Let $$ A=
\begin{pmatrix}
1 & 1 \\
2 & 1 \\
1 & 4
\end{pmatrix}
$$
The matrix $A$ has rank two (=number of independent columns), so it is full, hence has a left inverse which is
$$ A_L=
\begin{pmatrix}
-1 & 1 & 0\\
2 & -1 & 0\\
\end{pmatrix}
$$
(It is easy to check that $A_LA=I_{2 \times 2}$).
$A$ has three invertible square sub-matrices:
$ A_{1,2}=
\begin{pmatrix}
1 & 1 \\
2 & 1
\end{pmatrix}
$
$ A_{1,3}=
\begin{pmatrix}
1 & 1 \\
1 & 4
\end{pmatrix}
$
$ A_{2,3}=
\begin{pmatrix}
2 & 1 \\
1 & 4
\end{pmatrix}
$.
Each of the three sub-matrices is invertible, but generally it may happen that there are non-invertible sub-matrices, as the following example shows:
Let $$ B=
\begin{pmatrix}
1 & 0 \\
1 & 0 \\
0 & 1
\end{pmatrix}
$$
$ B_{1,2}=
\begin{pmatrix}
1 & 0 \\
1 & 0
\end{pmatrix}
$
$ B_{1,3}=
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
$
$ B_{2,3}=
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
$.
Here $B_{1,3}$ is not invertible.
Claim 1: Let $k$ be a field of characteristic zero and let $A$ be an $m \times n$ matrix over $k$, $m > n$.
Assume that $A$ is left invertible (hence all $m$ columns of $A$ are $k$-linearly independent).
Then there exists an $n \times n$ sub-matrix of $A$ which is invertible (equivalently, has an invertible determinant = a nonzero determinant, since we work over a field $k$).
Question 1: Is Claim 1 true? It seems true and follows from the assumption that $A$ has rank $m$.
Claim 2: Let $R$ be a commutative ring and let $A$ be an $m \times n$ matrix over $k$, $m > n$.
Assume that $A$ is left invertible (hence all $m$ columns of $A$ are $R$-linearly independent, so they generate a free $R$-module of rank $m$).
Then there exists an $n \times n$ sub-matrix of $A$ which is invertible (equivalently, has an invertible determinant = the determinant is an invertible element of $R$, see this).
Question 2: Is Claim 2 true? It seems true if Claim 1 is true or am I missing something?
Maybe the result of left invertibility is not equivalent to having full column rank $m$? (I think this result stull holds. see this).
Thank you very much! I apologize if my question is trivial.
Edit: Maybe all we can say is that there is square $n \times n$ sub-matrix $S$ of $A$ with non-zero determinant $d_S \in R$, but this does not imply that $S$ itself is an invertible matrix, since $d_s$ may not be invertible in $R$.
The following is not a counterexample: $A=
\begin{pmatrix}
t \\
t
\end{pmatrix}
$.
$A$ has rank one, $S=
\begin{pmatrix}
t
\end{pmatrix}
$ is not invertible, but $A$ itselt is not left invertible!
Best Answer
Over a commutative ring it is better to relate determinants to values taken in the image of your matrix [Cramer's Rule] and the minors do relate to surjectivity in the following way:
Since $A^T$ has a right inverse it is surjective and the proper claim is its $n\times n$ minors generate the unit ideal. [In a PID like $\mathbb Z$ you can also talk about a gcd of $1$.] With $R$ denoting the commutative ring, define $B:=A^T$ and we know $I_n = BC$ for some $C\in R^{m\times n}$.
Proof:
$1 = \det\big(I_n\big) = \det\big(BC\big) = \sum_{S} \det\big(B_{[n],S}\big)\det\big(C_{S,[n]}\big) = \sum_{S} \det\big(B_{[n],S}\big)\cdot \alpha_k$
by Cauchy-Binet [a polynomial identity that holds over $\mathbb C$ so it holds over every commutative ring. Ref e.g. https://en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula for notation clarifications though the OP has flipped the roles of $m$ and $n$.] You can of course transpose this at the end to talk about minors of $A$.