Lemma 52.12.7. Let A be a Noetherian ring. Let f \in \mathfrak a \subset A be an element of an ideal of A. Let M be a finite A-module. Assume
A is f-adically complete,
H^1_\mathfrak a(M) and H^2_\mathfrak a(M) are annihilated by a power of f.
Then with U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a) the map
\Gamma (U, \widetilde{M}) \longrightarrow \mathop{\mathrm{lim}}\nolimits \Gamma (U, \widetilde{M/f^ nM})
is an isomorphism.
Proof.
We may apply Lemma 52.3.2 to U and \mathcal{F} = \widetilde{M}|_ U because \mathcal{F} is a Noetherian object in the category of coherent \mathcal{O}_ U-modules. Since H^1(U, \mathcal{F}) = H^2_\mathfrak a(M) (Local Cohomology, Lemma 51.8.2) is annihilated by a power of f, we see that its f-adic Tate module is zero. Hence the lemma shows \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/f^ n \mathcal{F}) is equal to the usual f-adic completion of H^0(U, \mathcal{F}). Consider the short exact sequence
0 \to M/H^0_\mathfrak a(M) \to H^0(U, \mathcal{F}) \to H^1_\mathfrak a(M) \to 0
of Local Cohomology, Lemma 51.8.2. Since M/H^0_\mathfrak a(M) is a finite A-module, it is complete, see Algebra, Lemma 10.97.1. Since H^1_\mathfrak a(M) is killed by a power of f, we conclude from Algebra, Lemma 10.96.4 that H^0(U, \mathcal{F}) is complete as well. This finishes the proof.
\square
Comments (0)