Exercise 2, § 3

Let $\Gamma$ be a totally ordered group and $k$ a field.

(a) Let $A, B$ be two subsets of $\Gamma$ well ordered by the induced ordering; show that the set $A + B$ is well ordered and that, for all $y \in A + B$, the set of ordered pairs $(a, \beta) \in A \times B$ such that $a + \beta = y$ is finite. (To prove the first assertion, argue by reductio ad absurdum by showing that otherwise it would be possible to define a strictly decreasing sequence $(y_i)$ of elements of $A + B$ such that $\gamma_n = a_n + \beta_n$, where $a_n \in A, \beta_n \in B$, the sequence $(a_n)$ is strictly increasing and the sequence $(\beta_n)$ is strictly decreasing; conclude using Set Theory, Chapter 111, § 4, Exercise 3.)

(b) Let $S(\Gamma, k)$ be the set of families $x = (x_\alpha)_{\alpha \in \Gamma}$ such that $x_\alpha \in k$ for all $\alpha \in \Gamma$ and the set of $a$ for which $x_\alpha \neq 0$ is a well ordered subset of $\Gamma$. Show that $S(\Gamma, k)$ is an additive subgroup of $k^\Gamma$; moreover, if $x = (x_\alpha), y = (y_\alpha)$ are two elements of $S(\Gamma, k)$, the set $C$ of $a + \beta$ for the ordered pairs $(a, \beta)$ such that $x_\alpha \neq 0$ and $y_\alpha \neq 0$ is well ordered by (a) and, for all $y \in C$, the element $z_y = \sum_{\alpha + \beta = y} x_\alpha y_\beta$ is defined in $k$; if we write $z_\gamma = 0$ for $a \notin C$, the element $z = (z_\alpha) \in k^\Gamma$ belongs to $S(\Gamma, k)$; it is denoted by $xy$. Show that with this law of composition and addition on the product group $k^\Gamma$ the set $S(\Gamma, k)$ is a field; moreover, for all $x = (x_\alpha) \neq 0$ of $S(\Gamma, k)$, let $\lambda$ be the least of the $a \in \Gamma$ such that $x_\alpha \neq 0$; if we write $v(x) = \lambda$ (and $v(0) = +\infty$), show that $v$ is a valuation on $S(\Gamma, k)$ with $k$ as residue field and $\Gamma$ as value group. $v$ is called the canonical valuation on the field $S(\Gamma, k)$. The field $k$ is canonically identified with a subfield of $S(\Gamma, k)$. For all $a \in \Gamma$, let $u_a$ denote the element $(x_\lambda)_{\lambda \in \Gamma}$ such that $x_\alpha = 1, x_\lambda = 0$ for $\lambda \neq a$; for all non-zero $z \in S(\Gamma, k)$, there is then a unique ordered pair $(t, a) \in k \times \Gamma$ such that $v(z) = \alpha$ and $v(z - tu_a) > a$. $tu_a$ is called the dominant term of $z$.