Is a nonsingular matrix not the same as an invertible matrix? Over an arbitrary ring $R$, a matrix $A$ is said to be invertible if it has an inverse with entries in the same ring. This happens iff $\det A$ is a unit of $R$.
I've always thought that the terms "invertible" and "nonsingular" are synonymous. But I think the following problem (from Artin) suggests that they are not (at least over an arbitrary ring):

Let $\varphi: \mathbb Z^k\to \mathbb Z^k$ be a homomorphism given by multiplication by an integer matrix $A$. Show that the image of $\varphi$ is of finite index if and only if $A$ is nonsingular and that if so, then the index is equal to $|\det A|.$

If I understand correctly, in this context "nonsingular" means $\det A\ne 0$. And this is not the same as invertible since if $A$ is invertible, then $\varphi$ is bijective, and the image is the whole $\mathbb Z^k$.
So, am I correct in saying that a matrix $A$ over a ring $R$ is by definition nonsingular if $\det A\ne 0$? And being nonsingular does not imply being invertible (unless the underlying ring is a field)?
 A: Take the matrix
$$
\begin{pmatrix}
2 & 0 \\ 0 & 2
\end{pmatrix}.
$$
Over $\mathbb{R}$, this has inverse
$$
\begin{pmatrix}
\frac{1}{2} & 0 \\ 0 & \frac{1}{2}
\end{pmatrix}.
$$
But, it would have no inverse in $\mathbb{Z}$.  Yet, the kernel of the map would still be trivial, hence it is nonsingular.
A: A matrix $A$ is called right-singular if $Ax=0$ has non-trivial solutions, left-singular if $y^TA=0$ has non-trivial solutions, and singular if it is both left-singular and right-singular.
A matrix $A$ is called left-invertible if it has a left inverse, right-invertible if it has a right inverse, and invertible if it is a square matrix that has left and right inverses.
Determinant is not involved in the definitions of these two concepts. The usual definition of determinant does not apply in the first place if the ring is not commutative.
Suppose $A$ is a square matrix over a commutative ring $R$, so that you can speak of its determinant.


*

*$A$ is invertible if and only if $\det A$ is invertible in $R$. That is, $A$ is a unit in $M_n(R)$ if and only if $\det A$ is a unit in $R$. For the "only if" part, consider $\det(A)\det(A^{-1})=1$; for the "if" part, consider $A\operatorname{adj}(A)=\operatorname{adj}(A)A=\det(A)I$.

*$A$ is singular if and only if $\det A$ is a zero divisor in $R$. The "only if" part can be easily proven as follows. Suppose $Ax=0$ for some $x\ne0$. Without loss of generality we may assume that the first entry of $x$ is nonzero. Construct a matrix $B$ by replacing the first column of the identity matrix by $x$. Then $\det(B)\ne0$ but $\det(A)\det(B)=\det(AB)=0$ because the first column of $AB$ is zero. 
The proof of the "if" part is less trivial. See, e.g. chapter 9 of

Brown W.C. (1993), Matrices over Commutative Rings, Marcel Dekker, Inc., New York.

Note that $A$ can be singular even if $\det A\ne0$; e.g. over $R=\mathbb Z/4\mathbb Z$, we have $\pmatrix{3&1\\ 1&1}\pmatrix{2\\ 2}=0$ but $\det\pmatrix{3&1\\ 1&1}=2\ne0$. While the determinant ($=2$) in this case is nonzero, it is a zero divisor.

*It follows from (1) and (2) that $A$ is invertible only if $A$ is nonsingular (a more direct proof: $Ax=0\Rightarrow x=A^{-1}Ax=0$). The converse is not true. E.g. over $R=\mathbb Z$, every integer (which is a $1\times1$ matrix) $\ge2$ are nonsingular but not invertible.

*Right-invertibility, left-invertibility and invertibility are equivalent. So are right-singularity, left-singularity and singularity. The statement in (2) can thus be rephrased as "$A$ is a zero divisor in $M_n(R)$ if and only if $\det A$ is a zero divisor in $R$."

