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Proof. Let R be a normal domain, and let S \subset R be a multiplicative subset. Suppose g is an element of the fraction field of R which is integral over S^{-1}R. Let P = x^ d + \sum _{j < d} a_ j x^ j be a polynomial with a_ i \in S^{-1}R such that P(g) = 0. Choose s \in S such that sa_ i \in R for all i. Then sg satisfies the monic polynomial x^ d + \sum _{j < d} s^{d-j}a_ j x^ j which has coefficients s^{d-j}a_ j in R. Hence sg \in R because R is normal. Hence g \in S^{-1}R. \square


Comments (1)

Comment #10066 by Joe Lamond on

If is a normal domain, and is a multiplicative set containing , then is the zero ring, which is not even an integral domain. I suppose the zero ring is, strictly speaking, a normal ring, since it is vacuously true that the localizations at prime ideals are normal domains, but I wonder if this was really intended.

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  • 4 comment(s) on Section 10.37: Normal rings

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