Let $p$ be an odd prime, $\zeta$ a primitive $p^2$th root of unity and $\alpha = \sqrt[p]{p} \zeta$. Then,
(1) Calculate $[\mathbb{Q}(\zeta, \alpha): \mathbb{Q}] $ and $[\mathbb{Q}(\alpha): \mathbb{Q}]$.
(2) Is $\mathbb{Q}(\alpha) / \mathbb{Q}$ Galois, or not?
(3) Find the number of the intermediate fields of $\mathbb{Q}(\zeta, \alpha) / \mathbb{Q}$ of the degree over $\mathbb{Q}$ is $p^2$.

I can't understnad (3). Here is what I have tried:

$1+X^{p} + \cdots + X^{p(p-1)}$ is irreducible over $\mathbb{Q}$ by Eisenstein, and $\zeta$ is its root. So $[\mathbb{Q}(\zeta): \mathbb{Q}] = p(p-1) $.
And next, $\sqrt[p]{p} \mapsto \sqrt[p]{p} \zeta^i$ ($i = 0, \cdots, p-1$) are elements of $\operatorname{Gal}(\mathbb{Q}(\zeta, \alpha): \mathbb{Q}(\zeta))$, $\mathbb{Q}(\zeta, \alpha) = \mathbb{Q}( \zeta, \sqrt[p]{p})$, and $\sqrt[p]{p}^p \in \mathbb{Q}$. So $[\mathbb{Q}(\zeta, \alpha): \mathbb{Q}(\zeta)] = p$.
So $[\mathbb{Q}(\zeta, \alpha): \mathbb{Q}] = p^2(p-1)$.
Next, $\alpha$ is a root of $X^{p(p-1)} + p X^{p(p-2)} + \cdots + p^{p-1}$, so $[\mathbb{Q}(\alpha): \mathbb{Q}] \le p(p-1)$, and $[\mathbb{Q}(\zeta, \alpha): \mathbb{Q}(\alpha) ] \le p$. So $[\mathbb{Q}(\alpha): \mathbb{Q}] = p(p-1)$. This is (1).

For (2), since $\sqrt[p]{p} \zeta^2$, which is not in $\mathbb{Q}(\alpha)$, is a root of $X^{p(p-1)} + p X^{p(p-2)} + \cdots + p^{p-1}$, the minimal polynomial of $\alpha$ over $\mathbb{Q}$. So $\mathbb{Q}(\alpha) / \mathbb{Q}$ is not Galois.

For (3). let $\sigma_i : \sqrt[p]{p} \mapsto \sqrt[p]{p} \zeta^i$, $\tau_j : \zeta \mapsto \zeta^j$, for $0 \le i \le p-1, j \in (\mathbb{Z}/p^2\mathbb{Z})^*$, and $S = \{ \sigma_i \}, T = \{ \tau_j \}$.
Then $G$, the Galois group of $\mathbb{Q}(\zeta, \alpha) / \mathbb{Q}$, is $ST$.
Now $\tau_j \circ \sigma_i = \sigma_{ij} \circ \tau_j $ So $G = S \rtimes T$.
And $(\sigma_i \circ \tau_j)^{(p-1)} = 1$ iff ($j \neq 1$ and $j^{(p-1)} = 1$), or $\sigma_i \circ \tau_j = 1$.

Any help will be much appreciated!

  • 1
    $\begingroup$ (3) is asking how many subgroups of order $p-1$ does $G$ have ? The Galois group of $\Bbb{Q}(\zeta_{p^2}, p^{1/p^2})$ is $\text{Aff}_{p^2}$ the group of affine transformations $x \mapsto ax+b$ of $\Bbb{Z}/p^2\Bbb{Z}$ and $G = \text{Aff}_{p^2} / \langle x \mapsto x+p \rangle$ $\endgroup$ – reuns Jul 8 at 6:30
  • $\begingroup$ @reuns Please tell me in more detail. $\endgroup$ – agababibu Jul 9 at 8:05

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