Prove that $\chi_y = \prod\limits_{\sigma \in G} \left (X-\sigma(y) \right ).$ Let $L|K$ be a finite Galois extension with $[L:K]=n.$ Let $G=\text{Gal}\ (L|K).$ Then $\#G = n.$ Let $y \in L.$ Let us consider the $K$-linear map $T_y : L \longrightarrow L$ on $L$ defined by $$T_y(x)=yx,\ x \in L.$$ Let $M$ be the matrix representation of $T_y$ relative to some fixed $K$-ordered basis of $L.$ Then the characteristic polynomial $\chi_y$ of $y$ over $K$ is defined as $$\chi_y:= \chi_{T_y} = \det (XI_n - M).$$ Clearly $\chi_y$ is a monic polynomial with $\deg (\chi_y)=n.$ Now our instructor left for us as an exercise to show that $$\chi_y = \prod\limits_{\sigma \in G} \left (X-\sigma(y) \right).$$ What he had given to us as a machinary are as follows $:$
$(1)$ Let $\mu_y$ be the minimal polynomial of $y$ over $K.$ Then $$\mu_y = \prod\limits_{z \in Gy} \left (X-z \right)$$ where $Gy$ is the orbit of $y$ under the left action of $G$ on $L.$
$(2)$ $\chi_y = \mu_y^{\# G_y},$ where $G_y$ is the isotropy or stabilizer of $y$ under the left action of $G$ on $L.$
He proved both the above two results in his previous class. Now how do I deduce the above mentioned expression for $\chi_y$ with the help of above two results? I observed that if $F = \prod\limits_{\sigma \in G} \left (X-\sigma(y) \right )$ then $\deg (F) = \deg (\chi_y) =n.$ So If we can prove one of these two polynomials divide the other then due to the fact that both these two polynomials are monic they will be equal. But how do I prove that?
Any help regarding this will be highly appreciated. Thank you very much.
 A: Hint: Firstly remark if $deg(y)=n$, $\mu_y=\chi_y$. To see this, consider the basis $1,y,..,y^{n-1}$ and write the matrix of $T_y$ in this base.
Let $p=[L:K(y)]$ remark that $p=|G_x|$ Let $y_1,...,y_p$ a base of $[L:K(y)]$ remark that $y_1,y_1y,...,y_1y^{l-1},..,y_i,y_iy,..,y_i^{l-1},..$ is a basis of the $K$-vector space $L$, compute the matrix of $T_y$ in this basis and its characteristic polynomial.
A: I have an answer according to my idea. Let me give my argument regarding this. Let $L'$ be the splitting field of $F.$ I will prove that $\chi_y\ \big |\ F$ in $L'[X].$ Let $w \in Gy.$ Then $\exists$ $\rho \in G$ such that $\rho (y) = w.$ Let $\tau \in \rho G_y.$  Then $\exists$ $\rho' \in G_y$ such that $\tau = \rho \rho'.$ Then $\tau (y) = \rho (\rho'(y)) = \rho(y)=w.$ Since $\rho G_y \subseteq G$ so it follows that $$\prod\limits_{\sigma \in \rho G_y} (X - \sigma (y))\ \bigg |\ F\ \text {in}\ L'[X].$$ Since for $\sigma \in \rho G_y$  we have already proved that $\sigma (y) = w$ it follows that $$(X - w)^{\# \rho G_y}\ \bigg |\ F\ \text {in}\ L'[X].$$ Now $\# \rho G_y = \# G_y.$ So we have $$(X - w)^{\# G_y}\ \bigg |\ F\ \text {in}\ L'[X].$$ Since $w \in Gy$ was arbitrarily taken it follows that $$\begin{align*} \prod\limits_{z \in Gy} (X - z)^{\# G_y}\ \bigg |\ F\ \text {in}\ L'[X] \\ \implies \left ( \prod\limits_{z \in Gy} (X-z) \right )^{\# G_y}\ \bigg |\ F\ \text {in}\ L'[X] \\ \implies {\mu_y}^{\# G_y}\ \bigg | \ F\ \text {in}\ L'[X] \\ \implies \chi_y\ \big| \ F\ \text {in}\ L'[X] \end{align*}$$ as claimed.
