suppose with we need to show that so there exist integers such that therefore because the ideals of a

valuation ring are totally ordered. obviously suppose that then for some and hence which is impossible.

thus and so hence and therefore

if for some integer then because is prime. but that would be impossible because we're given that is "properly" contained in so we must have

Prove that if P is a prime ideal of V properly contained in A, then P is contained in B.