Exercise 20, § 1

An integral domain A is called Bezoutian (or a Bezout domain) if every finitely generated ideal of A is principal. Every Bezoutian Noetherian domain is a principal ideal domain. Every valuation ring is a Bezoutian domain (and therefore a Bezoutian domain is not therefore necessarily Noetherian nor completely integrally closed). (a) Show that every Bezoutian domain is integrally closed (cf. Exercise 6). If an integral domain is the union of a right directed family of Bezoutian subdomains, it is Bezoutian. If A is a Bezoutian domain, so is $S^{-1}A$ for every multiplicative subset S of A such that $0 \notin S$. (b) Let $v_i$ ($1 \leq i \leq n$) be independent valuations on a field K and $A_i$ the ring of the valuation $v_i$. Show that the intersection A of the $A_i$ is a Bezoutian domain. (If an ideal a of A is finitely generated, the set of $v_i(x)$ for $x \in a$ admits a least element $a_i$ in the value group of $v_i$; for all i, let $x_i \in a$ be such that $v_i(x_i) = a_i$. Using the Approximation Theorem (Chapter VI, § 7, no. 2, Theorem 1), show that there are elements $a_i \in A$ such that $x = \sum_{i=1}^n a_i x_i \in a$ satisfies the relations $v_i(x) = a_i$ for $1 \leq i \leq n$) If the $v_i$ are discrete valuations, A is a principal ideal domain. (c) Let K be an algebraically closed field of characteristic $\neq 2$ and let A be the subring of an algebraic closure of $K(X)$ generated by $K$ and two sequences of elements $(x_n), (1/x_n)$ ($1 \leq n < +\infty$), where $x_1 = X$ and $x_{n-1} = x_n^2$. Show that $A$ is Bezoutian (use (a)). If $\mathfrak{p}$ is a prime ideal $\neq 0$ of $A$, show that $\mathfrak{p}$ is generated by a sequence of elements of the form $x, -a$, where $a, \in K$, $a, \neq 0$ and $a_n^2 = a_{n-1}$ (consider for all $n$ the intersection $\mathfrak{p} \cap K[x_n, 1/x_n]$). Show that $\mathfrak{p}$ is not finitely generated.