Non-degenerate bilinear map on modules over an integral domain If $\varphi: R^m \times R^n \to R$ is a non-degenerate bilinear map and $R$ is an integral domain then we must have $m=n$.
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By "non-degenerate" bilinear map I mean that for every nonzero $m \in R^m$ there is an $n \in R^n$ such that $f(m,n)\neq 0$. The reverse also holds: for all non-zero $n \in R^n$ there is an $m \in R^m$ with $f(m,n)\neq 0$.

Suppose that $n>m$ and let $e_i,f_j$ be a basis for $R^m, R^n$ respectively. Define the matrix $A\in R^{m\times n}$ by $A_{i,j}=\varphi(e_i,f_j)$ then the columns of the matrix are linear dependent in $K=Frac(R)$, so there is a vector $0 \neq v\in K^n$ such that $Av=0$.
since K is the fraction field, then you can find $r\in R$ such that all the entries of $0\neq r\cdot v$ are in $R$ (for example, let r by the multiplication of the denumerators). You now have $A(rv)=0$ so 
$$ \varphi (e_i , \sum_j (rv)_j f_j) = \sum_j (rv)_j \varphi(e_i,f_j) = (A(rv))_i = 0$$
$\sum_j (rv)_i f_j \neq 0$ since $rv\neq 0$ and the $f_j$ are a basis, so $\varphi$ is degenerate.
