Kernel of the induced map of the wedge product

Question

Let $A$ be a noetherian ring and let $M$ be a finitely generated $A$-module. Let $F$ be a free $A$-module and let $d: F \to M$ be a homomorphism which maps a basis of $F$ to a minimal set of generators of $M$. Consider the submodule $N$ of $\bigwedge^2 F$ which is generated by the elements $x \wedge y$ where $x \in \ker (d)$. It is clear that $N$ lies the kernel of the induced map $\hat{d}:\bigwedge^2 F \to \bigwedge^2M$. I am interested in the quotient $\ker(\hat{d}) / N$.

My question: Are there some results concerning this object? For example in the case where $A$ is a local ring? Is there perhaps some geometric interpretation of it in the case where $A$ is a finitely generated algebra over some field $k$?

Answer 1

You don't need most of your assumptions to ensure that your quotient is $0$. More generally, we have:

(1) If $A$ is any commutative ring, and $f : M \to N$ is a surjective homomorphism of $A$-modules, then the kernel of the induced map $\wedge f : \wedge M \to \wedge N$ is the left ideal generated by $\mathrm{ker} f$ in the exterior algebra $\wedge M$.

This ideal, of course, is a graded two-sided ideal, and its $2$-nd graded component (i.e., its intersection with $\wedge^2 M$) is precisely the $A$-linear span of all $x \wedge y$ with $x \in \mathrm{ker} f$ and $y \in M$.

You can probably find (1) proven in good algebra texts. The analogue of (1) for tensor powers is Theorem 2.19 in Keith Conrad's Tensor Products II, and (1) follows from said analogue by diagram chasing. I would also expect to see (1) in Bourbaki. (If everything else fails, it is Corollary 80 in my old writeup on tensor powers, but that is barely readable due to my attempts at formalism.)