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I have a question

Prove that : $f(x)=x^m+1$ is irreducible in $\mathbb{Q}[x]$ if only if $m=2^n$

Thanks for your helps!

user41499
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2 Answers2

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This is a consequence of the identity $$ x^{(2n+1)2^m} + 1 = (x^{2n+1} + 1) (x^{(2n+1)(2^{m} - 1)} - x^{(2n+1)(2^{m} - 2)} + \dots -1) \\= (x^{2n+1} + 1) \sum_{k=1}^{2^m}(-1)^{k-1} x^{(2n+1)(2^{m} - k)} $$

mookid
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It's not true. If you take $f(x)=x^{6}+1$, it's irreducible in $\mathbb{Q}[x]$ but $6$ is not a power of $2$. I think you want $f(x)$ irreducible iff $m=2n$.

Simone
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    $$x^6+1 = (x^2+1)(x^4-x^2+1)$$ – David P Oct 16 '14 at 08:02
  • $x^6+1$ is not irreducible. Any root would be a $12$-th root of unity and the $12$-th cyclotomic field has degree $\phi(12)$=4<6. – Sebastian Schoennenbeck Oct 16 '14 at 08:02
  • @SebastianSchoennenbeck ? – JP McCarthy Oct 16 '14 at 08:21
  • @ Jp McCarthy: Just the first argument that came to my mind. Were $f=x^6+1$ irreducible we would get a field of degree $6$ over $\mathbb{Q}$ by adjoining a root of $f$ to $\mathbb{Q}$. But on the other hand each root of $f$ is a $12$-th root of unity and hence lies in the $12$-th cyclotomic field over $\mathbb{Q}$ which only has degree $4$. – Sebastian Schoennenbeck Oct 16 '14 at 12:00