3 Cyclotomic fields

Lemma 3.1 ✓

For \(n\) any integer, \(\Phi _n\) (the \(n\)-th cyclotomic polynomial) is a polynomial of degree \(\varphi (n)\) (where \(\varphi \) is Euler’s Totient function).

Proof ▼

The proof is classical.

Lemma 3.2 ✓

For \(n\) any integer, \(\Phi _n\) (the \(n\)-th cyclotomic polynomial) is an irreducible polynomial .

Proof ▼

The proof is classical.

Lemma 3.3 ✓

Let \(\zeta _p\) be a \(p\)-th root of unity for \(p\) an odd prime, let \(\lambda _p=1-\zeta _p\) and \(K=\mathbb {Q}(\zeta _p)\). Then

\[ \Delta (\{ 1,\zeta _p,\dots ,\zeta _p^{p-2}\} )=\Delta (\{ 1,\lambda _p,\dots ,\lambda _p^{p-2}\} )=(-1)^{\frac{(p-1)}{2}}p^{p-2}. \]

Proof ▼

First note \([K:\mathbb {Q}]=p-1\).

Since \(\zeta _p=1-\lambda _p\) we at once get \(\mathbb {Z}[\zeta _p]=\mathbb {Z}[\lambda _p]\) (just do double inclusion). Next, let \(\alpha _i=\sigma _i(\zeta _p)\) denote the conjugates of \(\zeta _p\), which is the same as the image of \(\zeta _p\) under one of the embeddings \(\sigma _i: \mathbb {Q}(\zeta _p) \to \mathbb {C}\). Now by Proposition 2.7 we have

\begin{align*} \Delta (\{ 1,\zeta _p,\dots ,\zeta _p^{p-2}\} )=\prod _{i {\lt} j} (\alpha _i-\alpha _j)^2 & =\prod _{i {\lt} j} ((1-\alpha _i)-(1-\alpha _j))^2\\ & =\Delta (\{ 1,\lambda _p,\dots ,\lambda _p^{p-2}\} )\end{align*}

Now, by Proposition 2.9, we have

\[ \Delta (\{ 1,\zeta _p,\cdots ,\zeta _p^{p-2}\} )=(-1)^{\frac{(p-1)(p-2)}{2}}N_{K/\mathbb {Q}}(\Phi _p'(\zeta _p) ) \]

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

\[ \Phi _p'(x)=\frac{px^{p-1}(x-1)-(x^p-1)}{(x-1)^2} \]

therefore

\[ \Phi _p'(\zeta _p)=-\frac{p\zeta _p^{p-1}}{\lambda _p}. \]

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

\[ N_{K/\mathbb {Q}}(\Phi _p'(\zeta _p) )=\frac{N_{K/\mathbb {Q}}(p)N_{K\mathbb {Q}}(\zeta _p)^{p-1}}{N_{K/\mathbb {Q}}(-\lambda _p)}=(-1)^{p-1}p^{p-2}=p^{p-2} \]

Theorem 3.4 ✓

Let \(\zeta _p\) be a \(p\)-th root of unity for \(p\) an odd prime, let \(\lambda _p=1-\zeta _p\) and \(K=\mathbb {Q}(\zeta _p)\). Then \(\mathcal{O}_K=\mathbb {Z}[\zeta _p]=\mathbb {Z}[\lambda _p]\).

Proof ▼

We need to prove is that \(\mathcal{O}_K=\mathbb {Z}[\zeta _p]\). The inclusion \(\mathbb {Z}[\zeta _p] \subseteq \mathcal{O}_K\) is obvious. Let now \(x \in \mathcal{O}_K\). By Lemma 2.13 and Proposition 3.3, there is \(k \in \mathbb {N}\) such that \(p^k x \in \mathbb {Z}[\zeta _p]\). We conclude by Lemma 2.14.

Lemma 3.5 ✓

Let \(\alpha \) be an algebraic integer all of whose conjugates have absolute value one. Then \(\alpha \) is a root of unity.

Proof ▼

Lemma 1.6 of [ Was82 ] .

Lemma 3.6 ✓

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

Proof ▼

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

Lemma 3.7 ✓

Any unit \(u\) in \(\mathbb {Z}[\zeta _p]\) can be written in the form \(\beta \zeta _p^k \) with \(k\) an integer and \(\beta \in \mathbb {R}\).

Proof ▼

See the Lean proof.

Lemma 3.8 ✓

Let \(p\) be an odd prime, \(\zeta _p\) a primitive \(p\)-th root of unity and let \(K=\mathbb {Q}(\zeta _p)\). Then for any \(i,j \in {0,\dots ,p-1}\) with \(i \ne j\), there exists a unit \(u \in \mathcal{O}_K^\times \) such that \(\zeta _p^i-\zeta _p^j = u (\zeta _p-1)\).

Proof ▼

This is Ex \(34\) in chapter 2 of [ Mar18 ] .

Lemma 3.9 ✓

Let \(R\) be a Dedekind domain, \(p\) a prime and \(\mathfrak {a},\mathfrak {b},\mathfrak {c}\) ideals such that

\[ \mathfrak {a}\mathfrak {b}=\mathfrak {c}^p \]

and suppose \(\mathfrak {a},\mathfrak {b}\) are coprime. Then there exist ideals \(\mathfrak {e},\mathfrak {d}\) such that

\[ \mathfrak {a}=\mathfrak {e}^p \qquad \mathfrak {b}=\mathfrak {d}^p \qquad \mathfrak {e}\mathfrak {d}=\mathfrak {c} \]

Proof ▼

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