minimal free matrix induces zero on residue ext

Given: The objects and hypotheses in the statement. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

The balanced Ext bifunctor: Assume the Axiom of Dependent Choice. Let $\mathcal{A}$ be an abelian category with enough projectives and enough injectives, and fix supplied projective and injective resolution data on all objects of $\mathcal{A}$. For each $n\ge 0$, define ${\mathrm{Ext}}_{\mathcal{A}}^n(M,N)$ to mean either ${\mathrm{Ext}}_P^n(M,N)$ or ${\mathrm{Ext}}_I^n(M,N)$, identified by the natural comparison isomorphism already proved. This notation is justified by the comparison theorem, its independence of comparison data, its two-variable naturality, and its change-of-resolution cocycle law; it is not a definition by equality of the two complexes.

Module categories have enough injectives: Assume the Axiom of Choice. For every unital ring $R$ and every left $R$-module $M$, there is an injective left $R$-module $I$ and a monomorphism $M\to I$. Thus left $R$-modules have enough injectives. For commutative $R$, one explicit functorial target is $J(M)={\left(R^{(M^{\vee })}\right)}^{\vee }\cong {\prod }_{\varphi \in M^{\vee }}{R}^{\vee },$ where $X^{\vee }={\mathrm{Hom}}_{\mathbb{Z}}(X,\mathbb{Q}\mathrm{/}\mathbb{Z})$; the embedding is $M\to M^{\vee \vee }\to J(M)$. Here $X^{\vee }$ is a left $R$-module by $(r\varphi )(x)=\varphi (rx)$.