Homomorphism of free modules $A^m\to A^n$ Let's $\varphi:A^m\to A^n$ is a homomorphism of free modules over commutative (associative and without zerodivisors) unital ring $A$. Is it true that $\ker\varphi\subset A^m$ is a free module?
Thanks a lot!

Here is a counterexample which is in some sense universal for the case $m = 2, n = 1$. Let $R = k[x, y, z, w]/(xy - zw)$ ($k$ a field). This is an integral domain because $xy - zw$ is irreducible. The homomorphism $R^2 \to R$ given by
$$(m, n) \mapsto (xm - zn)$$
has a kernel which contains both $(y, w)$ and $(z, x)$. If the kernel is free then it must be free on these two elements by degree considerations, but $x(y, w) = (zw, xw) = w(z, x)$ is a relation between them. 
