Proposition 15

Let $A$ be a filtered ring whose filtration is exhaustive and such that every principal ideal of $A$ is closed under the topology defined by the filtration. If the associated graded ring $\mathrm{gr}(A)$ (Chapter III, § 2, no. 3) is a completely integrally closed domain, then $A$ is a completely integrally closed domain.

Let $(A_n)_{n \in \mathbf{Z}}$ be the filtration defined on $A$; as $\bigcap_{n \in \mathbf{Z}} A_n$, is the closure of the ideal (0) (Chapter III, § 2, no. 5), the hypothesis implies first that the filtration $(A_n)$ is separated and, as $\mathrm{gr}(A)$ is an integral domain, so then is $A$ (Chapter III, § 2, no. 3, Corollary to Proposition 1). Let $x = b/a$ be an element of the field of fractions $K$ of $A$ ($a \in A, b \in A$) for which there exists an element $d \neq 0$ of $A$ such that $dx^n \in A$ for all $n \geq 0$. We must prove that $b \in A a$ and, as by hypothesis the ideal $A a$ is closed, it is sufficient to show that, for all $n \in \mathbf{Z}, b \in A a + A$. As the filtration of $A$ is exhaustive, there exists an integer $q \in \mathbf{Z}$ such that $b \in A a + A_q$. It will therefore suffice to prove that the relation $b \in A a + A_q$ implies $b \in A a + A_{q+1}$.

Suppose then that $b = a y + z$ where $y \in A, z \in A_{q+1}$. Then by hypothesis $dx^n \in A$ for all $n \geq 0$, whence we obtain immediately $d(x - y)^n \in A$ for all $n \geq 0$; in other words, $dz^n = a^n t_n$ where $t_n \in A$ for all $n \geq 0$. We can obviously limit our attention to the case where $z \neq 0$. Let $v$ denote the order function on $A$ (Chapter III, § 1, no. 2) and let us write $v(d) = n_1,\ v(z) = n_2 \geq m,\ v(a) = n_3$; let $d', z', a'$ be the respective images of $d, z, a$ in $A_{n_1}/A_{n_1+1},\ A_{n_2}/A_{n_2+1},\ A_{n_3}/A_{n_3+1}$. For all $n \geq 0,\ v(dz^n) = n_1 + nn_2$ (Chapter III, § 2, no. 3, Proposition 1) and hence the canonical image in $\mathrm{gr}(A)$ of $dz^n$ is $d'{z'}^n$; similarly it is seen that the canonical image in $\mathrm{gr}(A)$ of $a^n t_n$ is of the form ${d'}^n t'_n$ where $t'_n \in \mathrm{gr}(A)$ and, as $a' \neq 0$ we deduce that, for all $n \geq 0,\ d'(z'/a')^n \in \mathrm{gr}(A)$. The hypothesis that $\mathrm{gr}(A)$ is completely integrally closed therefore implies the existence of an $s' \in \mathrm{gr}(A)$ such that $z' = a's'$; decomposing $s'$ into a sum of homogeneous elements, it is further seen (since $z'$ and $a'$ are homogeneous) that $s'$ may be assumed to be homogeneous, that is the image of an element $s \in A$; then $v(as) = v(z) = n_2$ and $z \equiv as \pmod{A_{n_2+1}}$; as $n_2 \geq m, a fortiori$ $z \equiv as \pmod{A_{m+1}}$, hence $b \equiv a(y+s) \pmod{A_{m+1}}$.