Lemma 52.6.3. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $K \in D(\mathcal{O})$. The rule which associates to $U$ the set $\mathcal{I}(U)$ of sections $f \in \mathcal{O}(U)$ such that $T(K|_ U, f) = 0$ is a sheaf of ideals in $\mathcal{O}$.

Proof. We will use the results of Lemma 52.6.2 without further mention. If $f \in \mathcal{I}(U)$, and $g \in \mathcal{O}(U)$, then $\mathcal{O}_{U, gf}$ is an $\mathcal{O}_{U, f}$-module hence $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_{U, gf}, K|_ U) = 0$, hence $gf \in \mathcal{I}(U)$. Suppose $f, g \in \mathcal{O}(U)$. Then there is a short exact sequence

$0 \to \mathcal{O}_{U, f + g} \to \mathcal{O}_{U, f(f + g)} \oplus \mathcal{O}_{U, g(f + g)} \to \mathcal{O}_{U, gf(f + g)} \to 0$

because $f, g$ generate the unit ideal in $\mathcal{O}(U)_{f + g}$. This follows from Algebra, Lemma 10.24.2 and the easy fact that the last arrow is surjective. Because $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}( - , K|_ U)$ is an exact functor of triangulated categories the vanishing of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_{U, f(f + g)}, K|_ U)$, $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_{U, g(f + g)}, K|_ U)$, and $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_{U, gf(f + g)}, K|_ U)$, implies the vanishing of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_{U, f + g}, K|_ U)$. We omit the verification of the sheaf condition. $\square$

There are also:

• 2 comment(s) on Section 52.6: Derived completion on a ringed site

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).