Lemma 76.25.2 (Normalization commutes with smooth base change). Let S be a scheme. Let
\xymatrix{ Y_2 \ar[r] \ar[d] & Y_1 \ar[d]^ f \\ X_2 \ar[r]^\varphi & X_1 }
be a fibre square of algebraic spaces over S. Assume f 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_1 and X_2 \times _{X_1} X_1' \to X_1 is integral (Morphisms of Spaces, Lemma 67.45.5). Hence we get a morphism h : X_2' \to X_2 \times _{X_1} X_1' by the universal property of Morphisms of Spaces, Lemma 67.48.5. 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_2}, then X_2 \times _{X_1} X_1' is the relative spectrum of \varphi ^*\mathcal{A}' as the construction of the relative spectrum commutes with arbitrary base change. By Cohomology of Spaces, Lemma 69.11.2 we know that f_{2, *}\mathcal{O}_{Y_2} = \varphi ^*f_{1, *}\mathcal{O}_{Y_1}. Hence the result follows from Lemma 76.25.1.
\square
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