Exercise 5, § 1

Let $\mathbf{A}$ be an integral domain, $K$ its field of fractions and $U$ the subgroup of $K^*$ consisting of the invertible elements of $\mathbf{A}$; consider the group $K^*/U = G$ as ordered by the relation derived by passing to the quotient from the relation $x^{-1}y \in \mathbf{A}$ in $K^*$ (Algebra, Chapter VI, § 1, no. 5).

(a) For a fractional ideal $a \neq 0$ of $K$ to be divisorial, it is necessary and sufficient that the image of $a - \{0\}$ in $G$ be a major set in $G$ (Algebra, Chapter VI, § 1, Exercise 30); for every fractionalideal $a \neq 0$ of $K$, the image of $a - \{0\}$ is the least major set containing the image of $a - \{0\}$. If $\mathbf{A}$ is a Krull domain, $G$ is an ordered group isomorphic to a directed subgroup of an ordered group of the form $\mathbf{Z}^{(1)}$.

(b) Show that for a homomorphism $w$ of $G$ to a totally ordered group $\Gamma$ to be such that the restriction to $A - \{0\}$ of the composite mapping $K^* \to K^*/U \xrightarrow{w} \Gamma$ is a valuation, it is necessary and sufficient that $w$ be an increasing homomorphism.

(c) Let $H$ be the directed subgroup of the product ordered group $\mathbf{Z} \times \mathbf{Z}$ consisting of the ordered pairs $(s_1, s_2)$ such that $s_1 + s_2 \equiv 0$ (mod. 2). Show that there exists no integral domain $\mathbf{A}$ such that $K^*/U$ is an ordered group isomorphic to $H$ (deduce from (b) that such a domain would necessarily be a Krull domain, but Proposition 9 of no. 5 would not hold for this ring (cf. Exercise 22)).