Endomorphism of a rings

Question

I was trying to prove that a surjective Endomorphism $f:A \to A$ of a noetherian ring is also injective. I would like to know why this argument is not correct? $A/\rm{Ker}f \cong \rm{Im}f=A \Rightarrow Kerf=\{0\}$

Answer 1

Your argument cannot be correct, since it doesn't use the noetherian hypothesis; this is because for a non-noetherian ring the statement is wrong:
Consider the polynomial ring $A = \mathbb Z[x_1,x_2,x_3,\dotsc]$ in infinitely many variables. It is clearly not noetherian. Now, consider the ring homomorphism $f\colon A\to A$ given on variables by $f(x_1) = 0$ and $f(x_n) = x_{n-1}$ for $n>1$. It is surjective, but not injective.

Answer 2

It doesn't work because there isn't any general principle of ring theory according to which $R/I$ should be isomorphic to $R$ only if $I=0$.

If $f$ were surjective but not injective, then the sequence of ideals $$0\subseteq\ker( f)\subseteq \ker (f^2)\subseteq \ker (f^3)\subseteq\cdots$$ would be strictly increasing, which is inconsistent with noetherianity of $A$.