# How to prove $x^{2k}+1$ is irreducible over the rational numbers

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?

• 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 $\Bbb R[x]$, $$k\ge2\implies x^{2^k}+1=\left(x^{2^{k-1}}-\sqrt2 x^{2^{k-2}}+1\right)\left(x^{2^{k-1}}+\sqrt2 x^{2^{k-2}}+1\right)$$ – user228113 Aug 19 '16 at 16:49
• – Watson Aug 19 '16 at 16:51

$x^n + 1$ 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 $\mathbf Q[x]$.

• This is exactly what I want! Another stupid question, why $x^k+1$ divides $x^n+1$? – user330928 Aug 19 '16 at 16:38
• 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 $x \to x^{2^{v_2(n)}}$. – Starfall Aug 19 '16 at 16:44

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.

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