# Integral closure of pull-back of an ideal via a birational, finite morphism of rings

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 .

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

• 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 ...
– user
Oct 15 '19 at 21:59
• I have now edited my question to reflect this new question
– user
Oct 15 '19 at 22:36
• @user I've updated my post to respond to the edit. Oct 15 '19 at 23:56