**Notations**

We denote by $|S|$ the number of elements of a finite set $S$.

Let $A$ be a ring.
We denote by $A^{\times}$ the group of invertible elements of $A$.

Let $f(X) \in \mathbb{Z}[X]$ and $p$ be a prime number.
We denote by $\bar f(X)$ the reduction of $f(X)$ (mod $p$).

**Fixed symbols**

We fix the following symbols.

Let $l$ be a prime number(possibly 2).

Let $h \geq 1$ be an integer.

Let $\zeta$ be a primitive $l^h$-th root of unity in $\mathbb{C}$.

Let $K = \mathbb{Q}(\zeta)$.

**Lemma 1**
Let $\Omega$ be an algebraically closed field of characteristic $p$($p$ may be 0).
Suppose $p \neq l$.
Let $S$ be the set of roots of $X^{l^h} - 1$ in $\Omega$.
Then $S$ is a cyclic subgroup of $\Omega^{\times}$ of order $l^h$.
Let $P$ be the set of generators of $S$.
Then $|P| = l^{h-1}(l - 1)$.
Let $\Phi(X) = \prod_{\omega\in P}(X - \omega)$.
Then $\Phi(X) = (x^{l^{h-1}})^{l-1} + (x^{l^{h-1}})^{l-2} +\cdots+ x^{l^{h-1}} + 1$.

Proof:
Since $(X^{l^h} - 1)' = l^hX^{l^h-1}$ and $l \neq p$, $X^{l^h} - 1$ has no multiple roots in $\Omega$. Hence $|S| = l^h$.
It is well known that $S$ is a cyclic subgroup of $\Omega^{\times}$.
Let $P$ be the set of generators of $S$.
Then $|P| = l^h - l^{h-1} = l^{h-1}(l - 1)$.
$\Phi(X) = (X^{l^h} - 1)/(X^{l^{h-1}} - 1) = (X^{l^{h-1}})^{l-1} + (X^{l^{h-1}})^{l-2} +\cdots+ X^{l^{h-1}} + 1$.
**QED**

**Definition**
$\Phi(X)$ of Lemma 1 is called the cyclotomic polynomial of order $l^h$.

**Lemma 2**
Let $\zeta', \zeta$ be primitive $l^h$-th roots of unity in $\mathbb{C}$.
Then $(1 - \zeta')/(1 - z)$ is a unit of $K$.

Proof:
There exists an integer $a \geq 1$ such that $\zeta' = \zeta^a$.
Hence $(1 - \zeta')/(1 - \zeta) = \zeta^{a-1} +\cdots+ \zeta + 1$.
Hence $(1 - \zeta')/(1 - \zeta)$ is an algebraic integer in $K$.
Similarly $(1 - \zeta)/(1 - \zeta')$ is an algebraic integer in $K$.
Hence $(1 - \zeta')/(1 - \zeta)$ is a unit of $K$.
**QED**

**Lemma 3**
The following assertions hold.

(1) The cyclotomic polynomial $\Phi(X)$ of order $l^h$ is irreducible in $\mathbb{Q}[X]$.

(2) $\mathfrak{l} = (1 - \zeta)$ is a prime ideal of degree 1 of $K$.

(3) $l = \mathfrak{l}^{l^{h-1}(l - 1)}$.

Proof:
Let $P$ be the set of primitive $l^h$-th roots of unity in $\mathbb{C}$.
By Lemma 1, $\Phi(X) = \prod_{\zeta\in P}(X - \zeta) = (X^{l^{h-1}})^{l-1} + (X^{l^{h-1}})^{l-2} +\cdots+ X^{l^{h-1}} + 1$.
Substituting $X = 1$, we get $l = \prod_{\zeta\in P}(1 - \zeta)$. By Lemma 2, $(l) = \mathfrak{l}^{l^{h-1}(l - 1)}$, where $\mathfrak{l} = (1 - \zeta)$.
This proves (3).

Hence $l^{h-1}(l - 1) \leq [K : \mathbb{Q}]$.
On the other hand, since $\Phi(\zeta) = 0$, $[K : \mathbb{Q}] \leq l^{h-1}(l - 1)$.
Hence $l^{h-1}(l - 1) = [K : \mathbb{Q}]$.
Hence $\Phi(X)$ is irreducible.
This proves (1).

