The Stacks project

Lemma 29.53.10. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes. Suppose that $Y = Y_1 \amalg Y_2$ is a disjoint union of two schemes. Write $f_ i = f|_{Y_ i}$. Let $X_ i'$ be the normalization of $X$ in $Y_ i$. Then $X_1' \amalg X_2'$ is the normalization of $X$ in $Y$.

Proof. In terms of integral closures this corresponds to the following fact: Let $A \to B$ be a ring map. Suppose that $B = B_1 \times B_2$. Let $A_ i'$ be the integral closure of $A$ in $B_ i$. Then $A_1' \times A_2'$ is the integral closure of $A$ in $B$. The reason this works is that the elements $(1, 0)$ and $(0, 1)$ of $B$ are idempotents and hence integral over $A$. Thus the integral closure $A'$ of $A$ in $B$ is a product and it is not hard to see that the factors are the integral closures $A'_ i$ as described above (some details omitted). $\square$

Comments (2)

Comment #8499 by on

Suggested alternative proof: since is quasi-compact quasi-separated, so is the composite . Hence, we can take the normalization of in . To check that is the normalization of in , it suffices to see that the factorization of satisfies parts (1) and (2) of Lemma 29.53.4. Part (2) can be verified using the universal property of the factorization of . We see part (1), i.e., that is integral: On the one hand, is affine, by Schemes, Lemma 26.6.8. On the other hand, let be open affine. Denote to the inverse image in . The map is integral. The induced map on global sections by is , which is integral (it suffices to show that the ideals and of are integral over , for ).

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  • 3 comment(s) on Section 29.53: Relative normalization

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