Ramified primes in a cyclotomic number field of a prime power order

Let $l$ be a prime number(even or odd), $n \geq 1$ an interger. Let $\zeta$ be a primitive $l^n$-th root of unity in $\mathbb{C}$. Let $K = \mathbb{Q}(\zeta)$.

Is the following proposition true? If yes, how would you prove this?

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

Motivation How a prime number is decomposed in $K$ is fundamental in algebraic number theory. For example, it has a relation with the quadratic reciprocity law.

Effort We consider the case $n = 1$. By the first link below, a prime number $p \neq l$ is unramified. By the third and the fourth link below, $l$ is ramified.

Related question:

Decomposition of a prime number in a cyclotomic field

On the ring generated by an algebraic integer over the ring of rational integer

Special units of the cyclotomic number field of an odd prime order $l$

Decomposition of $l$ in a subfield of a cyclotomic number field of an odd prime order $l$

• Yes, the result is correct. Refer to Algebraic Number Theory by Neukirch page 61-63. – William Jul 26 '12 at 0:32
• This is definitely done in Jim Milne's notes, and it should be in any book on algebraic number theory. The proof I've seen uses the third link you gave. – Dylan Moreland Jul 26 '12 at 0:42
• @DylanMoreland The third link can be used only for the proof that $l$ is ramified. It cannot be used to show that other prime numbers are unramified. – Makoto Kato Jul 26 '12 at 3:07
• @DylanMorland You know there are several answers to a question. I guess Milne's proof is not the only one. It's nice to have several proofs, here. – Makoto Kato Jul 27 '12 at 5:50
• Please let me know the reason for the downvotes. Unless you make it clear, it's hard to improve my question. – Makoto Kato Jul 28 '12 at 6:44

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