# 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 $f(m,n)\neq 0$.

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First you should be clear what your definition of nondegenerate bilinear map is in this context. I would say that it should at least include the condition that you want to "show" in your second paragraph, as well as the other condition with the roles of $R^m$ and $R^n$ reversed. If you assume that, then you are reduced to showing a standard fact about injections of free $R$-modules... –  Pete L. Clark Feb 7 '11 at 6:31
Thank you, and done. –  user6560 Feb 7 '11 at 6:52
You switch from $\varphi$ to $f$ in your notation for the pairing. Also, I would choose other notation than $m \in R^m$ and $n \in R^n$. –  Matt E Feb 7 '11 at 9:03
Commenting so that this will be visible at the top: This question is Problem 1.7 on the homework for Math 620, at the University of Buffalo. As user "Student" points out, every question user6560 has asked is a homework question from that course. math.buffalo.edu/~badzioch/MTH620/Homework_files/hw1.pdf –  David Speyer Feb 13 '11 at 19:27

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.