Lemma 47.15.3. Let $A$ be a Noetherian ring. Let $K \in D^ b_{\textit{Coh}}(A)$. Let $\mathfrak m$ be a maximal ideal of $A$. If $H^ i(K)/\mathfrak m H^ i(K) \not= 0$, then there exists a finite $A$-module $E$ annihilated by a power of $\mathfrak m$ and a map $K \to E[-i]$ which is nonzero on $H^ i(K)$.

Proof. Let $I$ be the injective hull of the residue field of $\mathfrak m$. If $H^ i(K)/\mathfrak m H^ i(K) \not= 0$, then there exists a nonzero map $H^ i(K) \to I$. Since $I$ is injective, we can lift this to a nonzero map $K \to I[-i]$. Recall that $I = \bigcup I[\mathfrak m^ n]$, see Lemma 47.7.2 and that each of the modules $E = I[\mathfrak m^ n]$ is of the desired type. Thus it suffices to prove that

$\mathop{\mathrm{Hom}}\nolimits _{D(A)}(K, I) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _{D(A)}(K, I[\mathfrak m^ n])$

This would be immediate if $K$ where a compact object (or a perfect object) of $D(A)$. This is not the case, but $K$ is a pseudo-coherent object which is enough here. Namely, we can represent $K$ by a bounded above complex of finite free $R$-modules $K^\bullet$. In this case the $\mathop{\mathrm{Hom}}\nolimits$ groups above are computed by using $\mathop{\mathrm{Hom}}\nolimits _{K(A)}(K^\bullet , -)$. As each $K^ n$ is finite free the limit statement holds and the proof is complete. $\square$

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