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 10.122.3). Consider the diagram of topological spaces
By Algebra, Theorem 10.123.12 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 10.41.6 and 10.36.22. We conclude that \mathop{\mathrm{Spec}}(B'/IB') \supset \phi (T) is open and closed. By Algebra, Lemma 10.24.3 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 10.23.1. Finally, it follows that C'_1 = C_1 and the proof is complete. \square
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