}$ for some $n$. By flat base change (Cohomology of Schemes, Lemma 30.5.2) we have

\[ K = H^0(X_\eta , \mathcal{O}_{X_\eta }) \cong H^0(X_\eta , \mathcal{L}_\eta ) = H^0(X, \mathcal{L}) \otimes _ A K \]

where $K$ is the fraction field of $A$. Thus $n = 1$. Pick a generator $s \in H^0(X, \mathcal{L})$. Let $\mathfrak m \subset A$ be the maximal ideal. Then $\kappa = A/\mathfrak m = \mathop{\mathrm{colim}}\nolimits A/\pi $ where this is a filtered colimit over nonzero $\pi \in \mathfrak m$ (here we use that $A$ is a valuation ring). Thus $X_0 = \mathop{\mathrm{lim}}\nolimits X \times _ S \mathop{\mathrm{Spec}}(A/\pi )$. If $s|_{X_0}$ is zero, then for some $\pi $ we see that $s$ restricts to zero on $X \times _ S \mathop{\mathrm{Spec}}(A/\pi )$, see Limits, Lemma 32.4.7. But if this happens, then $\pi ^{-1} s$ is a global section of $\mathcal{L}$ which contradicts the fact that $s$ is a generator of $H^0(X, \mathcal{L})$. Thus $s|_{X_0}$ is not zero. Let $Z(s) \subset X$ be the zero scheme of $s$. Since $s|_{X_0}$ is not zero and since $X_0$ is integral, we see that $Z(s)_0 \subset X_0$ is an effective Cartier divisor. Since $f$ is proper and $S$ is local, every point of $Z(s)$ specializes to a point of $Z(s)_0$. Thus by Divisors, Lemma 31.18.9 part (3) we see that $Z(s)$ is a relative effective Cartier divisor, in particular $Z(s) \to S$ is flat. Hence if $Z(s)$ were nonemtpy, then $Z(s)_\eta $ would be nonempty which contradicts the fact that $s|_{X_\eta }$ is a trivialization of $\mathcal{L}_\eta $. Thus $Z(s) = \emptyset $ as desired. $\square$


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