is not an element of $(Y)$, and so $\sigma(f)$ is a unitary polynomial modulo $(Y)$. Therefore, $\sigma(f)$ is $X_n$-distinguished of degree $s$.

Let $\omega$ be a Weierstrass polynomial of degree $s$ in $X_n$. Then for all $d \geq 0$

1.  $Y^dF[X; Y, \lambda]/Y^d\omega F[X; Y, \lambda]$ is a finite free $F[X_1, \cdots, X_{n-1}; Y, \lambda]$-module, and

2.  $Y^dF[X_1, \cdots, X_{n-1}; Y, \lambda][X_n]/Y^d\omega F[X_1, \cdots, X_{n-1}; Y, \lambda][X_n]$ is isomorphic to\
    $Y^dF[X; Y, \lambda]/Y^d\omega F[X; Y, \lambda]$.

Suppose that $g$ is an element of $Y^dF[X; Y, \lambda]$, then $g = Y^dh$ for some element $h$ in $F[X; Y, \lambda]$. Since $\omega$ is $X_n$-distinguished, there exists a unique element $q$ in $F[X; Y, \lambda]$, and a unique polynomial $r$ in $F[X_1, \cdots, X_{n-1}; Y, \lambda][X_n]$ with $\deg_{X_n} (r) < s$, such that $h = q\omega + r$, so $g = qY^d\omega + Y^dr$. Therefore, $g \equiv Y^dr \mod Y^d\omega F[X; Y, \lambda]$, so the set $\{Y^d, Y^dX_n, \cdots, Y^dX_n^{s-1}\}$ forms a generating set of $Y^dF[X; Y, \lambda]/Y^d\omega F[X; Y, \lambda]$ over the ring $F[X_1,\cdots, X_{n-1}; Y, \lambda]$. The natural map $$Y^dF[X_1, \cdots, X_{n-1}; Y, \lambda][X_n] \rightarrow Y^dF[X; Y, \lambda]/Y^d\omega F[X; Y, \lambda]$$ is thus surjective. The kernel of this map is $Y^d 