I've already shown that the degree of the splitting field of $x^p-2$ over $\mathbb{Q}$ is $p(p-1)$ as follows:
$x^p-2$ has roots $\sqrt[p]{2}\omega_{k}$ for $k=0,1,…,p-1$, where the $\omega_{k}$ are the $p^{th}$ roots of unity.
$[\mathbb{Q}(\sqrt[p]{2}):\mathbb{Q}] = p$ since $x^p-2$ is irreducible,
$[\mathbb{Q}(\sqrt[p]{2}, \omega_{k}):\mathbb{Q}(\sqrt[p]{2})] = p-1$, since $\omega_{k}$ satisfies $x^{p-1}+x^{p-2}+…+x+1$ which is irreducible over $\mathbb{Q}$ by substituting y+1 = x and using Eisenstein. Since gcd$(p, p-1)=1$ it is also irreducible over $\mathbb{Q}(\sqrt[p]{2})$.
$\implies[\mathbb{Q}(\sqrt[p]{2}, \omega_{k}):\mathbb{Q}] = p(p-1)$.
Now onto $x^{p^2} -2$, which has roots $\sqrt[p^2]{2}\omega_{r}$ for $r=0,1,…,p^2-1$.
Since $x^{p^2} -2$ is irreducible over $\mathbb{Q} \implies [\mathbb{Q}(\sqrt[p^2]{2}):\mathbb{Q}] = p^2$.
Next, I need to find the degree of $[\mathbb{Q}(\sqrt[p^2]{2}, \omega_{r}):\mathbb{Q}(\sqrt[p^2]{2})]$.
Using the same method as above, $\omega_{r}$ satisfies $x^{p^2-1}+x^{p^2-2}+…+x+1$, which is irreducible over $\mathbb{Q}$. However in this case, gcd$(p^2, p^2-1)$ need not be 1.
So is this polynomial reducible? How can I find $[\mathbb{Q}(\sqrt[p^2]{2}, \omega_{r}):\mathbb{Q}(\sqrt[p^2]{2})]$?
Best Answer
An argument that depends on a bit of algebraic number theory, so may not be very useful to you.
The prime ideal $(2)$ is totally ramified in $\mathbb{Q}(\root {p^2}\of 2)/\mathbb{Q}$, but it is unramified in $\mathbb{Q}(\omega)/\mathbb{Q}$. Therefore the two extension fields are linearly disjoint, and we get that $$ [\mathbb{Q}(\omega, \root {p^2}\of 2):\mathbb{Q}]= [\mathbb{Q}(\root {p^2}\of 2):\mathbb{Q}]\cdot [\mathbb{Q}(\omega):\mathbb{Q}]. $$ Therefore $$ \Phi_{p^2}(x)=x^{p(p-1)}+x^{p(p-2)}+\cdots+x^{2p}+x^p+1 $$ remains irreducible over $\mathbb{Q}(\root {p^2}\of 2)$.