The Stacks project

Proof. The distinguished triangle $H^{-1}(K)[1] \to K \to H^0(K)[0] \to H^{-1}(K)[2]$ determines an exact sequence

Thus (2) implies that $I^{2c}$ annihilates $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$ for every $R$-module $N$. Assuming (1) we immediately see that $H^0(K)$ is $I^ c$-projective. On the other hand, we may choose an injective map $H^{-1}(K) \to N$ for some injective $R$-module $N$. Then this map is the image of an element of $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$ by the vanishing of the $\mathop{\mathrm{Ext}}\nolimits ^2$ in the sequence and we conclude $H^{-1}(K)$ is annihilated by $I^ c$.

Assume $R$ is Noetherian and $H^ i(K)$ is a finite $R$-module for $i = -1, 0$. By Lemma 15.84.5 we see that (3) is equivalent to (1) and (2). Also, if (3) holds then for $f \in I$ the multiplication by $f$ on $H^0(K)$ factors through a projective module, which implies that $H^0(K)_ f$ is a summand of a projective $R_ f$-module and hence itself a projective $R_ f$-module. Choosing $f_1, \ldots , f_ s$ to be generators of $I$ we find the equivalent conditions (1), (2), and (3) imply (5). Of course (5) trivially implies (4).

Assume (4). Since $H^{-1}(K)$ is a finite $R$-module and $I$-power torsion we see that $I^{c_1}$ annihilates $H^{-1}(K)$ for some $c_1 \geq 0$. Choose a short exact sequence

which determines an element $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_ R(H^0(K), M)$. For any $f \in I$ we have $\mathop{\mathrm{Ext}}\nolimits ^1_ R(H^0(K), M)_ f = \mathop{\mathrm{Ext}}\nolimits ^1_{R_ f}(H^0(K)_ f, M_ f)$ by Lemma 15.65.4. Hence if $H^0(K)_ f$ is projective, then a power of $f$ annihilates $\xi$. We conclude that $\xi$ is annihilated by $(f_1, \ldots , f_ s)^{c_2}$ for some $c_2 \geq 0$. Since $V(f_1, \ldots , f_ s) \subset V(I)$ we have $\sqrt{I} \subset (f_1, \ldots , f_ s)$ (Algebra, Lemma 10.17.2). Since $R$ is Noetherian we find $I^{c_3} \subset (f_1, \ldots , f_ s)$ for some $c_3 \geq 0$ (Algebra, Lemma 10.32.5). Hence $I^{c2c3}$ annihilates $\xi$. This in turn says that $H^0(K)$ is $I^{c_2c_3}$-projective (as multiplication by $a \in I$ which annihilate $\xi$ factor through $R^{\oplus r}$). Hence taking $c = \max (c_1, c_2c_3)$ we see that (2) holds. $\square$