Let $A$ be a ring and let $F$ be a finitely generated, free $A$-module. Let $\alpha: F \to \textrm{Hom}\_A (F, A)$ be a skew-symmetric homomorphism, i.e. $\alpha(x)(y)=-\alpha(y)(x)$ for all $x,y \in F$. Now let $M$ be a submodule of $F$ such that $\alpha(M) \subseteq M^{\perp} =\{ f \in \textrm{Hom}\_A (F, A): M \subseteq \ker(f)\}$.


If $A$ is a field, then we can decompose $\alpha=\beta-\beta^\*$ where $\beta: F \to \textrm{Hom}\_A (F, A)$ satisfies $M \subseteq \ker(\beta)$ and $\beta^\*$ is the adjoint homomorphism. I am interested in the case where $A$ is the polynomial ring in several variables over some field. Are there some conditions on $\alpha$ and $M$ known, such that we can decompose $\alpha$ in such a way? For example, if $M$ is a direct summand of $F$, it is possible, but I need something more general.



Decomposition of skew-symmetric maps
Let us denote $N^\vee:=\text{Hom}\_A(N,A)$,
$S^\perp:=\{f\in N^\vee\mid fS=0\}\subset N^\vee$, and
$\hat S:=S^{\perp\perp}\subset N^{\vee\vee}$ for any $A$-module $N$ and its submodule $S\subset N$. If $N$ is finitely generated free, then $N^{\vee\vee}=N$, $S\subset\hat S$, and $S^\perp=\hat S^\perp$.


In these terms, $\alpha\in F^\vee\otimes\_AF^\vee$ is antisymmetric (i.e., $\tau\alpha=-\alpha$) with $\alpha(M,M)=0$ and
$\beta\in M^\perp\otimes\_AF^\vee\subset F^\vee\otimes\_AF^\vee$ (i.e., $\beta(M,F)=0$) so that $\beta^\*=\tau\beta$, where $\tau$ permutes the $F^\vee$'s in $F^\vee\otimes\_AF^\vee$.


There is a tautological criterion: $\alpha=\beta-\beta^\*$ iff
$\alpha\in M^\perp\otimes\_AF^\vee+F^\vee\otimes\_AM^\perp$.


$\Longrightarrow$ is immediate.


$\Longleftarrow$ if $\alpha\in M^\perp\otimes\_AF^\vee+F^\vee\otimes\_AM^\perp$ is antisymmetric, then $\alpha=\beta\_1+\tau\beta\_2$ for some
$\beta\_1,\beta\_2\in M^\perp\otimes\_AF^\vee$ and
$\beta\_1+\tau\beta\_2=-\tau\beta\_1-\beta\_2$. So, $\alpha=\beta-\beta^\*$, where $\beta:=\frac12(\beta\_1-\beta\_2)$.


In particular, $\alpha(\hat M,\hat M)=0$ is a necessary condition.


Most practical can be the sufficient condition
$(M\otimes\_AM)^\perp=M^\perp\otimes\_AF^\vee+F^\vee\otimes\_AM^\perp$.


