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).
$\square$

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## Comments (1)

Comment #8499 by ElĂas Guisado on

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