Endomorphism of a rings 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? $A/\rm{Ker}f \cong \rm{Im}f=A \Rightarrow Kerf=\\{0\\}$

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.