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 $k$?



Kernel of the induced map of the wedge product
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*](http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod2.pdf), 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](http://mit.edu/~darij/www/algebra/tensorext.pdf), but that is barely readable due to my attempts at formalism.)


