Determinant for a basis of a free module It is known that a set of $n$ vectors in a vector space of dimension $n$ is a basis if and only if the determinant of the matrix of coordinates of the vectors in some basis is not zero. In this case the rows (columns) of the matrix are linearly independent, and it is sufficient for the set to be a basis.
However, the rule is not working for an arbitrary free module. For example, in $\mathbb Z^2$ elements $(1,2)$ and $(3,2)$ are linearly independent, each of them can be a part of a basis since $\gcd(1,2) = 1$ and $\gcd(3,2) = 1$, but two of them together do not form a basis since $1$ is not a linear combination of $2$.
Checking the determinant ($\pm 4$) we can find that it is not zero, but it is also not a unit, and the matrix of coordinates is not invertible.
If I understand it correctly, a set $S$ of $n$ elements is a basis of a free module of rank $n$ if and only if every element of the module can be obtained as a linear combination of elements of $S$. In other words, there must be a solution $X$ for the matrix equation $XM = V$ for any element $V$ of the module, where $M$ is the matrix of coordinates of $S$.
If the determinant of $M$ is a unit, then $M$ is invertible, and $X = VM^{-1}$ for every element $V$.
Thus, if the determinant of $M$ is a unit, the set $S$ is a basis. Is this correct?
What about the converse? If a set of elements is a basis of a free module of rank $n$, does it mean the determinant is a unit? For which types of rings it is true?
 A: Suppose $z_1,z_2,\dots,z_m$ are elements of $\mathbb{Z}^n$ (represented as columns). Let's show that they're linearly independent if and only if they're so when thought of as elements in $\mathbb{Q}^n$.
Linear independence in $\mathbb{Q}^n$ implies linear independence in $\mathbb{Z}^n$: this is obvious.
Linear independence in $\mathbb{Z}^n$ implies linear independence in $\mathbb{Q}^n$. Suppose $\alpha_1z_1+\dots+\alpha_mz_m=0$, with $\alpha_1,\dots,\alpha_n\in\mathbb{Q}$. It is not restrictive to assume $\alpha_i=\beta_i/t$, where $\beta_i\in\mathbb{Z}$ and $t\in\mathbb{Z}$. Then $\beta_1z_1+\dots+\beta_mz_m=0$ and we're done.
The elements $z_1,\dots,z_m$ are a generating set of $\mathbb{Z}^m$ if and only if the matrix $M=\begin{bmatrix} z_1 & \dots & z_m\end{bmatrix}$ has a right inverse with integer coefficients. The proof is easy.
In the particular case of $m=n$, the set being a basis exactly means that $\det M=\pm1$.
For what types of rings does it hold? For every integral domain, but the condition on the determinant should be replaced by “$\det M$ is invertible in the ring”. Use the field of fractions of the ring.
