I am trying to prove the following statement:
Let $f:A\to B$ be a surjective $k$-algebra homomorphism. Let $A$ and $B$ be local Noetherian rings with same finite Krull dimension. Then $f$ is an isomorphism if $A$ is an integral domain.
My understanding:
The math.SE discussion here implies that $f$ is certainly a ring isomorphism, but for that we don't need $A$ and $B$ to be local Noetherian rings.
Now, how can we use the fact of being local Noetherian rings to say that they are isomorphic as $k$-algebras? I have looked into the following two arguments so far:
Best Answer
If $f$ is a bijective $k$-algebra homomorphism, then it is already an isomorphism of $k$-algebras, as you can check that $f^{-1}$ is also a $k$-algebra homomorphism in that case. In particular, since $f$ is hypothesized to be a $k$-algebra homomorphism, once you know that $f$ is an isomorphism of rings, you're already done! I suspect the hypothesis that $A$ is Noetherian and local is given merely to ensure that $\dim A$ is finite. To see why this is the case, let $\mathfrak{m}$ be the maximal ideal of $A$. Then $\dim A=\operatorname{ht}\mathfrak{m}$; since $A$ is Noetherian, $\mathfrak{m}$ is finitely generated, and thus by the Hauptidealsatz $\operatorname{ht}\mathfrak{m}$ is finite, whence $\dim A$ is finite as well, as desired. Note that a Noetherian ring need not have finite dimension in general.
Now, in the link you've posted, Georges Elencwajg gives a geometric argument for why the result follows when $\dim A$ is finite, but perhaps things will be clearer with a more algebraic argument. Since $f$ is surjective, we know that $B\cong A\big/\ker f$, so in particular the prime ideals of $B$ are in bijective correspondence with the prime ideals of $A$ containing $\ker f$. Thus, since $\operatorname{ht}\mathfrak{m}$ is finite, let $$\ker f\leqslant\mathfrak{p}_0<\dots<\mathfrak{p}_n=\mathfrak{m}$$ be any strictly ascending chain of prime ideals containing $\ker f$ of maximal possible length; then in particular $\dim B=n$. (Why?) Since $\dim A=\dim B$ by hypothesis, this chain must also be a strictly ascending chain of prime ideals of $A$ of maximal possible length. But $A$ is a domain, so if $\mathfrak{p}_0$ were non-zero, then we could add a link $0=\mathfrak{p}_{-1}<\mathfrak{p}_0$ to the chain, contradicting maximality. So $\mathfrak{p}_0$ must be zero, which in particular forces $\ker f$ to be zero, and hence gives the desired result.
Exactly the same argument shows the more general result that, if $f:A\to B$ is a surjective homomorphism of $k$-algebras, where $\dim A=\dim B$ is finite and $A$ is a domain, then $f$ is an isomorphism of $k$-algebras. In particular, we do not need $A$ to be Noetherian or local, we only need it to be of finite dimension.