I know a proof of the following theorem using determinants.
For some reason, I'd like to know a proof without using them.
Theorem
Let $A$ be a commutative ring.
Let $E$ and $F$ be finite free modules of the same rank over $A$.
Let $f:E → F$ be a surjective $A$-homomorphism.
Then $f$ is an isomorphism.
Best Answer
You can show that every commutative ring is stably finite (see Lam's Lectures on Modules and Rings first 10 pages or so) which means that if $R^n\cong R^n\oplus N$, then $N=0$.
If you have a surjection $f:M\rightarrow M'$, then $M/\ker(f)\cong M'$, but $M'$ being projective implies that $0\rightarrow \ker(f)\rightarrow M\rightarrow M/\ker{f}\rightarrow 0$ splits, and so $M\cong \ker(f)\oplus M'$, whence $\ker{f}=\{0\}$.