Let f: M --> M be a surjective R-module homomorphism where M is a Noetherian left R-module.
We show that ker(f) = 0. For indirect proof, suppose ker(f) is nonzero.
We have an increasing chain of R-modules in M:
ker(f) < ker(f2) < ker(f3) < . . . < ker(fn) < ker(fn + 1) < . . . .
We claim that since ker(f) is nonzero, these inclusions are proper (hence contradicting the Noetherian property).Since ker(f) is nonzero, we may choose a nonzero element y in ker(f). Since f is surjective, there exists some x in M such that f(x) = y. So f(y) = f(f(x)) = 0. Note that x is not in ker(f) since f(x) = y, and y is nonzero. However, x is in ker(f2) since f(f(x)) = 0. So ker(f) is properly contained in ker(f2).
For induction, suppose that the inclusions ker(f) < . . . < ker(fn) (n > 2) are proper. Choose another nonzero y in ker(fn) \ ker(fn-1). Since f is surjective, it follows that y = f(x) for some x in M. So fn(y) = fn(f(x)) = 0. Note that x is in ker(fn + 1) but not in ker(fn) since fn(x) = 0 implies that y is in ker(fn - 1). So ker(fn) is properly contained in ker(fn + 1).