Exercise 7, § 4

A local integral domain A is said to be of dimension 1 if A is not a field and there is no prime ideal of A distinct from (0) and the maximal ideal m of A. For every ideal $a$ of A, let A: $a$ denote the sub-A-module of the field of fractions of A which is the transporter of $a$ in A (which always contains A). In the following A is assumed to be of dimension 1.

(a) Show that, if A is strongly Laskerian (Chapter IV, § 2, Exercise 28), then $A : m \neq A$. (In the opposite case, note that $A : m^r = A$ for every integer $r > 0$; on the other hand, every ideal $\neq 0$ of A contained in m is m-primary and so in particular is every principal ideal $Ax \subset m$ (where $x \neq 0$); note finally that $m^h \neq m^k$ for $h \# k$ (Chapter IV, § 2, Exercise 29 (d)).

(b) Show that for a strongly Laskerian local integral domain A of dimension 1 the following properties are equivalent:

(α) A is completely integrally closed. (β) The maximal ideal m is principal.

(γ) Every m-primary ideal is of the form $m^k$. (δ) The ideal $m$ is invertible (Chapter II, § 5, no. 6), which is equivalent to $m \cdot (A : m) = A$. (ε) $A$ is a discrete valuation ring. (To show that (α) implies (δ), use (a) and observe that if $m \cdot (A : m)^k = m$ for all $k > 0$, $A$ is not completely integrally closed (cf. Chapter VII, § 1, Exercise 4). To see that (6) implies (y), note that for every ideal $a \neq 0$ contained in $m$ we may write $a = ma_1$, where $a_1 \neq 0$ and use Chapter IV, § 2, Exercise 29 (d). Finally, to see that (y) implies (β), observe that $m \neq m^2$ and that (γ) implies that $m/m^2$ is a 1-dimensional vector $(A/m)$-space.) (c) Show, in the notation of Chapter III, § 3, Exercise 15 (b), that the local ring $A$ is a Noetherian integral domain of dimension 1 but is not integrally closed. (d) Let $K$ be an algebraically closed field and $A$ the subring of the field $K(X, Y)$ of rational functions in two indeterminates over $K$ consisting of the elements $f \in K(X, Y)$ such that 0 is substitutable for $X$ in $f$ and $f(0, Y)$ belongs to $K$. Show that $A$ is a strongly Laskerian integrally closed local integral domain of dimension 1 but that it is not completely integrally closed. (If $B$ is the polynomial ring $K[X, Y]$, $p$ the prime ideal $BX$ of $B$, note that $A$ is the subring of $B_p$ equal to $K + pB_p$.) (*)