# A surjective homomorphism between finite free modules of the same rank

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 $\ker{f}=\{0\}$.