$r$ in $F[X_1, \cdots, X_{n-1}; Y, \lambda][X_n]$ such that $\deg_{X_n}( r) < s$ and $X_n^s = qf + r$. We let $\omega = X_n^s - r$. $\omega = qf$ is clearly $X_n$-distinguished of degree $s$. Since the $X_n$-degree of $\omega \mod (Y)$ is the same the $X_n$-degree of $f \mod (Y)$, we see that $q$ is a unit. Set $e = q^{-1}$, yielding $f = e\cdot \omega$. Uniqueness is immediate from the uniqueness of the division algorithm.

A Weierstrass polynomial is a monic $X_n$-distinguished polynomial in $F[X_1, \cdots, X_{n-1}; Y, \lambda][X_n]$.

Let $\sigma$ be the map such that $\sigma(X_i) = (X_i + X_j^d)$ with $d\geq 1$, and $\sigma(X_j) = X_j$ for all $j \neq i$. Then $\sigma$ is a well-defined automorphism of $F[X; Y, \lambda]$.

Without loss of generality, we can assume that $i = n$ and $j = 1$. Let $f(X,Y) = \sum_{\mu} f_\mu(Y)X^\mu = \sum_{\mu} f_\mu(Y)X_1^{\mu_1}\cdots X_n^{\mu_n}$. Observe that, $$\begin{aligned}
 \sigma(f) &=& \sum_\mu f_\mu(Y)X_1^{\mu_1}\cdots X_{n-1}^{\mu_{n-1}}(X_n + X_1^d)^{\mu_n} \\
&=& \sum_\mu f_\mu(Y)X_1^{\mu_1}\cdots X_{n-1}^{\mu_{n-1}}\sum_{j=0}^{\mu_n} \binom{\mu_{n}}{j}X_1^{dj}X_n^{\mu_n-j}.\end{aligned}$$ The quantity $dj + \mu_n - j= (d-1)j +\mu_n$ is maximal when $j = \mu_n$. Therefore, the above converges if $$\begin{aligned}
 \sum_\mu f_\mu(Y)X_1^{\mu_1 + d\mu_n}\cdots X_{n-1}^{\mu_{n-1}}X_n^{\mu_n}\end{aligned}$$ converges. Choose $c$ so that $|f_\mu(Y)|p^{c\cdot \mu}$ converges to 