4
$\begingroup$

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 $m=n$.

edit:

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

  • 8
    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. http://www.math.buffalo.edu/~badzioch/MTH620/Homework_files/hw1.pdf2011-02-13

1 Answers 1

4

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.

  • 0
    This is a great answer. I hardly remembered that the ring of fractions of an integral domain must be a field, so this reduces to a vectorspace problem. Cheers!2011-02-07