In their paper What makes a complex exact?, Buchsbaum and Eisenbud define an ideal $I(\varphi, M)$, where $M$ is a module over a commutative ring, and $\varphi: F \to G$ is a homomorphism of projective $R$-modules. I'm not sure that I understand the definition correctly. First they define $I(\varphi)$ to be the image of the induced morphism $$(\Lambda^k G)^* \otimes \Lambda^kF \to R, \tag{$*$}$$ for $k = \operatorname{rank}(\varphi)$, that is the smallest $k$ such that $0 \neq \Lambda^k \varphi: \Lambda^k F \to \Lambda^k G$. Then they write
Similarly, if $M$ is an $R$-module, we define $I(\varphi, M)$ to be the image of the above map where $k = \operatorname{rank}(\varphi, M)$.
So is $\operatorname{rank}(\varphi, M)$ really the only property of $M$ that enters into the definition of $I(\varphi, M)$? Or do I have to tensorize something in $(*)$ by $M$? This is somehow also suggested in this question, though it's not the point of discussion there.
So I think that the image of $(\Lambda^k G)^* \otimes \Lambda^kF \to R$ is the same as the image of $(\Lambda^k G \otimes M)^* \otimes (\Lambda^kF \otimes M)\to R$, at least if $k = \operatorname{rank}(\varphi, M)$. The reason here is that the canonical map $M^* \otimes M \to R$ is surjective.