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Thanks for your answer Ram. But I will change my question , because I want a more directly answer. And I'll write exactly what I want.

Edited question:

Let's consider a finite Galois extension $L/K$. And let's take $\alpha \in L$. Let $ G = Gal(L,K)$ . Let's consider the conjugates of $\alpha$ i.e $ \sigma(\alpha) $ where $\sigma \in G $. Let's call the different roots of the minimal polynomial of $\alpha$ by $m_{\alpha}(x) \in K[x] $ by $ \alpha_1 , ... \alpha_r $. Since $L/K$ is Galois, we know that these roots are contained in the set of all the conjugates of $\alpha$.

Let $ d= [K(\alpha):K]$ . What I want to prove is that each $\alpha_i $ appears exacly $d$ times in the conjugates.

share|cite|improve this question This might help – Ram Nov 15 '12 at 16:04
Well thanks! yes this might help and it's very interesting! But I want also a direct proof, not searching equivalent definitions to prove what I want! Someone has a direct proof of this? – Daniel Nov 15 '12 at 18:31
@Ram: When pointing someone to an 11-page document it is helpful to indicate which part you have in mind (most of that file is not about the question being asked). – KCd Nov 15 '12 at 19:30
@Daniel: The group $G$ acts transitively on the roots $\alpha_1,\dots,\alpha_r$, and the size of the stabilizer subgroup of $\alpha_1$, say, is $[L:K(\alpha_1)]$ by Galois theory. Therefore the orbit-stabilizer formula says the length $r$ of the orbit is $\#G/[L:K(\alpha_1)] = [L:K]/[L:K(\alpha_1)] = [K(\alpha_1):K]$, so $r = d$. By the way, your use of the term "conjugate" is incorrect. A conjugate is any $\sigma(\alpha)$, but they don't have multiplicity. For example, in ${\mathbf Q}(i,\sqrt{2})$, $\sqrt{2}$ has just 2 ${\mathbf Q}$-conjugates, not 4, even though the Galois group has size 4. – KCd Nov 15 '12 at 19:31
@KCd Great answer!! Thanks for the help :D!!!! – Daniel Nov 15 '12 at 20:06

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