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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.

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    $\begingroup$ "For example, if $M$ is a direct summand ..., but I need something more general." How much more general? The result is not true in full generality. Let $A$ be $k[x,y]$. Let $F$ be $A^{\oplus 2}$ with basis ($\mathbf{e},\mathbf{f})$. Let $\alpha(\mathbf{e})$ equal $\mathbf{f}$, and let $\alpha(\mathbf{f})$ equal $-\mathbf{e}$. Let $M$ be the kernel of the $A$-module homomorphism $\phi:A^{\oplus 2}\to A$, $\phi(\mathbf{e})=x$, $\phi(\mathbf{f})=y$. If there were such $\beta$, it would factor through $\phi$. Now reduce modulo $\langle x,y \rangle$ for a contradiction. $\endgroup$
    – Jason Starr
    Commented Feb 12, 2014 at 18:07

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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 $(M\otimes_AM)^\perp=M^\perp\otimes_AF^\vee+F^\vee\otimes_AM^\perp$.

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