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I guess this is a very stupid question but I am stuck. Obviously $x^{2k}+1$ has no root in $\mathbb Q$, but I guess this does not imply it must be irreducible. Is there some very easy way to prove this?

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    $\begingroup$ The implication $$p(x)\text{ has no roots in }\Bbb K\implies p(x)\text{ is irreducible}$$ is true when $\deg p=2$ or $\deg p=3$. For higher degrees, it fails quite often. For instance, all polynomials of degree $\ge 2$ are reducible in $\Bbb R$, though the ones of even degree might not have roots. Classic example: in $\Bbb R[x]$, $$k\ge2\implies x^{2^k}+1=\left(x^{2^{k-1}}-\sqrt2 x^{2^{k-2}}+1\right)\left(x^{2^{k-1}}+\sqrt2 x^{2^{k-2}}+1\right)$$ $\endgroup$ – user228113 Aug 19 '16 at 16:49
  • $\begingroup$ Related: math.stackexchange.com/questions/976277 $\endgroup$ – Watson Aug 19 '16 at 16:51
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$ x^n + 1 $ is irreducible in $ \mathbf Q[x] $ precisely when $ n $ is a power of 2. Indeed, if $ n $ is not a power of 2 then write $ n = 2^{v_2(n)} k $ with $ k $ odd, then $ x^{2^{v_2(n)}} + 1 $ divides $ x^n + 1 $. To show the converse, note that $ [\mathbf Q(\zeta_{2^k}) : \mathbf Q] = \varphi(2^k) = 2^{k-1} $, so the $ 2^k $th cyclotomic polynomial is of degree $ 2^{k-1} $, however we have $ x^{2^k} - 1 = (x^{2^{k-1}} - 1)(x^{2^{k-1}} + 1) $ which implies that the $ 2^k $th cyclotomic polynomial is actually $ x^{2^{k-1}} + 1 $, and hence this polynomial is irreducible for $ k \geq 0 $.

Another argument can be obtained by noting that the prime $ 2 $ is totally ramified in the ring of integers $ \mathbf Z[\zeta_n] $, and therefore we might guess that $ x^n + 1 $ becomes Eisenstein at $ 2 $ with an appropriate variable transformation. This guess is correct: the binomial coefficient $ C(n, k) $ is always divisible by $ 2 $ for $ 1 \leq k \leq n-1 $ when $ n $ is a power of $ 2 $, therefore $ (x+1)^n + 1 $ is Eisenstein at $ 2 $, and thus irreducible in $ \mathbf Q[x] $.

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  • $\begingroup$ This is exactly what I want! Another stupid question, why $x^k+1$ divides $x^n+1$? $\endgroup$ – user330928 Aug 19 '16 at 16:38
  • $\begingroup$ It doesn't - my bad. I fixed that part of the argument. The idea is that you can use the elementary factorization identity $ x^d + 1 = (x + 1)(x^{d-1} - x^{d-2} \ldots + 1) $ which holds for $ d $ odd, and substitute $ x \to x^{2^{v_2(n)}} $. $\endgroup$ – Starfall Aug 19 '16 at 16:44
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This is false. For instance, $$x^{12}+1= (x^4+1) (x^8-x^4+1)$$

The factorization of $x^{2k}+1$ appears in the factorization of $x^{4k}-1$, which is given by cyclotomic polynomials.

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  • $\begingroup$ Cool, so is there any criterion to judge for which k this is true? $\endgroup$ – user330928 Aug 19 '16 at 16:30
  • $\begingroup$ Like how do I know $x^4+1$ and $x^8+1$ is irreducible or not? $\endgroup$ – user330928 Aug 19 '16 at 16:30
  • $\begingroup$ But your example is the case k=6...... $\endgroup$ – user330928 Aug 19 '16 at 16:32
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    $\begingroup$ @lhf $x^k+1$ is irreducible precisely for $k$ a power of $2$. In particular, $x^8+1$ is irreducible. $\endgroup$ – Wojowu Aug 19 '16 at 16:32

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