Lemma 10.115.7. Let R \to S be an injective finite type ring map. Assume R is a domain. Then there exists an integer d and a factorization
R \to R[y_1, \ldots , y_ d] \to S' \to S
by injective maps such that S' is finite over R[y_1, \ldots , y_ d] and such that S'_ f \cong S_ f for some nonzero f \in R.
Proof.
Pick x_1, \ldots , x_ n \in S which generate S over R. Let K be the fraction field of R and S_ K = S \otimes _ R K. By Lemma 10.115.4 we can find y_1, \ldots , y_ d \in S such that K[y_1, \ldots , y_ d] \to S_ K is a finite injective map. Note that y_ i \in S because we may pick the y_ j in the \mathbf{Z}-algebra generated by x_1, \ldots , x_ n. As a finite ring map is integral (see Lemma 10.36.3) we can find monic P_ i \in K[y_1, \ldots , y_ d][T] such that P_ i(x_ i) = 0 in S_ K. Let f \in R be a nonzero element such that fP_ i \in R[y_1, \ldots , y_ d][T] for all i. Then fP_ i(x_ i) maps to zero in S_ K. Hence after replacing f by another nonzero element of R we may also assume fP_ i(x_ i) is zero in S. Set x_ i' = fx_ i and let S' \subset S be the R-subalgebra generated by y_1, \ldots , y_ d and x'_1, \ldots , x'_ n. Note that x'_ i is integral over R[y_1, \ldots , y_ d] as we have Q_ i(x_ i') = 0 where Q_ i = f^{\deg _ T(P_ i)}P_ i(T/f) which is a monic polynomial in T with coefficients in R[y_1, \ldots , y_ d] by our choice of f. Hence R[y_1, \ldots , y_ d] \subset S' is finite by Lemma 10.36.5. Since S' \subset S we have S'_ f \subset S_ f (localization is exact). On the other hand, the elements x_ i = x'_ i/f in S'_ f generate S_ f over R_ f and hence S'_ f \to S_ f is surjective. Whence S'_ f \cong S_ f and we win.
\square
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