Lemma 10.50.4. Let A be a valuation ring with maximal ideal \mathfrak m and fraction field K. Let x \in K. Then either x \in A or x^{-1} \in A or both.
Proof. Assume that x is not in A. Let A' denote the subring of K generated by A and x. Since A is a valuation ring we see that there is no prime of A' lying over \mathfrak m. Since \mathfrak m is maximal we see that V(\mathfrak m A') = \emptyset . Then \mathfrak m A' = A' by Lemma 10.17.2. Hence we can write 1 = \sum _{i = 0}^ d t_ i x^ i with t_ i \in \mathfrak m. This implies that (1 - t_0) (x^{-1})^ d - \sum t_ i (x^{-1})^{d - i} = 0. In particular we see that x^{-1} is integral over A, and hence x^{-1} \in A by Lemma 10.50.3. \square
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