Exercise 3, § 3

Let $K$ be a field, $w$ a valuation on $K$, $\Gamma$ the value group of $w$ and $k$ the residue field of $w$. We form the corresponding field $S(\Gamma, k)$ and preserve the notation of Exercise 2. For all $t \in k$, let $t^0$ be an element of the class $t$ in the valuation ring of $w$; for all $a \in \Gamma$, let $u_a^0$ be an element of $K$ such that $w(u_a^0) = a$.

(a) Let $M$ be a subset of $S(\Gamma, k)$ containing the elements $tu_a$ for every ordered pair $(t, a) \in k \times \Gamma$; suppose that there exists a bijection $x \mapsto x^0$ of $M$ onto a subset $M^0$ of $K$ and that the following conditions are fulfilled: (1) $(tu_a)^0 = t^0 u_a^0$ for every ordered pair $(t, a) \in k \times \Gamma$; (2) if $x = (x_\alpha)$ belongs to $M$ and $C_x$ is the well-ordered subset of $\Gamma$ consisting of the $a$ such that $x_\alpha \neq 0$, then for every segment $D$ of $C_x$ the element $(x_\lambda)_{\lambda \in D}$ belongs to $M$; (3) if $x, y$ are two elements of $M$ and $tu_a$ is the dominant term of $x - y$, then $w(x^0 - y^0 - t^0 u_a^0) > w(x^0 - y^0)$. If $z'$ is an element of $K$ not belonging to $M^0$, show that there exists an element $z \notin M$ in $S(\Gamma, k)$ such that the mapping which coincides with $x \mapsto x^0$ on M and maps $z'$ to $z$ also satisfies the above conditions. (Note that, if $w(z') = \alpha$, there exists a $t^0$ such that

write $z_\alpha = t u_\alpha$, then show by transfinite induction that the $z_\beta$ of index $\beta > \alpha$ may be determined so that $z = (z_\lambda)_{\lambda \in \Gamma}$ solves the problem.) (b) Deduce from (a) that $\mathrm{Card}(K) \leq \mathrm{Card}\ S(\Gamma, k)$.