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I would like to if the polynomials of the form $x^{2^{n}}+1$ are irreducible over $\mathbb{Q}$ and in that case if there is some "easy" proof for that (where easy means not using a big theory like Galois).

Edward Jiang
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iago
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  • Rational Root Test? – Jesse P Francis Jun 03 '15 at 11:16
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    @JessePFrancis That wouldn't show irreducibility, only that it doesn't have any linear factors. For example, $(x^2+1)^2$ doesn't have any rational roots, but that doesn't mean it's irreducible. – Hayden Jun 03 '15 at 11:17
  • @Hayden, but in this particular case involving only two terms? – Jesse P Francis Jun 03 '15 at 11:20
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    @JessePFrancis I don't know which particular case you're referring to, the OP's polynomial or my example. You can use Rational Root Theorem to show a polynomial is irreducible when the degree is 2 or 3, but for any higher degree you would need to do more than just that. If in the OP's example, $n=1$, then yes, Rational Root Theorem shows irreducibility, but in mine it doesn't (hence me giving the example). – Hayden Jun 03 '15 at 11:24
  • @Hayden, thank you. I meant OP's polynomial! Anyway, I feel its inappropriate to discuss it here, hence asked a new question – Jesse P Francis Jun 03 '15 at 12:21

1 Answers1

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Let $f(x)=x^{2^n}+1$. Note that if $f(x+1)$ is irreducible, then so is $f(x)$.

We have:

$$f(x+1)=\left(x^{2^n} + \binom{2^n}{1}x^{2^n-1}+\dots+\binom{2^n}{2^n-1}x+1\right)+1$$

Note that $2$ divides all the coefficients except that of $x^{2^n}$, and $4$ does not divide the constant coefficient, $2$. Thus the polynomial is irreducible by Eisenstein's criterion.

sbares
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