how can one compute the minimal polynomial of $\alpha^5=1, \alpha \neq1$ over $\mathbb{Q}$?
thanks.
minimal polynomial of $\alpha^5=1$ over Q
1
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galois-theory
minimal-polynomials
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0Hint: clearly, it is a factor of $x^5-1$. – 2017-02-05
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2$\Phi_5(x)=x^4+x^3+x^2+x+1$ is an irreducible polynomial by Eisenstein criterion (wrt $p=5$) applied to $\Phi_5(x+1)$. Done. – 2017-02-05
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0I understand this polynomial is irreducible. But how did you know this was the right polynomial @JackD'Aurizio – 2017-02-05
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1@usere5225321: it is $\frac{x^5-1}{x-1}$, hence it vanishes at $x=\alpha$. – 2017-02-05
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0So if instead i had $\alpha ^p=1, \alpha \neq 1$, the answer would be $\phi_p(x)$? (for p prime of course) – 2017-02-05
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0@usere5225321: exactly. – 2017-02-05
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0“How did you know?” Sometimes it’s the case that we know because someone else told us. Not everything is obvious at first glance, that’s why we get pleasure by imparting our knowledge and experience to others. – 2017-02-06