Since $[K : \mathbb{Q}] = l^{h-1}(l - 1)$, $\mathfrak{l}$ is a prime ideal of degree 1 by (3). This proves (2).
**QED**

**Lemma 4**
The notations are the same as Lemma 3.

$(\Phi'(\zeta)) = \mathfrak{l}^{l^{h-1}(h(l - 1) - 1)}$

Proof:
$X^{l^h} - 1 = \Phi(X)(X^{l^{h-1}} - 1)$.
Taking derivatives of the bothe sides, we get
$l^hX^{l^h-1} = \Phi'(X)(X^{l^{h-1}} - 1) + l^{h-1}\Phi(X)X^{l^{h-1}-1}$.
Substituting $X = \zeta$, we get $l^h\zeta^{l^h-1} = \Phi'(\zeta)(\zeta^{l^{h-1}} - 1)$.

Since $\zeta^{l^{h-1}}$ is a primitive $l$-th root of unity,
$(l) = (\zeta^{l^{h-1}} - 1)^{l-1}$ in $\mathbb{Q}(\zeta^{l^{h-1}})$.
Since $(l) = \mathfrak{l}^{l^{h-1}(l - 1)}$ in $K$,
$(\zeta^{l^{h-1}} - 1) = \mathfrak{l}^{l^{h-1}}$ in $K$.

Hence $\mathfrak{l}^{hl^{h-1}(l - 1)} = \Phi'(\zeta)\mathfrak{l}^{l^{h-1}}$.
Hence $(\Phi'(\zeta)) = \mathfrak{l}^{l^{h-1}(h(l - 1) - 1)}$.
**QED**

**Lemma 5**

The notations are the same as Lemma 3.

Let $d$ be the discriminant of $\Phi(X)$.
Then $|d| = l^{l^{h-1}(h(l - 1) - 1)}$.

Proof:
By Lemma 4, $(\Phi'(\zeta)) = \mathfrak{l}^{l^{h-1}(h(l - 1) - 1)}$.
Taking the norms of the both sides, we get $|d| = |N_{K/\mathbb{Q}}(\Phi'(\zeta))| = N(\mathfrak{l})^{l^{h-1}(h(l - 1) - 1)}$.
Since $N(\mathfrak{l}) = l$, $|d| = l^{l^{h-1}(h(l - 1) - 1)}$.
**QED**

**Lemma 6**
$\mathbb{Z}[\zeta]$ is the ring of algebraic integers of $K$.

Proof:
This follows from Lemma 3, Lemma 5 and the proposition of my answer to this question.

**Lemma 7**
Let $\Phi(X)$ be the cyclotomic polynomial of order $l^h$ in $\mathbb{Q}[X]$.
Let $p$ be a prime number such that $p \neq l$.
Let $X^{l^h} - 1 \in \mathbb{Z}[X]$.
Since $(X^{l^h} - 1)' = l^hX^{l^h-1}$, $X^{l^h} - 1$ has no multiple irreducible factor mod $p$.
Since $X^{l^h} - 1 = \Phi(X)(X^{l^{h-1}} - 1)$,
$\Phi(X)$ has no multiple irreducible factor mod $p$.
Let $\Phi(X) \equiv f_1(X)\cdots f_r(X)$ (mod $p$), where $f_i(X)$ are distinct monic irreducible polynomials mod $p$.
Let $f$ be the smallest positive integer such that $p^f \equiv 1$ (mod $l^h$).
Then the degree of each $f_i(X)$ is $f$.

Proof:
Let $F = \mathbb{Z}/p\mathbb{Z}$.
Let $\Omega$ be the algebraic closure of $F$.
Let $S$ be the set of roots of $X^{l^h} - 1$ in $\Omega$.
Since $(X^{l^h} - 1)' = l^hX^{l^h-1}$, $X^{l^h} - 1$ has no multiple roots in $\Omega$.
Hence $|S| = l^h$.
It is well known that $S$ is a cyclic subgroup of $\Omega^{\times}$.
Let $P$ be the set of generators of $S$.
$|P| = l^h - l^{h-1} = l^{h-1}(l - 1)$.
Let $\bar \Phi(X) = \prod_{\omega\in P}(X - \omega) \in F[X]$.
Then $\bar \Phi(X) = (X^{l^{h-1}})^{l-1} + (X^{l^{h-1}})^{l-2} +\cdots+ X^{l^{h-1}} + 1$.
Hence $\bar \Phi(X) = \bar f_1(X)\cdots\bar f_r(X)$.

