3 Cyclotomic fields

Lemma 3.1 ✓

For $n$ any integer, $Φ_{n}$ (the $n$ -th cyclotomic polynomial) is a polynomial of degree $φ (n)$ (where $φ$ is Euler’s Totient function).

Proof ▼

The proof is classical.

Lemma 3.2 ✓

For $n$ any integer, $Φ_{n}$ (the $n$ -th cyclotomic polynomial) is an irreducible polynomial .

Proof ▼

The proof is classical.

Lemma 3.3 ✓

Let $ζ_{p}$ be a $p$ -th root of unity for $p$ an odd prime, let $λ_{p} = 1 - ζ_{p}$ and $K = Q (ζ_{p})$ . Then

Δ ({1, ζ_{p}, \dots, ζ_{p}^{p - 2}}) = Δ ({1, λ_{p}, \dots, λ_{p}^{p - 2}}) = (- 1)^{\frac{(p - 1)}{2}} p^{p - 2} .

Proof ▼

First note $[K : Q] = p - 1$ .

Since $ζ_{p} = 1 - λ_{p}$ we at once get $Z [ζ_{p}] = Z [λ_{p}]$ (just do double inclusion). Next, let $α_{i} = σ_{i} (ζ_{p})$ denote the conjugates of $ζ_{p}$ , which is the same as the image of $ζ_{p}$ under one of the embeddings $σ_{i} : Q (ζ_{p}) \to C$ . Now by Proposition 2.7 we have

\begin{aligned} Δ ({1, ζ_{p}, \dots, ζ_{p}^{p - 2}}) = \prod_{i < j} (α_{i} - α_{j})^{2} & = \prod_{i < j} ((1 - α_{i}) - (1 - α_{j}))^{2} \\ = Δ ({1, λ_{p}, \dots, λ_{p}^{p - 2}}) \end{aligned}

Now, by Proposition 2.9, we have

Δ ({1, ζ_{p}, \dots, ζ_{p}^{p - 2}}) = (- 1)^{\frac{(p - 1) (p - 2)}{2}} N_{K / Q} (Φ_{p}^{'} (ζ_{p}))

Since $p$ is odd $(- 1)^{\frac{(p - 1) (p - 2)}{2}} = (- 1)^{\frac{(p - 1)}{2}}$ . Next, we see that

Φ_{p}^{'} (x) = \frac{p x^{p - 1} (x - 1) - (x^{p} - 1)}{(x - 1)^{2}}

therefore

Φ_{p}^{'} (ζ_{p}) = - \frac{p ζ_{p}^{p - 1}}{λ_{p}} .

Lastly, note that $N_{K / Q} (ζ_{p}) = 1$ , since this is the constant term in its minimal polynomial. Similarly, we see $N_{K / Q} (λ_{p}) = p$ . Putting this all together, we get

N_{K / Q} (Φ_{p}^{'} (ζ_{p})) = \frac{N_{K / Q} (p) N_{K Q} (ζ_{p})^{p - 1}}{N_{K / Q} (- λ_{p})} = (- 1)^{p - 1} p^{p - 2} = p^{p - 2}

Theorem 3.4 ✓

Let $ζ_{p}$ be a $p$ -th root of unity for $p$ an odd prime, let $λ_{p} = 1 - ζ_{p}$ and $K = Q (ζ_{p})$ . Then $O_{K} = Z [ζ_{p}] = Z [λ_{p}]$ .

Proof ▼

We need to prove is that $O_{K} = Z [ζ_{p}]$ . The inclusion $Z [ζ_{p}] \subseteq O_{K}$ is obvious. Let now $x \in O_{K}$ . By Lemma 2.13 and Proposition 3.3, there is $k \in N$ such that $p^{k} x \in Z [ζ_{p}]$ . We conclude by Lemma 2.14.

Lemma 3.5 ✓

Let $α$ be an algebraic integer all of whose conjugates have absolute value one. Then $α$ is a root of unity.

Proof ▼

Lemma 1.6 of [ Was82 ] .

Lemma 3.6 ✓

Let $p$ be a prime, $K = Q (ζ_{p})$ $α \in K$ such that there exists $n \in N$ such that $α^{n} = 1$ , then $α = \pm ζ_{p}^{k}$ for some $k$ .

Proof ▼

If $n$ is different to $p$ then $K$ contains a $2 p n$ -th root of unity. Therefore $Q (ζ_{2 p n}) \subset K$ , but this cannot happen as $[K : Q] = p - 1$ and $[Q (ζ_{2 p n}) : Q] = φ (2 n p)$ .

Lemma 3.7 ✓

Any unit $u$ in $Z [ζ_{p}]$ can be written in the form $β ζ_{p}^{k}$ with $k$ an integer and $β \in R$ .

Proof ▼

See the Lean proof.

Lemma 3.8 ✓

Let $p$ be an odd prime, $ζ_{p}$ a primitive $p$ -th root of unity and let $K = Q (ζ_{p})$ . Then for any $i, j \in 0, \dots, p - 1$ with $i \neq j$ , there exists a unit $u \in O_{K}^{\times}$ such that $ζ_{p}^{i} - ζ_{p}^{j} = u (ζ_{p} - 1)$ .

Proof ▼

This is Ex $34$ in chapter 2 of [ Mar18 ] .

Lemma 3.9 ✓

Let $R$ be a Dedekind domain, $p$ a prime and $a, b, c$ ideals such that

a b = c^{p}

and suppose $a, b$ are coprime. Then there exist ideals $e, d$ such that

a = e^{p} b = d^{p} e d = c

Proof ▼

It follows from the unique decomposition of ideals in a Dedekind domain. □