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). $\square$

Comment #8499 by on

Suggested alternative proof: since $Y_i\to Y$ is quasi-compact quasi-separated, so is the composite $f_i:Y_i\to Y\to X$. Hence, we can take the normalization $X_i'\to X$ of $X$ in $Y_i$. To check that $X'_1\amalg X'_2\to X$ is the normalization of $X$ in $Y$, it suffices to see that the factorization $Y_1\amalg Y_2\to X'_1\amalg X'_2\to X$ of $f$ satisfies parts (1) and (2) of Lemma 29.53.4. Part (2) can be verified using the universal property of the factorization $Y_i\to X'_i\to X$ of $f_i$. We see part (1), i.e., that $\nu:X'_1\amalg X'_2\to X$ is integral: On the one hand, $\nu$ is affine, by Schemes, Lemma 26.6.8. On the other hand, let $U=\operatorname{Spec} A\subset X$ be open affine. Denote $U_i'=\operatorname{Spec} A_i'$ to the inverse image in $X_i'$. The map $A\to A_i'$ is integral. The induced map on global sections by $\nu^{-1}(U)=U_1'\amalg U_2'\to U$ is $A\to A_1'\times A_2'=A'$, which is integral (it suffices to show that the ideals $I_1=A_1'\times\{0\}$ and $I_2=\{0\}\times A_2'$ of $A'$ are integral over $A$, for $I_1+I_2=A'$).

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