Exercise 13.4.3

Proof.

(a)

Assume that $h=y+{\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^r{\beta }_i{u}_i$ is reducible, i.e., $h=\mathit{ge}$ with $g,e\in L[u_1,...,u_n,y]$. In the ring $L[u_1,...,u_n][y]$ the polynomial $h$ is clearly irreducible as every polynomial of degree 1. Then $g$ or $e$ is in $L[u_1,...,u_n]$. In case $g\in L[u_1,...,u_n]$, the comparison of coefficients in $y$ gives $g\mid 1$, and then $g$ is a unit in $L[u_1,\dots ,u_n,y]$.

Hence the assumption is wrong and $h$ is irreducible in $L[u_1,...,u_n,y]$.

(b)

The equality $h=\mathit{gq}$, where $h,g,q\in L[u_1,\dots ,u_n,y]$, is also true in the ring $L(u_1,\dots ,u_n)[y]=K[y]$, where $K=L(u_1,\dots ,u_n)$ is a field, so that $K[y]$ is an Euclidean ring. Write $k=F(u_1,\dots ,u_n)$.

The division in the Euclidean ring $k[y]$ gives $q_1,r_1\in k[y]$ such that $h=g{q}_1+r_1$. The two equalities $f=\mathit{gq}=q_{1}g+r_1$ in $K[y]$, and the unicity of the division in this ring proves $q=q_1\in F(u_1,\dots ,u_n)[y]$. Moreover $q\in L[u_1,\dots ,u_n,y]$ is a polynomial, therefore $q\in F[u_1,\dots ,u_n,y]$.

(c)

By the proof of Theorem 13.4.2, $$h=\underset{\mu \in G_f}{\prod }\left(y-\underset{i=1}{\overset{n}{\prod }}{u}_i{\alpha }_{\mathit{\mu \sigma }(i)}\right)(=\tilde{h}).$$

Let $\tau \in {\sigma }^{-1}{G}_f\sigma$. There exists $\nu \in G_f$ such that $\tau ={\sigma }^{-1}{\nu }^{-1}\sigma$. Then

$$\begin{array}{llll}\hfill \tau \cdot h& =\underset{\mu \in G_f}{\prod }\left(y-\underset{i=1}{\overset{n}{\prod }}{u}_{({\sigma }^{-1}{\nu }^{-1}\sigma )(i)}{\alpha }_{(\mathit{\mu \sigma })(i)}\right)& \hfill & \\ \hfill & =\underset{\mu \in G_f}{\prod }\left(y-\underset{j=1}{\overset{n}{\prod }}{u}_{({\sigma }^{-1}{\nu }^{-1})(j)}{\alpha }_{\mu (j)}\right)(j=\sigma (i))& \hfill & \\ \hfill & =\underset{\mu \in G_f}{\prod }\left(y-\underset{k=1}{\overset{n}{\prod }}{u}_k{\alpha }_{(\mathit{\mu \nu \sigma })(k)}\right)(k=({\sigma }^{-1}{\nu }^{-1})(j))& \hfill & \\ \hfill & =\underset{\lambda \in G_f}{\prod }\left(y-\underset{k=1}{\overset{n}{\prod }}{u}_k{\alpha }_{(\mathit{\lambda \sigma })(k)}\right)(\lambda =\mathit{\mu \nu })& \hfill & \\ \hfill & =\underset{\mu \in G_f}{\prod }\left(y-\underset{i=1}{\overset{n}{\prod }}{u}_i{\alpha }_{(\mathit{\mu \sigma })(i)}\right)& \hfill & \\ \hfill & =h.& \hfill & \end{array}$$

Therefore $\tau \in G$. We have proved ${\sigma }^{-1}{G}_f\sigma \subset G$.