Let $R,S$ be Noetherian normal rings (i.e. they are locally normal domains at every prime ideal, so in particular they are reduced). Let $f: R \to S$ be a ring homomorphism such that via this map, $S$ is a finitely generated $R$-module (i.e. $f$ is a finite) and also assume that the induced morphism of schemes $f^{\#}:$ Spec $(S) \to $ Spec $(R)$ is Birational ( https://stacks.math.columbia.edu/tag/01RN) i.e. $f^{\#}$ induces bijection between the set of generic points of irreducible components of the schemes (i.e. induces a bijection among the minimal prime ideals of the rings) and for every generic point $P \in$ Spec $(S)$ of an irreducible component of Spec $(S)$ (i.e. whenever $P$ is a minimal prime ideal of $S$) , $f$ induces an isomorphism of local rings $R_{f^{-1}(P)} \to S_P$ .

My question is:

Is it true that for every ideal $I$ of $S$, we have that $\overline {f^{-1}(I) }=f^{-1}(\overline I)$ ?

If this is not true in general, what if we also assume $R,S$ are finitely generated $k$-algebras for some field $k$ and $f$ is a $k$-algebra homomorphism?

Notation: $\bar I$ denotes the integral closure of an ideal.

Initially I asked the question when $R, S$ are integral domain i.e. Spec$(R)$ and Spec $(S)$ are irreducible, in which case, it has been answered by Mohan and KReiser by noting $f$ is actually an isomorphism in that case .

  • 1
    $\begingroup$ If $f$ is finite, birational and $R$ is integrally closed, doesn't it make $f$ an isomorphism? $\endgroup$
    – Mohan
    Oct 15 '19 at 16:24
  • $\begingroup$ @Mohan: thanks for catching that ... see my new edited question, I now don't require my schemes to be irreducible ... $\endgroup$
    – user
    Oct 15 '19 at 22:37

Old answer when $R,S$ were domains:

As Mohan surmises in his comment, $f$ is an isomorphism. This is because $f:R\to S$ expresses $S$ as a finite (thus integral) extension of $R$ inside $K=Frac(R)=Frac(S)$, and $R$ is integrally closed, so $R=S$. One may now quickly verify from the definition that the property you wish for holds.

New answer now that $R,S$ are not required to be domains:

Any normal noetherian ring is a finite product of domains. On the scheme side, this means that a noetherian normal scheme is a finite disjoint union of it's irreducible (and thus connected) components. By the definition of birationality, we see that there's a unique component of $S$ over each component of $R$, so after base change along the inclusion of this component (and the fact that finite morphisms are preserved under base change) we see we're back at the previous situation and $f$ must also be an isomorphism.

  • $\begingroup$ Ah I see ... that makes sense ... do you know what happens if I keep everything same but instead I weaken the hypothesis on $R,S$ and assume they are not Integral domain i.e. I don't assume Spec $(R)$ and Spec $(S)$ to be irreducible (of course they will still be reduced as they are normal) ... one subtlety that would happen now is that we cannot express the Birationally of $f^{\#}:$ Spec $(S)\to$ Spec $(R)$ so easily anymore ... $\endgroup$
    – user
    Oct 15 '19 at 21:59
  • $\begingroup$ I have now edited my question to reflect this new question $\endgroup$
    – user
    Oct 15 '19 at 22:36
  • $\begingroup$ @user I've updated my post to respond to the edit. $\endgroup$
    – KReiser
    Oct 15 '19 at 23:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.