Exercise 11, § 1

A divisor $d$ in an integral domain $A$ is called finitely generated if it is of the form $\operatorname{div}(a)$, where $a$ is a finitely generated fractional ideal (which is not necessarily divisorial; cf. (b)).

(a) Show that, if $A$ is a Krull domain, every divisor of $D(A)$ is of the form $\operatorname{div}(a)$, where $a = Ax + Ay$, $x, y$ being elements $\neq 0$ of the field of fractions of $A$ (use Corollary 2 to Proposition 9 of no. 5).

(b) Let $K$ be a field, $A = K[X_n]_{n \in \mathbf{N}}$ the ring of polynomials over $K$ in a denumerable set of indeterminates, which is a Krull ring (Exercise 8). Let $A'$ be the subring of $A$ generated by the unit element and the monomials $X_i X_j$ for all pairs $(i, j)$ (equivalently, $A'$ is the set of polynomials all whose terms have even degree). If $L$ and $L'$ are the fields of fractions of $A$ and $A'$, show that $A' = A \cap L'$, hence $A'$ is a Krull ring. Let $p$ be the ideal in $A'$ generated by the products $X_i X_j$ for all $i \geq 0$; show that $p$ is a prime ideal of height 1 (hence divisorial) but that it k not finitely generated.