Charactristic polynomial of a F-linear transformation with respect to Galois group

Question

Let $K$ be a Galois extension of $F$, and let $a \in K$. Let $L_a : K \to K$ be the $F$-linear transformation defined by $L_a(b)=ab$. Show that the characteristic polynomial of $L_a$ is $\prod_{\sigma \in \operatorname{Gal}(L/K)}(x- \sigma (a))$ and the minimal polynomial of $L_a$ is $\min(F,a).$

Answer 1

I think I have got the answer. let $r=[F(a):F]$, and $H=Gal(K/F(a))$, let $\tau_1,\ldots \tau_r$be left coset representatives of $H$ in $G=Gal(K/F)$ and $|G|=n$.

We can show that $min(F,a)=\prod_{i=1}^{r}(x-\tau_i(a))$, because every element in $G$ has to send $\tau_i(a)$ to a certain $\tau_j(a)$, and this conjugation is bijective, thus $\prod_{i=1}^{r}(x-\tau_i(a))$ is fixed under any element of $G=Gal(K/F)$,and since $K/F $ is Galois, so $F$ is the fixed field of $G$, we get $\prod_{i=1}^{r}(x-\tau_i(a))\in F[x]$, but it has a root $a$, so $min(F,a)$ divides $\prod_{i=1}^{r}(x-\tau_i(a))$, and they have the same degree, so$ min(F,a)=\prod_{i=1}^{r}(x-\tau_i(a))$.

Now $\prod_{\sigma \in G}(x-\sigma(a))=min(F,a)^{\frac{n} {r}}$. $K/F(a)$ is also Galois because it is both normal and separable.Suppose $b_1,\ldots, b_{\frac{n} {r}}$ is a basis for $K/F(a)$, then $a^{i}\cdot b_j$,where $i\in\{1,...,r-1\},j \in \{1,..., \frac{n} {r}\}$ is a basis for $K/F$, and for fixed $j$, we have the characteristic polynomial of $L_a$ is $min(F,a)$, thus for the whole set of basis, we have a block matrix with $\frac{n} {r}$ blocks, each have characteristic polynimial $min(F,a)$, therefore the characteristic polynomial on $K$ is the $\frac{n} {r}$th power of $min(F,a)=\prod_{\sigma \in G}(x-\sigma(a))$. QED