0$. Further, we assume that $\lambda(0) = 0$ and $ \lambda(\infty) = \infty$, because normalizing $\lambda$ this way does not change $F[X; Y, \lambda]$. Without loss of generality we also assume that $\lambda$ is strictly increasing. Finally, we assume that $F[X; Y, \lambda]$ has only one $Y$ variable. We call an element $$\begin{aligned}
 g = \sum_{d=0}^\infty g_d(X_1, \cdots, X_{n-1}, Y)X_n^d\end{aligned}$$ in $F[X; Y, \lambda]$ $X_n$-distinguished of degree $s$ if $g_s$ is a unit in $F[X_1,\cdots, X_{n-1};Y,\lambda]$, and ${\text{ord}}_Y(g_d) \geq 1$ for all $d> s$. The main result of this paper is the following

Under the above assumptions, we have

1.  (Euclidean Algorithm) *Suppose that $g$ is $X_n$-distinguished of degree $s$ in $F[X; Y, \lambda]$, and that $f$ is an element of $F[X; Y, \lambda]$. Then there exist unique elements, $q$ in $F[X; Y, \lambda]$, and $r$ in the polynomial ring $F[X_1, \cdots, X_{n-1}; Y, \lambda][X_n]$ with ${\rm deg}_{X_n}(r) <s$, such that $f = qg + r$.*

2.  (Weierstrass Factorization) *Let $g$ be $X_n$-distinguished of degree $s$. Then there exists a unique monic polynomial $\omega$ in $F[X_1, \cdots, X_{n-1}; Y, \lambda][X_n]$ of degree $s$ in $X_n$ and a unique unit $e$ in $F[X ; Y, \lambda]$ such that $g = e\cdot 