Proving that $x^{2^n} + 1$ is irreducible in $\mathbb Q[x]$

Question

$x^{2^n} + 1$ is irreducible in $\mathbb Q[x]$.

I've been working on this and this is my process: I would like to use Eisenstein's criterion so I considered the substitution $y=x-1$. So $$x^{2^n}+1=(y+1)^{2^n}+1=\sum_{k=0}^{2^n}{2^n \choose k}y^{2^n-k}+1$$ $$={2^n \choose 0}y^{2^n}+{2^n \choose 1}y^{2^n-1}+\ldots+{2^n \choose 2^{n}-1}y+2$$ Take $p=2$. Since $p$ doesn't divide $1=a_{2^n}$ and $p^2=4$ doesn't divide $2=a_0$ it only remains to show that every ${2^n \choose j}$ for $1\leq j \leq 2^n-1$ is divisible by $2$ but I haven't succed showing the latter. Any ideas? Also, is my process correct? Thanks in advance

It is the cyclotomic polynomial whose roots are the primitive roots of unity of order $2^{n+1}$. Those are irreducible over $\Bbb{Q}$. — Jyrki Lahtonen, Sep 11 '18 at 14:21
Related question Are these polynomials irreducible over $\Bbb Z$? (part c)) — Sil, Sep 15 '18 at 07:52

score 3 · Answer 1 · answered Sep 11 '18 at 14:18

If you meant to say that $x^{2^n}+1$ is irreducible in $\mathbb{Q}[x]$, then yes, it is indeed irreducible. You are almost there with Eistenstein's Criterion approach. To show that $\displaystyle\binom{2^n}{j}$ is even for $j=1,2,\ldots,2^n-1$, you may observe that $$(y+1)^2\equiv y^2+1\pmod{2}\,.$$ Then, use induction on $n$ to verify that $$(y+1)^{2^n}\equiv y^{2^n}+1\pmod{2}\,.$$

Proving that $x^{2^n} + 1$ is irreducible in $\mathbb Q[x]$

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