Do Euclidean rings admit $b$-adic representations ?

Let $(R,f)$ be an Euclidean domain with Euclidean function $f: R\setminus \lbrace 0 \rbrace \to \mathbb{Z}_+$. Given a fixed non-unit $b$ and $x \neq 0$, we obtain by division with remainder $x_0,y_1$ s.t. $x=x_0 + y_1b$. Repeating this process we find $x_0,...,x_n,y_{n+1}$ s.t. $$x=x_0 + x_1b + \cdots + x_nb^n + y_{n+1}b^{n+1}$$ with $x_i=0$ or $f(x_i) < f(b)$ for $i=0,...,n$.

a) Does this process necessarily terminate, i.e. has each $x\neq 0$ a $b$-adic represenation $$x=x_0 + x_1b + \cdots + x_nb^n$$ with $x_i=0$ or $f(x_i) < f(b)$ for $i=0,...,n$ ?

b) If not, what are counter-examples ?

c) If not, what are additional conditions on $f$ such that a) holds ?

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This page on the meta site might interest you. – Did Nov 19 '12 at 8:37

Let $(R,f)=(k[[t]],\deg)$ be the power series ring in one variable over a field, where the degree denotes minimum degree of a monomial appearing in an expansion (i.e. "order of vanishing" at $t=0$). Let $x=1+t+t^2+\cdots,$ and $b=t.$ Then we can express $x$ as $1+t(1+t+t^2+\cdots),$ i.e. $x=x_0+y_1b$ with $x_0=1,y_1=x,b=t.$ We also see immediately that $f(x_0)=\deg(1)=0<f(b)=\deg(t)=1.$
The process then continues with $y_1=x,$ going on forevermore. That is, we get $x_i=1$ for all $i\in\Bbb N.$
Well, that's interesting. On the one hand it shows that the process doesn't need to terminate, on the other hand, there might still be $t$-adic representations, since $f(x)=0 < f(t)$, whence $x=x+0\cdot t$ is a $t$-adic representation of $x$. – Ralph Nov 16 '12 at 10:39