3 Cyclotomic fields

lemma 3.1

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

Theorem 3.2

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 \(\mathcal{O}_K=\mathbb {Z}[\zeta _p]=\mathbb {Z}[\lambda _p]\) moreover

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

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 2.6 we have

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

Now, by Proposition 2.8, 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

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

So the last thing we need to prove is that \(\mathcal{O}_K=\mathbb {Z}[\zeta _p]\). From the calculation we just did, the only prime dividing the discriminant is \(p\), therefore Lemma 2.12 tells us the only prime we need to check is \(p\). But from Lemma 2.13 we know that dividing by \(p\) wont give us any new integral elements, so this must be an integral basis which give the result.

lemma 3.3

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

lemma 3.4

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 3.5

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

lemma 3.6

Let \(p\) be a prime and \(n=p^k\). Then

\[ p=u(1-\zeta _n)^{\varphi (n)} \]

where \(u \in \mathbb {Z}[\zeta _n]^{\times }\).

lemma 3.7

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

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