\lambda]_c$. This Euclidean algorithm will then produce Weierstrass factorization for $X_n$-distinguished elements, which will allow us to deduce that $F[X; Y, \lambda]$ is noetherian and factorial.

Let $$\begin{aligned}
 g &=& \sum_{k=0}^\infty g_k(X,Y)X_n^k\end{aligned}$$ be $X_n$-distinguished of degree $s$ in $F[X; Y, \lambda]_c$. Then every $f$ in $F[X; Y, \lambda]_c$ can be written uniquely in the form $$\begin{aligned}
 f = qg + r\end{aligned}$$ where $q$ is and element of $F[X; Y, \lambda]_c$ and $r$ is a polynomial in $F[X_1, \cdots, X_{n-1}; Y, \lambda]_c[X_n]$, with $\deg_{X_n}(r) < s$. Further, if $f$ and $g$ are polynomials in $X_n$, then so are $q$ and $r$.

Let $\alpha$, $\tau$ be projections given by, $$\begin{aligned}
 \alpha : \sum_{k=0}^\infty g_k(X, Y)X_n^k &\mapsto& \sum_{k=0}^{s-1} g_k(X,Y)X_n^k\\
\tau : \sum_{k=0}^\infty g_k(X,Y)X_n^k &\mapsto& \sum_{k=s}^\infty g_k(X,Y)X_n^{k-s} \\\end{aligned}$$ We see that $\tau(w)$ and $\alpha(w)$ are elements of $F[X;Y,\lambda]_c$, and that $\tau(wX_n^{s}) = w$. It is also clear that $\tau(w) = 0$ if, and only if, $\deg_{X_n}( w) < s$, for all $w$ in $F[X; Y, \lambda]_c$.

Such $q$ and $r$ exist if, and only if, $\tau(f) = \tau(qg)$. Thus, we must solve, $$\begin{aligned}
 \tau(f) = \tau(q\alpha(g)) + \tau(q\tau(g)X_n^{s}) = \tau(q\alpha(g)) + q\tau(g).\end{aligned}$$ Note that $\tau(g)$ is invertible, trivially, because it is congruent to a 