Lemma 10.50.5. Let A \subset K be a subring of a field K such that for all x \in K either x \in A or x^{-1} \in A or both. Then A is a valuation ring with fraction field K.
Proof. If A is not K, then A is not a field and there is a nonzero maximal ideal \mathfrak m. If \mathfrak m' is a second maximal ideal, then choose x, y \in A with x \in \mathfrak m, y \not\in \mathfrak m, x \not\in \mathfrak m', and y \in \mathfrak m'. Then neither x/y \in A nor y/x \in A contradicting the assumption of the lemma. Thus we see that A is a local ring. Suppose that A' is a local ring contained in K which dominates A. Let x \in A'. We have to show that x \in A. If not, then x^{-1} \in A, and of course x^{-1} \in \mathfrak m_ A. But then x^{-1} \in \mathfrak m_{A'} which contradicts x \in A'. \square
Comments (2)
Comment #8361 by Et on
Comment #8966 by Stacks project on
There are also: