Proposition 2

Let A be a local ring whose maximal ideal is a principal ideal Ap. If $\bigcap_{n=1}^{\infty} Ap^n = (0)$ (for example if A is Noetherian, cf. Chapter III, § 3, no. 2, Corollary to Proposition 5), the only ideals of A are (0) and the Ap^n; then either p is nilpotent or A is a valuation ring.

Let A be filtered by the Ap^n and let v denote the corresponding order function (Chapter III, § 2, no. 2). As

the relation $v(x) = +\infty$ implies $x = 0$. Let a be an ideal $\neq (0)$ of A and a an element of a at which v takes its least value; let us write $v(a) = s$ ($s \neq +\infty$). Then $a \subset Ap^s$. In particular, there exists $u \in A$ such that $a = up^s$; as $a \notin Ap^{s+1}$, $u \notin Ap$; hence u is invertible and $p^s \in A$ $a \subset a$. It follows that $a = Ap^s$, whence our first assertion. It is also seen that every element $a \neq 0$ of A may be written in the form $a = up^{v(a)}$ where u is invertible. If $a' = u'p^{v(a')}$ ($u'$ invertible) is another non-zero element of A, then $aa' = uu'p^{v(a)+v(a')}$; hence, if p is not nilpotent, $aa' \neq 0$ and A is an integral domain. Then, as the set of ideals of A is totally ordered by inclusion, we conclude that A is a valuation ring (Theorem I (e)).

For example, if $p$ is a prime number, the local ring $\mathbf{Z}_{(p)}$ is a valuation ring. Let $B = K[X_1, \ldots, X_n]$ be the polynomial ring in $n$ indeterminates over a field $K$; the ideal $BX_1$ is prime, since $B/BX_1$ is isomorphic to $K[X_2, \ldots, X_n]$; hence $B_{BX_1}$ is a valuation ring; it is composed of rational functions $PQ^{-1}$, where $P$ and $Q$ are polynomials and $Q(0, X_2, \ldots, X_n) \neq 0$.

More generally, we shall see that, if $F$ is an extremal element of

$B_{BF}$ is a valuation ring (cf. Chapter VII, § 3, no. 5). \*

The ring of formal power series $K[[T]]$ in one indeterminate over a field $K$ is a Noetherian local integral domain whose maximal ideal is principal; it is therefore a valuation ring. On the other hand the ring $K[[T_1, T_2]]$ of formal power series in two indeterminates, which is a Noetherian local integral domain is not a valuation ring, for neither of the elements $T_1, T_2$ is a multiple of the other.