# Galois extension - minimum polynomial

Let $$K$$ be a Galois extension of $$F$$ and let $$a\in K$$. Let $$n=[K:F]$$, $$r=[F(a):F]$$, $$G=\text{Gal}(K/F)$$ and $$H=\text{Gal}(K/F(a))$$.

We symbolize with $$\tau_1, \ldots , \tau_r$$ the representatives of the left cosets of $$H$$ in $$G$$.

1. Show that $$\displaystyle{\min (F,a)=\prod_{i=1}^r\left (x-\tau_i(a)\right )}$$.

2. Show that $$\displaystyle{\prod_{\sigma \in G}\left (x- \tau_i(a)\right )=\min (F,a)^{n/r}}$$, where $$\tau_i$$ is the unique representative such that $$\sigma \in \left[ \tau_i \right]$$.



Could you give me a hint how we could show these two points? I don't really have an idea.

For the first point, try to show that if $$i \neq j$$, then $$\tau_i(a) \neq \tau_j(a)$$.
For the second one, try to show that if $$\sigma, \tau$$ belong to the same class modulo $$H$$, then $$\sigma(a) = \tau(a)$$
• For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ? – Mary Star Nov 24 '18 at 8:38
• Yes, since $\tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $\tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above – Bilo Nov 24 '18 at 14:57
• Yes, for sure. First show that if $\sigma,\tau$ belong to the same class modulo $H$, then $\sigma(a)=\tau(a)$; Since $\vert G:H \vert = r$, each coset contains exacly $\frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then \begin{align} \prod_{\sigma \in G}(x-\tau_i(a)) &= \prod_{\sigma \in C_1}(x-\tau_1(a)) \cdots \prod_{\sigma \in C_r}(x-\tau_r(a)) \\ &= (x-\tau_1(a))^{n/r} \cdots (x-\tau_r(a))^{n/r} \\ &= (\prod_{1}^r(x-\tau_i(a)) )^{n/r} \\ &= min(F,a)^{n/r} \end{align} Hope it's clear. – Bilo Nov 24 '18 at 17:19