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The Stacks project

Lemma 37.19.2 (Normalization commutes with smooth base change). Let

\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


Comments (2)

Comment #7687 by nkym on

I was wondering if was meant to be the integral closure of instead of .


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