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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 $\sigma \in \left[ \tau_i \right]$.

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

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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 $\sigma(a) = \tau(a)$

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    $\begingroup$ Do I have to write a full proof? $\endgroup$ – Bilo Nov 23 '18 at 20:39
  • $\begingroup$ For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ? $\endgroup$ – Mary Star Nov 24 '18 at 8:38
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    $\begingroup$ 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 $\endgroup$ – Bilo Nov 24 '18 at 14:57
  • $\begingroup$ Ah ok!! Could you explain to me also further the second point? $\endgroup$ – Mary Star Nov 24 '18 at 16:41
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    $\begingroup$ 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. $\endgroup$ – Bilo Nov 24 '18 at 17:19

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