Lemma 49.15.5. Let $A$ be a Noetherian ring. Let $f \in \mathfrak a$ be an element of an ideal of $A$. Let $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$. Assume
$A$ has a dualizing complex and is complete with respect to $f$,
$A_ f$ is $(S_2)$ and for every minimal prime $\mathfrak p \subset A$, $f \not\in \mathfrak p$ and $\mathfrak q \in V(\mathfrak p) \cap V(\mathfrak a)$ we have $\dim ((A/\mathfrak p)_\mathfrak q) \geq 3$.
Then the completion functor
\[ \textit{Coh}(\mathcal{O}_ U) \longrightarrow \textit{Coh}(U, I\mathcal{O}_ U), \quad \mathcal{F} \longmapsto \mathcal{F}^\wedge \]
is fully faithful on the full subcategory of finite locally free objects.
Proof.
We will show that Lemma 49.15.4 applies. Assumption (1) of Lemma 49.15.4 holds. Observe that $\text{cd}(A, (f)) \leq 1$, see Local Cohomology, Lemma 48.3.3. Since $A_ f$ is $(S_2)$ we see that every associated prime $\mathfrak p \subset A$, $f \not\in \mathfrak p$ is a minimal prime. Thus we get assumption (2) of Lemma 49.15.4. If $\mathfrak p \subset A$, $f \not\in \mathfrak p$ satisfies $V(\mathfrak p) \cap V(I) \subset V(\mathfrak a)$ and if $\mathfrak q \in V(\mathfrak p) \cap V(f)$ is a generic point, then $\dim ((A/\mathfrak p)_\mathfrak q) = 1$. Then we obtain $\dim (A_\mathfrak p) \geq 2$ by looking at the minimal primes $\mathfrak p_0 \subset \mathfrak p$ and using that $\dim ((A/\mathfrak p_0)_\mathfrak q) \geq 3$ by assumption. Thus $\text{depth}(A_\mathfrak p) \geq 2$ by the $(S_2)$ assumption. This verifies assumption (3) of Lemma 49.15.4 and the proof is complete.
$\square$
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