Invertible matrices in commutative rings 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!)
 A: 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 :)
A: The following is less elementary than using determinants since it uses
Zorn's lemma.  It may be more direct and simpler than using Nakayama's
lemma.
Let $B$ be the left inverse.  Only the implication that $BA = $1 implies
$AB = 1$ (or just that A has a right inverse) is nontrivial.
First suppose that $R$ is a field. Let $f$ be the endomorphism of $R^n$
corresponding to $A$.  By elementary linear algebra, $f$ is injective
iff it is surjective.  This can be proved without using determinants,
using dimensions or by solving linear equations.  This gives the result when
translated back to matrices.  ($BA = 1$ implies $f$ injective implies $f$
both injective and surjective implies $f$ invertible implies $A$ invertible.)
Next suppose that $R$ is an integral domain.  Apply the previous case to
the quotient field.  The matrix identity $BA = 1$ ascends to the field
and (by the previous case) implies $AB = 1$ over the field.  The latter identity descends to $R$.
Next, for general $R$, let $\mathfrak p$ be any prime ideal of $R$.  Apply
the previous case to $R/\mathfrak p$. It says that $AB = 1$ over
$R/\mathfrak p$.  Equivalently, all of the coefficients of $N = 1-AB$ are
in $\mathfrak p$. Since $\mathfrak p$ was arbitrary, all of these
coefficients are nilpotent.  The last step used Zorn's lemma via basic commutative algebra.
$N$ itself is nilpotent since large powers of $N$ have a large power of
at least one coefficient of $N$ in each monomial term in each of their coefficients.  So $AB = 1-N$ is invertible with inverse
$C=1+N+N^2+\ ...\ +N^k$ for sufficiently large $k$.  $BC$ is a right
inverse for $A$.  This completes the proof.
The final step can be done in another way by observing that $N$ is a
projection and a nilpotent projection must be $0$.  $N$ was $0$ all
along.
