Lemma 10.161.6. Let R be a Noetherian domain. If R[z, z^{-1}] is N-1, then so is R.
Proof. Let R' be the integral closure of R in its field of fractions K. Let S' be the integral closure of R[z, z^{-1}] in its field of fractions. Clearly R' \subset S'. Since K[z, z^{-1}] is a normal domain we see that S' \subset K[z, z^{-1}]. Suppose that f_1, \ldots , f_ n \in S' generate S' as R[z, z^{-1}]-module. Say f_ i = \sum a_{ij}z^ j (finite sum), with a_{ij} \in K. For any x \in R' we can write
x = \sum h_ i f_ i
with h_ i \in R[z, z^{-1}]. Thus we see that R' is contained in the finite R-submodule \sum Ra_{ij} \subset K. Since R is Noetherian we conclude that R' is a finite R-module. \square
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