Lemma 51.14.1. Let $A$ be a Noetherian ring. Let $T \subset \mathop{\mathrm{Spec}}(A)$ be a subset stable under specialization. Let $K$ be an object of $D_{\textit{Coh}}^+(A)$. Let $n \in \mathbf{Z}$. The following are equivalent

$H^ i_ T(K)$ is finite for $i \leq n$,

there exists an ideal $J \subset A$ with $V(J) \subset T$ such that $J$ annihilates $H^ i_ T(K)$ for $i \leq n$.

If $T = V(I) = Z$ for an ideal $I \subset A$, then these are also equivalent to

there exists an $e \geq 0$ such that $I^ e$ annihilates $H^ i_ Z(K)$ for $i \leq n$.

**Proof.**
This lemma is the natural generalization of Lemma 51.7.1 whose proof the reader should read first. Assume (1) is true. Recall that $H^ i_ J(K) = H^ i_{V(J)}(K)$, see Dualizing Complexes, Lemma 47.10.1. Thus $H^ i_ T(K) = \mathop{\mathrm{colim}}\nolimits H^ i_ J(K)$ where the colimit is over ideals $J \subset A$ with $V(J) \subset T$, see Lemma 51.5.3. Since $H^ i_ T(K)$ is finitely generated for $i \leq n$ we can find a $J \subset A$ as in (2) such that $H^ i_ J(K) \to H^ i_ T(K)$ is surjective for $i \leq n$. Thus the finite list of generators are $J$-power torsion elements and we see that (2) holds with $J$ replaced by some power.

Let $a \in \mathbf{Z}$ be an integer such that $H^ i(K) = 0$ for $i < a$. We prove (2) $\Rightarrow $ (1) by descending induction on $a$. If $a > n$, then we have $H^ i_ T(K) = 0$ for $i \leq n$ hence both (1) and (2) are true and there is nothing to prove.

Assume we have $J$ as in (2). Observe that $N = H^ a_ T(K) = H^0_ T(H^ a(K))$ is finite as a submodule of the finite $A$-module $H^ a(K)$. If $n = a$ we are done; so assume $a < n$ from now on. By construction of $R\Gamma _ T$ we find that $H^ i_ T(N) = 0$ for $i > 0$ and $H^0_ T(N) = N$, see Remark 51.5.6. Choose a distinguished triangle

\[ N[-a] \to K \to K' \to N[-a + 1] \]

Then we see that $H^ a_ T(K') = 0$ and $H^ i_ T(K) = H^ i_ T(K')$ for $i > a$. We conclude that we may replace $K$ by $K'$. Thus we may assume that $H^ a_ T(K) = 0$. This means that the finite set of associated primes of $H^ a(K)$ are not in $T$. By prime avoidance (Algebra, Lemma 10.15.2) we can find $f \in J$ not contained in any of the associated primes of $H^ a(K)$. Choose a distinguished triangle

\[ L \to K \xrightarrow {f} K \to L[1] \]

By construction we see that $H^ i(L) = 0$ for $i \leq a$. On the other hand we have a long exact cohomology sequence

\[ 0 \to H^{a + 1}_ T(L) \to H^{a + 1}_ T(K) \xrightarrow {f} H^{a + 1}_ T(K) \to H^{a + 2}_ T(L) \to H^{a + 2}_ T(K) \xrightarrow {f} \ldots \]

which breaks into the identification $H^{a + 1}_ T(L) = H^{a + 1}_ T(K)$ and short exact sequences

\[ 0 \to H^{i - 1}_ T(K) \to H^ i_ T(L) \to H^ i_ T(K) \to 0 \]

for $i \leq n$ since $f \in J$. We conclude that $J^2$ annihilates $H^ i_ T(L)$ for $i \leq n$. By induction hypothesis applied to $L$ we see that $H^ i_ T(L)$ is finite for $i \leq n$. Using the short exact sequence once more we see that $H^ i_ T(K)$ is finite for $i \leq n$ as desired.

We omit the proof of the equivalence of (2) and (3) in case $T = V(I)$.
$\square$

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