I'm trying to understand module classes that are defined as the kernels of higher Ext functors (e.g., arising [here](https://www.sciencedirect.com/science/article/pii/S0168007207000632?via%3Dihub); as this paper suggests, I'm coming at this problem outside of module theory). One way to do this, of course, is to understand the higher Ext functors and then specialize to the case when they become zero. However, I'm wondering if there's a more direct route?


That is, in the $n=1$ case, $Ext^1(M, N)$ correspond to the extensions of $M$ by $N$ under a natural notion of equivalence, and the equivalence class of $0$ is represented the trivial extension, i.e., $M + N$.


Is there a similar construction for $n\geq 2$? Is there a canonical/trivial longer exact sequence that we can always build and that will represent the equivalence class of 0 in $Ext^n(M,N)$?



Zeros of higher Ext functors
Elements of $\operatorname{Ext}^i(M,N)$ for $i \geq 1$ can be represented by *Yoneda extensions*: exact sequences $$E = \big(0 \to N \to Z\_{i-1} \to \ldots \to Z\_0 \to M \to 0\big)$$
modulo the equivalence relation $E \sim E'$ if there exists another Yoneda extension $E''$ with morphisms of chain complexes $E \leftarrow E'' \to E'$ that are the identity on $M$ and $N$. See for instance [Tag [06XP](https://stacks.math.columbia.edu/tag/06XP)] for details.


The idea is that $E$ gives a quasi-isomorphism $\sigma\_{\leq 0} E \stackrel\sim\to M$ (where $\sigma$ is the brutal truncation [Tag [0118](https://stacks.math.columbia.edu/tag/0118)]), so the composition
$$M \cong \sigma\_{\leq 0} E \twoheadrightarrow \sigma\_{\leq -i} E = N[i]$$
gives the required morphism $M \to N[i]$ in the derived category. If you want to produce the zero morphism, you want $\sigma\_{\leq 0} E$ to be the direct sum complex
$$\big(0 \to N \stackrel{\text{id}}\to N \to 0 \to \ldots \to 0\big) \oplus \big(0 \to 0 \ldots \to 0 \to M\big),$$
which is for instance the case when
$$E = \big( 0 \to N \stackrel{\text{id}}\to N \to 0 \to \ldots \to 0 \to M \stackrel{\text{id}}\to M \to 0 \big).$$
(If $i = 1$, then we see that the middle $M$ and $N$ sit in the same degree, so we should really consider the split extension $0 \to N \to N \oplus M \to M \to 0$.)


