Proposition 2

Let $A$ be an integral domain. For $A$ to be factorial, it is necessary and sufficient that there exist a subset $P$ of $A$ such that every $a \in A - \{0\}$ may be written uniquely in the form $a = u \prod_{p \in P} p^{n(p)}$, where $u \in U$ and the $n(p)$ are positive integers which are zero except for a finite number of them.

If $P$ satisfies this condition, clearly all its elements are extremal and every extremal element of $A$ is associated with a unique element of $P$. Recall that $P$ is then called a representative system of extremal elements of $A$ (Algebra, Chapter VII, $§ 1$, no. 3, Definition 2).

Suppose always that $A$ is factorial. It has been seen (no. 2, Theorem 1) that the group $\mathscr{P}^*$ is a lattice. We may therefore apply the results of Algebra, Chapter VI, $§ 1$, nos. 9 and 13. In particular, every element of $K^*$ may be written in an essentially unique way in the form of an irreducible fraction. Any two elements $a, b$ of $K^*$ have a g.c.d. and a l.c.m.; if $a = u \prod_{p \in P} p^{n(p)}$ and

are decompositions of $a$ and $b$ as products of extremal elements, then:

where $w, w'$ are in U. We recover, in particular, the results of Algebra, Chapter VIII, § 1, no. 3.

For all $p \in P$, the mapping $a \mapsto n(p)$ is a discrete valuation $v_p$ on K whose ring is obviously $A_{A_p}$. It follows from Theorem 1(e) that the $v_p$ are just the essential valuations of A and that the ideals $Ap$ ($p \in P$) are just the prime ideals of A of height 1.