Lemma 28.7.9. Let X be an integral normal scheme. Then \Gamma (X, \mathcal{O}_ X) is a normal domain.
The ring of functions on a normal scheme is normal.
Proof. Set R = \Gamma (X, \mathcal{O}_ X). It is clear that R is a domain. Suppose f = a/b is an element of its fraction field which is integral over R. Say we have f^ d + \sum _{i = 0, \ldots , d - 1} a_ i f^ i = 0 with a_ i \in R. Let U \subset X be a nonempty affine open. Since b \in R is not zero and since X is integral we see that also b|_ U \in \mathcal{O}_ X(U) is not zero. Hence a/b is an element of the fraction field of \mathcal{O}_ X(U) which is integral over \mathcal{O}_ X(U) (because we can use the same polynomial f^ d + \sum _{i = 0, \ldots , d - 1} a_ i|_ U f^ i = 0 on U). Since \mathcal{O}_ X(U) is a normal domain (Lemma 28.7.4), we see that f_ U = (a|_ U)/(b|_ U) \in \mathcal{O}_ X(U). It is clear that f_ U|_ V = f_ V whenever V \subset U \subset X are nonempty affine open. Hence the local sections f_ U glue to an element g \in R = \Gamma (X, \mathcal{O}_ X). Then bg and a restrict to the same element of \mathcal{O}_ X(U) for all U as above, hence bg = a, in other words, g maps to f in the fraction field of R. \square
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