Proposition 5

Let $f \in B$ be a series whose reduced series is non-zero. Let $s$ denote its reduced order and $M$ the sub-A-module of $B$ with basis $\{1, X, \ldots, X^{s-1}\}$. Then $B$ is the direct sum of $M$ and $f$ is not a divisor of zero in $B$.

(a) We show that $f \mid B \cap M = (0)$. Suppose that there is a relation:

We show that the $b_i$ (and hence the $r_j$) are all zero, which will prove in particular that $f$ is not a divisor of zero in $B$. Since $A$ is Hausdorff, it suffices to show that $b_i \in m^n$ for all $i \geq 0$ and all $n \geq 0$. It is obvious for $n = 0$. We shall argue by double induction: we shall assume that $b_i \in m^{n-1}$ for all $i$ and $b_i \in m^n$ for $i < k$ and show that this implies that $b_k \in m^n$. For this, we write $f = \sum_{i=1}^{\infty} a_i X^i$ and compare the coefficients of $X^{s+k}$ in (3); then: The terms in the first bracket belong to $m^n$ since the $b_i \in m^n$ for $i < k$; similarly for those in the second, since the $b_i \in m^{n-1}$ for all $i$ and the $a_i \in m$ for $i \leq s-1$. Hence $b_k a_s \in m^n$ and, as $a_s$ is an invertible element of $A$, $b_k \in m^n$.

(b) We show that $f \mid B + M = B$. We write

it is an invertible element of $B$. Then if therefore we write $fg^{-1} - X^s = (f - X^s g) g^{-1} = -h$, the coefficients of $h$ belong to $m$. Then let $r$ be an element of $B$. By induction on $n$ we define a sequence $(q^{(n)})$ of elements of B: we take $q^{(0)}$ to be the unique series satisfying: writing $h = \sum_{i=0}^m h_i X^i$ and $q^{(n)} = \sum_{i=0}^\infty q_i^{(n)} X^i$, the $q_i^{(n)}$ are defined by: It follows immediately from (6) that: As $h_j \in m$ for all $j$, it also follows from (6), by induction on $n$, that $q_i^{(n)} \in m^n$ for all $i$ and all $n$. As A is complete, it follows that the series converges to an element $q$ of B. By (5) and (7), As M is closed, at the limit (8) gives $r \equiv (X^s - h)q \pmod{M}$, that is We may also use the results of Chapter III, § 2 to show the relation $B = fB + M$ (cf. Exercise 12). The method followed here has the advantage of being applicable to convergent series.