Exercise 5, § 4

Let $K$ be a field and $\mathfrak{M}$ the set of subrings of $K$ which are not fields. Show that a maximal element of $\mathfrak{M}$ is a valuation ring of height 1 of its field of fractions $L$ and that $K$ is an algebraic extension of $L$. The following conditions are equivalent: (a) $\mathfrak{M}$ is not empty. (b) $\mathfrak{M}$ admits a maximal element. (c) There exists a valuation of height 1 on $K$ (d) $K$ is not an algebraic extension of a finite prime field (cf. § 8, no. 1).