Exercise 4, § 2

It is said that an ordered field $K$ is euclidean if for every $x \geq 0$ in $K$, there exists $y \in K$ such that $x = y^2$.**

a) Let $K$ be an ordered euclidean field, $G$ a subfield of $K$ such that there exists no extension of $G$ contained in $K$, distinct from $G$ and comparable with $G$. Now prove that if $f$ is a place of $K$ with values in an ordered field $L$, algebraic extension of $G$, and reducing to the identity in $G$, then $f$ is equivalent to the canonical place of $K$ with respect to $G$. (One may consider that $f$ takes its values in a maximal ordered field $N$ algebraic extension of $L$. Observe that $G$ is euclidean, and that if $x \in G$, $y \in K$ and $x < y$, one has $f(y) = \infty$ or $f(y) > x$. If $A$ and $\mathfrak{p}$ are the ring and the ideal of the place $f$, now prove successively that one has $\mathfrak{p} \subset I(G)$, $F(G) \subset A$, $I(G) \subset \mathfrak{p}$ and $A \subset F(G)$, noting that $N$ is comparable with $G$).

b) Let $K$ be an euclidean ordered field, $f$ a place of $K$ with values in a maximal ordered field $L$, $A$ and $\mathfrak{p}$ the ring and ideal of $f$. Let $G$ be a maximal subfield among the subfields $E$ of $K$ such that $E \cap \mathfrak{p} = 0$. Now prove that $G \subset A$ and that $G$ is algebraically closed in $K$, and consequently euclidean. Prove that there exists no extension of $G$ contained in $K$, distinct from $G$ and comparable to $G$. Moreover, the subfield of $L$ generated by $f(A)$ is algebraic over $f(G)$ and $f$ is consequently equivalent to the canonical place of $K$ with respect to $G$.

c) Let $K$ be a maximal ordered field, $L$ a maximal ordered subfield of $K$, distinct from $K$ and such that $K$ is comparable to $L$ (Alg., Chap. VI, § 2, Exer. 15). Now prove that there exists no place $f$ of $K$ with values in $L$, reducing to the identity automorphism in $L$.