= c_0 + g\omega$. Write $q_i = a(c_0 + g\omega) = ac_0 + ag\omega$. However, by the uniqueness of the division algorithm, the same representation of the division algorithm which holds in $F[[Y]][X]$, holds in $F[X; Y, \lambda]$, and since $\deg_X (ac_0) < s$, we must have $ag = 0$. This is a contradiction. Therefore, the $q_i$ are irreducible in both rings. Write $f = eu\cdot Y^dq_1\cdots q_m$ uniquely as a product of irreducible factors and a unit. Continue by induction.

$F[X_1, \cdots, X_n; Y, \lambda]$ is noetherian.

Assume first that $n =1$. Let $I \subseteq F[X; Y, \lambda]$ be an ideal. Suppose that $d$ is the largest positive integer such that $I \subseteq Y^d F[X; Y,\lambda]$. Then every $f$ in $I$ is divisible by $Y^d$. Choose an element $f$ in $I$ such that ${\text{ord}}_Y f = d$. We can then write $f = e\cdot Y^d \omega $ for some unit $e$, and Weierstrass polynomial $\omega$. Consider the image of $I$ in $Y^dF[X; Y, \lambda]/Y^d\omega F[X; Y, \lambda] \simeq Y^dF[[Y]][X]/Y^d\omega F[[Y]][X]$; this is Noetherian. Therefore, we can pull back the finite list of generators for the image of $I$ and add $Y^d \omega$ to get a finite generating system for 