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I have a problem to prove that $x^5+4x^4+10x^2-6x+2$ is Irreducible polynomial over the field $\mathbb{Q}(7^{(1/7)})$ I try to use eisenstein criterion for the polynomial above $\mathbb{Z}[7^{(1/7)}]$ but I cant find the proper prime ideal which has to contain $2,4,6,10$. p also doesnt contain $1$ and $p^2$ doesnt contain $2$. thanks.

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Before applying Eisenstein's criterion, you have to be sure that if a polynomial is irreducble over $\mathbb{Z}[7^{1/7}]$, then it is irreduicble over $\mathbb{Q}[7^{1/7}]$. –  Rankeya Jan 23 '13 at 19:56

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We know that $f(x) = x^5+ 4x^4 + 10x^2 - 6x + 2$ is irreducible over $\mathbb{Z}$ by Eisentein's, hence over $\mathbb{Q}$. Let $\alpha$ be a root of $f(x)$ in $\mathbb{C}$. Then $\mathbb{Q}(\alpha)$ is a degree $5$ field extension of $\mathbb{Q}$. Now, consider the field $\mathbb{Q}(\alpha)(7^{1/7}) = \mathbb{Q}(7^{1/7})(\alpha)$. We know that $\mathbb{Q}(7^{1/7})$ is a degree $7$ field extension of $\mathbb{Q}$ (the polynomial $x^7 - 7$ is irreducible over $\mathbb{Z}$ by Eisenstein's, hence also irreducible over $\mathbb{Q}$). Thus, $7|[\mathbb{Q}(7^{1/7})(\alpha):\mathbb{Q}]$. And, we also see that $5|[\mathbb{Q}(7^{1/7})(\alpha):\mathbb{Q}] = [\mathbb{Q}(\alpha)(7^{1/7}):\mathbb{Q}]$.

Can you deduce from this that $[\mathbb{Q}(7^{1/7})(\alpha):\mathbb{Q}] = 35$? Then, you are basically done because $\mathbb{Q}(7^{1/7})(\alpha)$ is a degree $5$ field extension of $\mathbb{Q}(7^{1/7})$, and so $f(x)$ must be the minimal polynomial of $\alpha$ over $\mathbb{Q}(7^{1/7})$.

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I understood thats because (7,5)=1. thanks –  user56714 Jan 23 '13 at 20:15
    
And the fact that $[\mathbb{Q}(7^{1/7})(\alpha): \mathbb{Q}] \leq 35$. –  Rankeya Jan 24 '13 at 0:51

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