Does an injective endomorphism of a finitely-generated free R-module have nonzero determinant? Alternately, let $M$ be an $n \times n$ matrix with entries in a commutative ring $R$. If $M$ has trivial kernel, is it true that $\det(M) \neq 0$?  
This math.SE question deals with the case that $R$ is a polynomial ring over a field. There it was observed that there is a straightforward proof when $R$ is an integral domain by passing to the fraction field.
In the general case I have neither a proof nor a counterexample. Here are three general observations about properties that a counterexample $M$ (trivial kernel but zero determinant) must satisfy. First, recall that the adjugate $\text{adj}(M)$ of a matrix $M$ is a matrix whose entries are integer polynomials in those of $M$ and which satisfies 
$$M \text{adj}(M) = \det(M).$$
If $\det(M) = 0$ and $\text{adj}(M) \neq 0$, then some column of $\text{adj}(M)$ lies in the kernel of $M$. Thus:

If $M$ is a counterexample, then $\text{adj}(M) = 0$.

When $n = 2$, we have $\text{adj}(M) = 0 \Rightarrow M = 0$, so this settles the $2 \times 2$ case. 
Second observation: recall that by Cayley-Hamilton $p(M) = 0$ where $p$ is the characteristic polynomial of $M$. Write this as
$$M^k q(M) = 0$$
where $q$ has nonzero constant term. If $q(M) \neq 0$, then there exists some $v \in R^n$ such that $w = q(M) v \neq 0$, hence $M^k w = 0$ and one of the vectors $w, Mw, M^2 w,\dots, M^{k-1} w$ necessarily lies in the kernel of $M$. Thus if $M$ is a counterexample we must have $q(M) = 0$ where $q$ has nonzero constant term.
Now for every prime ideal $P$ of $R$, consider the induced action of $M$ on $F^n$, where $F = \overline{ \text{Frac}(R/P) }$. Then $q(\lambda) = 0$ for every eigenvalue $\lambda$ of $M$. Since $\det(M) = 0$, one of these eigenvalues over $F$ is $0$, hence it follows that $q(0) \in P$. Since this is true for all prime ideals, $q(0)$ lies in the intersection of all the prime ideals of $R$, hence

If $M$ is a counterexample and $q$ is defined as above, then $q(0)$ is nilpotent. 

This settles the question for reduced rings. Now, $\text{det}(M) = 0$ implies that the constant term of $p$ is equal to zero, and $\text{adj}(M) = 0$ implies that the linear term of $p$ is equal to zero. It follows that if $M$ is a counterexample, then $M^2 \mid p(M)$. When $n = 3$, this implies that
$$q(M) = M - \lambda$$
where $\lambda$ is nilpotent, so $M$ is nilpotent and thus must have nontrivial kernel. So this settles the $3 \times 3$ case. 
Third observation: if $M$ is a counterexample, then it is a counterexample over the subring of $R$ generated by the entries of $M$, so

We may assume WLOG that $R$ is finitely-generated over $\mathbb{Z}$. 


Lam's Exercises in modules and rings includes the following:

which tells us that your determinant is not a zero-divisor.
The paper where McCoy does that is [Remarks on divisors of zero, MAA Monthly 49 (1942), 286--295] If you have JStor access, this is at http://www.jstor.org/stable/2303094
There is a pretty corollary there: a square matrix is a zero-divisor in the ring of matrices over a commmutative ring iff its determinant is a zero divisor.
