Zeros of higher Ext functors

Question

I'm trying to understand module classes that are defined as the kernels of higher Ext functors (e.g., arising here; 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 $Ext^n(M,N)$?

Answer 1

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] 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]), 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 $0 \to N \to N \oplus M \to M \to 0$.)

Answer 2

If your category has enough projectives, choose a projective resolution of $M$, and write $\Omega^nM$ for the $n$th kernel in this resolution, say $$0 \to \Omega^nM \to P_{n-1} \to \cdots \to P_1\to P_0\to M \to 0.$$ If $\zeta$ is an element of $\mathop{\rm Ext}^n(M,N)$, then choose a cocycle $\hat\zeta\colon \Omega^nM\to N$ representing $\zeta$. If necessary, enlarge your projective resolution by adding an irrelevant exact sequence of projectives, to make $\hat\zeta$ surjective with kernel $X$, say. Then form the pushout $P_{n-1}/X$ of $$\begin{smallmatrix}\Omega^nM&\to&P_{n-1}\\\downarrow\\N\end{smallmatrix}$$ to obtain a sequence $$0 \to N \to P_{n-1}/X \to P_{n-2}\to\cdots \to P_0\to M \to 0.$$ The Yoneda class of this sequence is the element $\zeta$. It is weakly initial in the Yoneda class. The map $\Omega^nM\to N$ factors through the projective $P_{n-1}$ if and only if $\zeta=0$. Similarly, if there are enough injectives then the dual construction will give a representative of the Yoneda class that is weakly terminal. All others sit between these two extremes.