What is the Galois group of $\frac{x^{2n}+1}{x^2+1}$ over $\mathbb{Q}$?

Question

I am trying to compute the Galois group of $h_n(x^2) = \frac{x^{2n}+1}{x^2+1}$ where $n$ is odd $\ge 3$ and $h_n(x) = \frac{x^{n}+1}{x+1}$. As a first step I computed the Galois group of $h_n(x)$. We have $h_n(x) = \prod_{d|n,d>1} \phi_d(-x)$ where $\phi_d(x)$ is the $d$-cyclotomic polynomial. From this we get $h_n(-x)(x-1)= \phi_1(x)\prod_{d|n,d>1}\phi_d(x) = \prod_{d|n} \phi_d(x) = x^n-1$. Hence it follows that $Gal(h_n(x)) = Gal(h_n(-x)(x-1)) = Gal(x^n-1) = (\mathbb{Z}/n\mathbb{Z})^*$ Let $H$ be a finite group which operates on $n-1$ elements and let $G$ be another finite group. Then $|G\wr H| = |G|^{n-1} |H|$ where $\wr$ denotes the wreath product. My conjecture is that $Gal(h_n(x^2))$ has something to do with the wreath product. As a first guess one might say that: $Gal(h_n(x^2)) = C_2 \wr (\mathbb{Z}/(n\mathbb{Z}))^*$ from which it would follow that $|Gal(h_n(x^2))| = 2^{n-1}\varphi(n)$, where $\varphi(n)$ is the Euler-Phi function. But computations with GP-Pari show that when $n = 5$ then $|Gal(h_n(x^2))| = 2^3 \neq 2^6=2^{5-1}\varphi(5)$. Hence the conjecture can not be true. Has somebody an idea what the group $Gal(h_n(x^2))$ might be? (With or without proof).

Answer 1

Let $$\zeta=e^{\pi i/2n}$$ be a primitive complex root of unity of order $4n$.

Because $h_n(x^2)\mid x^{4n}-1$ all the zeros of $h_n(x^2)$ are some powers of $\zeta$. Because $n$ is odd, $n>1$, we see that $\zeta$ is a zero of $h_n(x^2)$.

Therefore the splitting field of $h_n(x^2)$ is the $4n$th cyclotomic field $K=\Bbb{Q}(\zeta)$. It is well known that $$ \operatorname{Gal}(K/\Bbb{Q})\simeq(\Bbb{Z}/4n\Bbb{Z})^* $$ is of order $\phi(4n)=2\phi(n)$. This matches with what GP-Pari told you.