Lemma 52.25.5. Assume $\text{depth}(A) \geq 3$ or equivalently $\text{depth}(A/fA) \geq 2$. Let $(\mathcal{L}, \mathcal{L}_0, \lambda )$ be an invertible coherent triple. Then

$\chi (\mathcal{L}, \mathcal{L}_0, \lambda ) = \text{length}_ A \mathop{\mathrm{Coker}}(\Gamma (U, \mathcal{L}) \to \Gamma (U_0, \mathcal{L}_0))$

and in particular this is $\geq 0$. Moreover, $\chi (\mathcal{L}, \mathcal{L}_0, \lambda ) = 0$ if and only if $\mathcal{L} \cong \mathcal{O}_ U$.

Proof. The equivalence of the depth conditions follows from Algebra, Lemma 10.72.7. By the depth condition we see that $\Gamma (U, \mathcal{O}_ U) = A$ and $\Gamma (U_0, \mathcal{O}_{U_0}) = A/fA$, see Dualizing Complexes, Lemma 47.11.1 and Local Cohomology, Lemma 51.8.2. Using Local Cohomology, Lemma 51.12.2 we find that $M = \Gamma (U, \mathcal{L})$ is a finite $A$-module. This in turn implies $\text{depth}(M) \geq 2$ for example by part (4) of Local Cohomology, Lemma 51.8.2 or by Divisors, Lemma 31.6.6. Also, we have $\mathcal{L}_0 \cong \mathcal{O}_{X_0}$ as $X_0$ is a local scheme. Hence we also see that $M_0 = \Gamma (X_0, \mathcal{L}_0) = \Gamma (U_0, \mathcal{L}_0|_{U_0})$ and that this module is isomorphic to $A/fA$.

By the above $\mathcal{F}' = \widetilde{M}$ is a coherent $\mathcal{O}_ X$-module whose restriction to $U$ is isomorphic to $\mathcal{L}$. The isomorphism $\lambda : \mathcal{L}/f\mathcal{L} \to \mathcal{L}_0|_{U_0}$ determines a map $M/fM \to M_0$ on global sections which is an isomorphism over $U_0$. Since $\text{depth}(M) \geq 2$ we see that $H^0_\mathfrak m(M/fM) = 0$ and it follows that $M/fM \to M_0$ is injective. Thus by definition

$\chi (\mathcal{L}, \mathcal{L}_0, \lambda ) = \text{length}_ A \mathop{\mathrm{Coker}}(M/fM \to M_0)$

which gives the first statement of the lemma.

Finally, if this length is $0$, then $M \to M_0$ is surjective. Hence we can find $s \in M = \Gamma (U, \mathcal{L})$ mapping to a trivializing section of $\mathcal{L}_0$. Consider the finite $A$-modules $K$, $Q$ defined by the exact sequence

$0 \to K \to A \xrightarrow {s} M \to Q \to 0$

The supports of $K$ and $Q$ do not meet $U_0$ because $s$ is nonzero at points of $U_0$. Using Algebra, Lemma 10.72.6 we see that $\text{depth}(K) \geq 2$ (observe that $As \subset M$ has $\text{depth} \geq 1$ as a submodule of $M$). Thus the support of $K$ if nonempty has dimension $\geq 2$ by Algebra, Lemma 10.72.3. This contradicts $\text{Supp}(M) \cap V(f) \subset \{ \mathfrak m\}$ unless $K = 0$. When $K = 0$ we find that $\text{depth}(Q) \geq 2$ and we conclude $Q = 0$ as before. Hence $A \cong M$ and $\mathcal{L}$ is trivial. $\square$

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