Injective linear map between modules If you have an injective linear map between two free modules of equal dimension, is the determinant of the matrix representing the map necessarily nonzero?  If not is there an obvious counterexample?  (Everything is over a multivariate polynomial ring over a field.)

Thanks!
Over an integral domain $D$ this is straightforward as Jack Schmidt says in the comments. Suppose $T : D^n \to D^n$ is a morphism such that $\det(T) = 0$. Then $T \otimes \text{Frac}(D) : \text{Frac}(D)^n \to \text{Frac}(D)^n$ has the same property, where $\text{Frac}(D)$ is the fraction field. Since we are now over a field, $T \otimes \text{Frac}(D)$ is not injective, and so it annihilates some nonzero vector in $\text{Frac}(D)^n$. Scaling this vector suitably puts it in $D^n$, and so $T$ is not injective.

Over a general commutative $R$ I have a proof for $n = 2$ (consider the adjugate) but it doesn't seem to generalize. I'll get back to you.
Let $T: A^n \to A^n$ be a morphism and let's use the same notation for its matrix, $T=(t_{ij})$. Let us assume that $det(T)=0$ and I will prove that $T$ fails to be injective. 

There must be an integer $k$ such that there exists a minor of rank $k$ with nonzero determinant and any minor of larger rank has determinant zero. Without loss of generality, we can assume that the minor sits in the first $k$ rows and first $k$ columns. If $k=0$, then $T=0$, which is trivial. Since our task is to find a non-zero vector $x=(x_1,\cdots,x_n)$ killed by $T$ and inparticular our vector can be chosen to have the form $(x_1,\cdots,x_{k+1},0,\cdots,0)$, it suffices to assume that n=k+1.

Having adopted these assumptions, since $TT^*=det(T)I=0$, we just have to find a vector $x\neq 0$ such that $T^*x=0$, where $T^*$ is the adjugate matrix of $T$. Now $T^*$ is obviously nonzero since the first $k\times k$ minor of $T$ has nonzero determinant. Hence, such an $x$ must exist(assuming that the ring in question has a unit) and we are home!