# The characteristic polynomial of $L_{a}$ is $\prod_{\sigma \in \mathrm{Gal}(K/F)}(x-\sigma(a))$ and the minimal polynomial of $L_{a}$ is $\min(F,a)$.

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

I proved in a previous question that: $$\prod_{\sigma \in \mathrm{Gal}(K/F)}(x-\sigma(a)) = \min(F,a)^{n/r}$$ and $$\min(F,a)=\prod_{1}^{r}(x-\tau_i(a))$$ where $r = [F(a):a]$ and $\tau_{1},...,\tau_{r}$ are left coset representatives of $\mathrm{Gal}(K/F(a))$ in $\mathrm{Gal}(K/F)$.

I couldn't solve it, but I found an answer that uses exactly that. However, I didn't understand all the details of that answer. Actually, from when it gets a base for $K/F$, the details got confusing for me. Also, I couldn't see the reason for minimal polynomial of $L_{a}$ to be $\min(F,a)$. Can someone help me? Thanks for the advance!

Let $L=L_a$, and let $f(X)\in F[X]$ be a polynomial. Then $$f(L):x\mapsto f(a)x.$$ Therefore $f(L)=0$ iff $f(a)=0$ iff $f$ is a multiple of the minimum polynomial $m$ of $a$. We can make $K$ into a vector space for the field $F'=F[X]/(m(X))$ by defining $$(g(x)+(m(X)))\cdot x=g(L)x=g(a)x.$$ Then $K$ has dimension $d=|K:F|/\deg(m)$ over $F'$. Then the characteristic polynomial of $L$ is the $d$-th power $m^d$ of its minimum polynomial $m$.
• $K$ is the direct sum of $d$ subspaces on which $L$ acts cyclically with minimum polynomial $m$. – Lord Shark the Unknown Jun 29 '18 at 14:19