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Let $L$ be a finite field extension of $F$ and let $α ∈ L$. Also, let $α_1 = α, α_2, · · · , α_r$ be the distinct elements of $L$ obtained by applying the elements of $Gal(L|F)$ to $α$. Let us consider the polynomial $h(x) = \prod^r_ {j=1} (x − α_j ) ∈ L[x]$. I wish to show that there is a positive integer $m$ such that $\prod_{σ∈Gal(L|F)}(x − σ(α)) = h^m(x)$

Not sure if it's relevant but I know from a textbook proof of "$F$ is the fixed field of $Gal (L/F)$ acting on $L$ implies $F ⊂ L$ is a normal separable extension" that this $h ∈ F[x]$ and that $h$ is irreducible over $F$, thus is the minimal polynomial of $α$ over F. The above formula for $h$ also shows that $h$ is separable and splits completely over L, but I don't know how to go about proving the desired result still hmm~

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  • $\begingroup$ This is basically restatement of the following. Denote $K$ as splitting field of $a$ over $F$. Now $Gal(L/K)\times Gal(K/F)\to Gal(L/F)$ is bijection and the morphism is obvious by composition. So every automorphism of $K/F$ extends $|Gal(L/K)|$ different ways. In particular, you see $a_1=\sigma(a)$ showing up exactly $|Gal(L/K)|$ times. This holds for any $a_i$ roots. So $m=[L:K]$ where $K$ is the splitting field of $a$ over $F$. $\endgroup$ – user45765 Oct 6 '18 at 17:56
  • $\begingroup$ Ohh does m relate to r? $\endgroup$ – Homaniac Oct 6 '18 at 18:33
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Since the action of $Gal(L/F)$ on $\{\alpha_1,\dots,\alpha_r\}$ is a group action, each of the $(x-\alpha_i)$ will appear the same number of times in the product.

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