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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