Let $A$ be a square matrix over a commutative ring $R$. Then $A$ has a left inverse iff it is invertible. Does there exist a elementary proof of this fact? (i.e. without using the determinant!)
[Math] Invertible matrices in commutative rings
abstract-algebracommutative-algebramatrices
Related Solutions
I found the answer in this book (in Section $6.4.14$, “Determinants, Ranks and Linear Equations”). I'd tried using a similar Laplace expansion myself but was missing the idea of using the largest dimension at which the minors are not all annihilated by the same non-zero element. I'll try to summarize the argument in somewhat less formal terms, omitting the tangential material included in the book.
Let $A$ be an $m\times n$ matrix over a commutative ring $R$. We want to find a condition for the system of equations $Ax=0$ with $x\in R^n$ to have a non-trivial solution. If $R$ is a field, various definitions of the rank of $A$ coincide, including the column rank (the dimension of the column space), the row rank (the dimension of the row space) and the determinantal rank (the dimension of the lowest non-zero minor). This is not the case for a general commutative ring. It turns out that for our present purposes a useful generalization of rank is the largest integer $k$ such that there is no non-zero element of $R$ that annihilates all minors of dimension $k$, with $k=0$ if there is no such integer.
We want to show that $Ax=0$ has a non-trivial solution if and only if $k\lt n$.
If $k=0$, there is a non-zero element $r\in R$ which annihilates all matrix elements (the minors of dimension $1$), so there is a non-trivial solution
$$A\pmatrix{r\\\vdots\\r}=0\;.$$
Now assume $0\lt k\lt n$. If $m\lt n$, we can add rows of zeros to $A$ without changing $k$ or the solution set, so we can assume $k\lt n\le m$. There is some non-zero element $r\in R$ that annihilates all minors of dimension $k+1$, and there is a minor of dimension $k$ that isn't annihilated by $r$. Without loss of generality, assume that this is the minor of the first $k$ rows and columns. Now consider the matrix formed of the first $k+1$ rows and columns of $A$, and form a solution $x$ from the $(k+1)$-th column of its adjugate by multiplying it by $r$ and padding it with zeros. By construction, the first $k$ entries of $Ax$ are determinants of a matrix with two equal rows, and thus vanish; the remaining entries are each $r$ times a minor of dimension $k+1$, and thus also vanish. But the $(k+1)$-th entry of this solution is non-zero, being $r$ times the minor of the first $k$ rows and columns, which isn't annihilated by $r$. Thus we have constructed a non-trivial solution.
In summary, if $k\lt n$, there is a non-trivial solution to $Ax=0$.
Now assume conversely that there is such a solution $x$. If $n\gt m$, there are no minors of dimension $n$, so $k\lt n$. Thus we can assume $n\le m$. The minors of dimension $n$ are the determinants of matrices $B$ formed by choosing any $n$ rows of $A$. Since each row of $A$ times $x$ is $0$, we have $Bx=0$, and then multiplying by the adjugate of $B$ yields $\det B x=0$. Since there is at least one non-zero entry in the non-trivial solution $x$, there is at least one non-zero element of $R$ that annihilates all minors of size $n$, and thus $k\lt n$.
Specializing to the case $m=n$ of square matrices, we can conclude:
A system of linear equations $Ax=0$ with a square $n\times n$ matrix $A$ over a commutative ring $R$ has a non-trivial solution if and only if its determinant (its only minor of dimension $n$) is annihilated by some non-zero element of $R$, that is, if its determinant is a zero divisor or zero.
The determinant of the identity is always the multiplicative identity of the underlying ring and consequently the multiplicative property of the determinant implies that the determinant of an invertible matrix is itself invertible.
In essence $$AB = I \implies \det(A)\det(B) = 1$$ so that $\det(A)$ and $\det(B)$ are units.
Best Answer
I would not call the following argument elementary, but maybe fundamental:
The matrix $A$ defines an endomorphism $f$ of the $R$-module $R^n$. When $f$ is left-invertible or invertible, it has zero kernel. Hence in proving the equivalence left-invertible $\Leftrightarrow$ invertible we can assume that we have an exact sequence
$0 \rightarrow R^n \stackrel{f}{\rightarrow} R^n \rightarrow C \rightarrow 0$
where $C$ is the cokernel of $f$. But then $f$ is left-invertible if and only if the above exact sequence splits. Hence $f$ is left-invertible if and only if $R^n \cong R^n \oplus C$. But this latter condition is true only when $C=0$. Hence $f$ is left-invertible if and only if it is an isomorphism.
Remark: There are quite a bit "non-elementary arguments" in actually establishing that $C=0$: tensor the equality $R^n \cong R^n \oplus C$ with $\kappa(m)=R_m/mR_m$, where $m$ is some maximal ideal, to obtain $\kappa(m)^n \cong \kappa(m)^n \oplus C_m/mC_m$. This is an isomorphism of finite dimensional $\kappa(m)$-vector spaces. Since the dimensions of left and right sides must be equal, we see that $C_m/mC_m$ must be zero. Then Nakayama's Lemma implies that $C_m=0$. But an $R$-module whose localization at every maximal ideal is zero must be the zero module.
Conclusion: If you are willing to take your scalars to be a ring instead of a field, then in general you need to resort to more sophisticated arguments. But in this particular case we have that the beautiful equation $adj(A) A = det(A) I$ still holds and we could have avoided the above commutative-algebraic arguments. I will leave it you to decide which one is more elementary :)