$\xymatrix{ Y_2 \ar[r] \ar[d]_{f_2} & Y_1 \ar[d]^{f_1} \\ X_2 \ar[r]^\varphi & X_1 }$

be a fibre square in the category of schemes. Assume $f_1$ is quasi-compact and quasi-separated, and $\varphi$ is smooth. Let $Y_ i \to X_ i' \to X_ i$ be the normalization of $X_ i$ in $Y_ i$. Then $X_2' \cong X_2 \times _{X_1} X_1'$.

Proof. The base change of the factorization $Y_1 \to X_1' \to X_1$ to $X_2$ is a factorization $Y_2 \to X_2 \times _{X_1} X_1' \to X_2$ and $X_2 \times _{X_1} X_1' \to X_2$ is integral (Morphisms, Lemma 29.44.6). Hence we get a morphism $h : X_2' \to X_2 \times _{X_1} X_1'$ by the universal property of Morphisms, Lemma 29.53.4. Observe that $X_2'$ is the relative spectrum of the integral closure of $\mathcal{O}_{X_2}$ in $f_{2, *}\mathcal{O}_{Y_2}$. If $\mathcal{A}' \subset f_{1, *}\mathcal{O}_{Y_1}$ denotes the integral closure of $\mathcal{O}_{X_1}$, then $X_2 \times _{X_1} X_1'$ is the relative spectrum of $\varphi ^*\mathcal{A}'$, see Constructions, Lemma 27.4.6. By Cohomology of Schemes, Lemma 30.5.2 we know that $f_{2, *}\mathcal{O}_{Y_2} = \varphi ^*f_{1, *}\mathcal{O}_{Y_1}$. Hence the result follows from Lemma 37.19.1. $\square$

Comment #7687 by nkym on

I was wondering if $\mathcal{A}'$ was meant to be the integral closure of $\mathcal{O}_{X_1}$ instead of $\mathcal{O}_{X_2}$.

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