Surjective endomorphism of a Noetherian ring is an automorphism
= Surjective endomorphism of a Noetherian ring is an automorphism
For a surjective <ring endomorphism> $f$ of a <Noetherian ring>, the ascending kernel chain $\ker f\subseteq\ker f^2\subseteq\cdots$ stabilizes. If $\ker f^m=\ker f^{m+1}$ and $f(a)=0$, choose $b$ with $f^m(b)=a$. Then $b\in\ker f^{m+1}=\ker f^m$, so $a=0$. Thus the <endomorphism> is a <ring automorphism>.