# the number of the intermediate fields of a field extension

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!

• (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 $G = \text{Aff}_{p^2} / \langle x \mapsto x+p \rangle$ – reuns Jul 8 at 6:30
• @reuns Please tell me in more detail. – agababibu Jul 9 at 8:05