Exercise 9, § 2

Let $\mathbf{A}$ be an integral domain. (a) Let $(p_i)_{1 \leq i \leq m}, (p'_j)_{1 \leq j \leq n}$ be two finite families of invertible prime ideals such that $p_1p_2\ldots p_m = p'_1p'_2\ldots p'_n$. Show that $m = n$ and that there exists a permutation $\pi$ of $[1, n]$ such that $p'_i = p_{\pi(i)}$ for all $i$ such that $1 \leq i \leq n$. (b) Let $p$ be a prime ideal of $\mathbf{A}$ and $a$ an element of $\mathbf{A} - q$; if $p \subset \mathbf{A}a + p^2$, show that $p = p(\mathbf{A}a + p)$; deduce that, if $p$ is invertible, then $\mathbf{A}a + p = \mathbf{A}$. (c) Show that, if every ideal of $\mathbf{A}$ is a product (not necessarily a priori unique) of prime ideals of $\mathbf{A}$, $\mathbf{A}$ is a Dedekind domain. (Show first that every invertible prime ideal $q$ is maximal, considering an element $a \in \mathbf{A} - q$ and decompositions of $\mathbf{A}a + p$ and $\mathbf{A}a^2 + q$ as products of prime ideals and applying (a) in the ring $A/p$ to show that $Aa^2 + p = (Aa + p)^2$; then use (b). Prove then that every prime ideal $p \neq 0$ is invertible, by decomposing as a product of prime ideals a principal ideal $Ab$ where $b \in p$, observing that the factors of this product are invertible and, by applying the above, show that $p$ is necessarily equal to one of these factors. Conclude with the aid of Exercise 8.)