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 $\mathbb{Z}$. 

 A: 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.
A: Here is an elementary proof of the fact that if the determinant $D$ of an $n \times n$ matrix M is a zero-divisor, then there is a nonzero vector $X$ such that $MX = 0$.
Let $a$ be a nonzero scalar such that $aD = 0$. Let $M'$ be a square submatrix of $M$ of maximum size $r$ such that $a \det M' \ne 0$. (If there is no such submatrix, let $r = 0$.) We have $r < n$ by the definition of $a$. After permuting the rows and columns of $M$ if necessary, we may assume that $M'$ is located in the top left corner of $M$. 
Let $M''$ be the $r \times (r + 1)$ matrix in the top left corner of $M$, and let $d_j$, for $j = 1, \dots, r+1$, be the minor of $M''$ obtained by deleting its $j$th column.
Now define $X = aX_0$, where
$$X_0 = (d_1, -d_2, \dots, (-1)^r d_{r+1}, 0, \dots, 0).$$
We have $ad_{r+1} = a\det M' \ne 0$, so $X \ne 0$. 
I claim that $MX = 0$. If $i > r$, then the $i$th coordinate of $MX_0$ is, up to sign, the minor of $M$ obtained from its first $r + 1$ columns and rows $1, 2, \dots, r, i$. By the definition of $r$, this minor becomes zero upon multiplication by $a$. When $i \leq r$, the the $i$th coordinate of $MX_0$ is zero because it is, up to sign, the determinant obtained by extending $M''$ with its own $i$th row. This proves the claim.
A: Yes, such  an injective morphism has non-zero determinant.  
Actually, if $M$ is a finitely generated free module over the commutative ring $R$ and $u:M\to M$ is an endomorphism, one has the precise equivalence: $$u \;\text {is injective}\iff \det(u) \; \text {is not a zero divisor in}\;  R.$$   The proof is based on the fact that  elements  $m_1,m_2, \ldots  ,m_n\in M$ form a linearly independent set iff there exists a non-zero $0\neq \lambda\in R$ with $\lambda (m_1\wedge m_2\wedge \ldots\wedge  m_n)=0\in \Lambda^nM$.  
You can find the details in Bourbaki, Algebra, III, §7, Proposition 3, page 524.
A: The argument from Bourbaki referenced in Georges Elencwajg's answer is exactly the kind of argument I was hoping would work so let me record it here, with some simplifications.
Proposition: Let $m_1, ... m_n$ be elements of some $R$-module $M$ which are linearly dependent. Then there exists nonzero $r \in R$ such that $r (m_1 \wedge ... \wedge m_n) = 0$ in $\Lambda^n(M)$. 
Proof. If $\sum r_i m_i = 0$ is a linear dependence, assume WLOG that $r_1 \neq 0$. Then $r_1 m_1 = - \sum_{i \ge 2} r_i m_i$, so by basic properties of the wedge product we find that $(r_1 m_1) \wedge m_2 \wedge ... \wedge m_n = 0$, so we can take $r = r_1$. 
Proposition: If $M$ is free then the converse holds. 
Proof. We proceed by induction on $n$. If $n = 1$ this is clear. In general, suppose that there exists a nonzero $r$ such that
$$r (m_1 \wedge ... \wedge m_n) = 0.$$
If $r (m_2 \wedge ... \wedge m_n) = 0$, then by the inductive hypothesis $m_2, ..., m_n$ are linearly dependent and we are done, so suppose otherwise. Then $r (m_2 \wedge ... \wedge m_n) \neq 0$. Hence, using the fact that $\Lambda^{n-1}(M)$ is also free (this is crucial!), we conclude that there exists an alternating $n-1$-form $f : M^{n-1} \to R$ such that
$$f(r m_2, ..., m_n) = s \neq 0$$
(because of the following fact: if $g$ is a nonzero vector in a free $R$-module $F$, then there exists some $R$-linear map $\alpha : F \to R$ such that $\alpha(g) \neq 0$).
But since $m_1 \wedge (rm_2) \wedge ... \wedge m_n = 0$, the alternating $n$-form
$$x_1 f(x_2, ..., x_n) - x_2 f(x_1, x_3, ..., x_n) \pm ...$$
necessarily vanishes on it, so
$$m_1 f(rm_2, ..., m_n) = m_1 s = rm_2 f(m_1, m_3, ..., m_n) \mp ...$$
and we conclude that the $m_i$ are linearly dependent. 
Corollary: Let $f : M \to N$ be an injective map of free modules. Then the induced map $\Lambda(f) : \Lambda(M) \to \Lambda(N)$ is also injective.
Proof. Suppose otherwise. Let $e_i, i \in I$ be an ordered basis of $M$. For a finite subset $S$ of $I$, let $e_S$ denote the wedge of all of the elements of $S$ (in the order determined by the order of $I$). If $\Lambda(f)$ is not injective, then let $\sum c_S e_S$ be some element of its kernel. Since the kernel is a two-sided ideal (because $\Lambda(f)$ is a ring homomorphism), we may freely take exterior products on either side (this is also crucial!). Now, if $\sum c_S e_S$ has at least two nonzero terms in it, then there exists $i$ such that $e_i$ appears in some term but not every term, so it follows that
$$\sum c_S e_S \wedge e_i \in \text{ker}(\Lambda(f))$$
as well. This is an element of the kernel with strictly fewer nonzero terms. Hence an element of the kernel with the minimal number of nonzero terms necessarily has a single nonzero term $c_S e_S$. Writing this as
$$c_S e_1 \wedge ... \wedge e_k$$
and applying $\Lambda(f)$ we get
$$c_S f(e_1) \wedge ... \wedge f(e_k) = 0.$$
By the above, it follows that $f(e_1), ... f(e_k)$ are linearly dependent, but this contradicts $f$ injective. 
Corollary: Let $f : R^n \to R^n$ be an endomorphism of a free module. Then $f$ is injective if and only if $\det(f)$ is not a zero divisor.
Proof. By the above, if $f$ is injective then $\Lambda^n(f) : \Lambda^n(R^n) \to \Lambda^n(R^n)$ is also injective. Since it acts by multiplication by $\det(f)$, we conclude that $\det(f)$ is not a zero divisor. If $f$ is not injective then some $\lambda_1 e_1 + \lambda_2 e_2 + \cdots + \lambda_n e_n$ lies in the kernel of $\Lambda(f)$, where not all $\lambda_i$ are $0$; thus, by taking exterior products we conclude that $\lambda_i e_1 \wedge ... \wedge e_n$ also lies in the kernel of $\Lambda(f)$ for all $i$, and therefore $\lambda_i \det(f) = 0$ for all $i$. 
