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Let $S$ be an integral Dedekind scheme.

Let $f:X\longrightarrow \mathbf{P}^1_{S}$ be a finite flat surjective morphism, where $X$ is an integral normal scheme.

Let $\eta$ be the generic point of $S$. Note that $f_\eta:X_\eta\longrightarrow \mathbf{P}^1_{K(S)}$ is a finite morphism of curves over $K(S)$.

Question. Is $f$ the normalization of $\mathbf{P}^1_S$ in the function field of $X_\eta$?

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up vote 2 down vote accepted

In your case the function field of $X_{\eta}$ is the same as the function field of $X$. Thus the following general remark answers your question affirmatively.

Assume $Y$ is an integral scheme and $L$ is an algebraic extension of the function field $K(Y)$ of $Y$. Let $\pi\colon X \to Y$ be an integral morphism of schemes such that $X$ is integral and normal and such that $\pi$ induces on function fields the extension $K(Y) \subset L = K(X)$. Then $X$ is the normalization of $Y$ in $L$. In fact this follows essentially from the definition of "normalization" and the fact that integral ring homomorphisms are stable under localization.

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Ok. So it's quite elementary actually. – Homalor Jul 28 '11 at 16:03

Yes. This follows from Zariski's Main Theorem (although there are probably more direct arguments in this case).

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Let me see if I got this. Let $\widetilde{X}\longrightarrow \mathbf{P}^1_S$ be the normalization. Then, by the universal property of normalization, there is a unique morphism $q:X\longrightarrow \widetilde{X}$. Since $q$ is birational and quasi-finite, Zariski's main theorem implies that it is an open immersion. Now, since $f$ was finite, we have that $q$ is finite. Therefore, it is closed. Therefore, $q$ is the identity morphism. – Homalor Jul 28 '11 at 14:49

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