Let $\omega$ be a root of \bar f_i(X) in $\Omega$.
Since $\omega$ is a root of $\bar \Phi(X)$, $\omega \in P$.
Let $E$ be the unique subfield of $\Omega$ such that $|E| = p^f$.
It is well known that $E^{\times}$ is a cyclic group.
Since $|E^{\times}| = p^f - 1$ and $l^h|p^f - 1$, $E^{\times}$ has a unique cyclic subgroup of order $l^h$.
Hence $\omega \in E$.

Let $L$ be a proper subfiled of $E$.
Let $[L : F] = p^r$.
Suppose $\omega \in L$.
Then $l^h|p^r - 1$, i.e. $p^r \equiv 1$ (mod $l^h$).
Since $r < f$, this is a contradiction.
Hence $E = F(\omega)$.
Hence the minimal polynomial of $\omega$ over $F$ has degree $f$.
Since $\bar f_i(X)$ is irreducible, $\bar f_i(X)$ is the minimal polynomial of $\omega$.
This completes the proof.
**QED**

**Proposition**
The only prime number which ramifies in $K$ is $l$, except $l = 2$ and $h = 1$.

Proof:
By Lemma 3, $l$ ramifies in $K$ except $l = 2$ and $h = 1$.

Let $p$ be a prime number such that $p \neq l$.
Let $\Phi(X) \equiv f_1(X)\cdots f_r(X)$ (mod $p$) be as in Lemma 7.
By this question, $P_i = (p, f_i(\zeta))$ is a prime ideal of $\mathbb{Z}[\zeta]$ lying over $p\mathbb{Z}$.
It is easy to see that $\mathbb{Z}[\zeta]/P_i$ is a finite extension of $\mathbb{Z}/p\mathbb{Z}$ of degree $f$.
By Lemma 6, $\mathbb{Z}[\zeta]$ is the ring of algebraic integers in $K$.
It is well known that each $P_i$ has the same ramification index $e$ and $l^{h-1}(l - 1) = efg$, where $g$ is the number of prime ideals of $\mathbb{Z}[\zeta]$ lying over $p\mathbb{Z}$.
Since $l^{h-1}(l - 1) = fr$, $r = g$ and $e = 1$.
Hence $p$ does not ramify in $K$.
This completes the proof.
**QED**

## Best Answer

According to Galois theory, the sub-field structure of $\mathbb{Q}(\omega_p)$ (where $\omega_p = e^{\frac{2\pi i}{p}}$) corresponds with the subgroup structure of $Aut(\mathbb{Q}(\omega_p))$, the automorphism group of $\mathbb{Q}(\omega_p)$.

Clearly, when $\sigma \in Aut(\mathbb{Q}(\omega_p))$, then $\left(\sigma(\omega_p)\right)^p = \sigma(\omega_p^p) = \sigma(1) = 1$, so $\sigma(\omega_p)$ is again a $p-th$ order root of unity, hence we must have $\sigma(\omega_p) = \omega_p^j$ for some $j \in \{1,2,...,p-1\}$.

Also, defining $\sigma_j$ by $\sigma_j(\omega_p) = \omega_p^j$, we see that $\sigma_j\left(\sigma_k(\omega_p)\right) = \sigma_j(\omega_p^k) = \omega_p^{jk}$, implying that $Aut(\mathbb{Q}(\omega_p))$ is isomorphic to $\mathbb{F}_p^*$, so it is cyclic, generated by $\sigma_g$ where $g$ is a generator of $\mathbb{F}_p^*$.

The subgroups of a cyclic group $<\negthickspace \sigma \negthickspace>$ are $<\negthickspace\sigma^e\negthickspace>$ where $e$ divides the group order, $p-1$ in this case. Thus, for example, for $p=13$ we get the following correspondence between subfields of $\mathbb{Q}(\omega_p)$ and subgroups of $\mathbb{F}_p^*$:

Where $p^*=\left(\frac{-1}{p}\right) p$, in the example $p^*= 13$.

This answers (1) and (2).

I leave (3) and (4) for you :-).