Lemma 15.11.5. Let $(A, I)$ be a pair. Let $A \to B$ be a finite type ring map such that $B/IB = C_1 \times C_2$ with $A/I \to C_1$ finite. Let $B'$ be the integral closure of $A$ in $B$. Then we can write $B'/IB' = C_1 \times C'_2$ such that the map $B'/IB' \to B/IB$ preserves product decompositions and there exists a $g \in B'$ mapping to $(1, 0)$ in $C_1 \times C'_2$ with $B'_ g \to B_ g$ an isomorphism.

**Proof.**
Observe that $A \to B$ is quasi-finite at every prime of the closed subset $T = \mathop{\mathrm{Spec}}(C_1) \subset \mathop{\mathrm{Spec}}(B)$ (this follows by looking at fibre rings, see Algebra, Definition 10.121.3). Consider the diagram of topological spaces

By Algebra, Theorem 10.122.12 for every $\mathfrak p \in T$ there is a $h_\mathfrak p \in B'$, $h_\mathfrak p \not\in \mathfrak p$ such that $B'_ h \to B_ h$ is an isomorphism. The union $U = \bigcup D(h_\mathfrak p)$ gives an open $U \subset \mathop{\mathrm{Spec}}(B')$ such that $\phi ^{-1}(U) \to U$ is a homeomorphism and $T \subset \phi ^{-1}(U)$. Since $T$ is open in $\psi ^{-1}(V(I))$ we conclude that $\phi (T)$ is open in $U \cap (\psi ')^{-1}(V(I))$. Thus $\phi (T)$ is open in $(\psi ')^{-1}(V(I))$. On the other hand, since $C_1$ is finite over $A/I$ it is finite over $B'$. Hence $\phi (T)$ is a closed subset of $\mathop{\mathrm{Spec}}(B')$ by Algebra, Lemmas 10.40.6 and 10.35.22. We conclude that $\mathop{\mathrm{Spec}}(B'/IB') \supset \phi (T)$ is open and closed. By Algebra, Lemma 10.23.3 we get a corresponding product decomposition $B'/IB' = C'_1 \times C'_2$. The map $B'/IB' \to B/IB$ maps $C'_1$ into $C_1$ and $C'_2$ into $C_2$ as one sees by looking at what happens on spectra (hint: the inverse image of $\phi (T)$ is exactly $T$; some details omitted). Pick a $g \in B'$ mapping to $(1, 0)$ in $C'_1 \times C'_2$ such that $D(g) \subset U$; this is possible because $\mathop{\mathrm{Spec}}(C'_1)$ and $\mathop{\mathrm{Spec}}(C'_2)$ are disjoint and closed in $\mathop{\mathrm{Spec}}(B')$ and $\mathop{\mathrm{Spec}}(C'_1)$ is contained in $U$. Then $B'_ g \to B_ g$ defines a homeomorphism on spectra and an isomorphism on local rings (by our choice of $U$ above). Hence it is an isomorphism, as follows for example from Algebra, Lemma 10.22.1. Finally, it follows that $C'_1 = C_1$ and the proof is complete. $\square$

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