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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 .

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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