2.3 Cyclotomic fields

One of the more interesting number fields are the ones we get by adjoining a root of unity to \(\mathbb {Q}\). In other words \(\mathbb {Q}(\zeta _n)\) where \(\zeta _n\) is a root of \(x^n-1\). From now on, when we write \(\zeta _n\) we mean a primitive \(n\)-root of unity, meaning \(n\) is the smallest non-zero integer such that \(\zeta _n^n=1\).

Lets look at the case \(n=p\) for \(p\) some prime number. Then from Example 1.2.16 we know that \(x^p-1\) is not irreducible, but

\[ \Phi _p(x)=1+x+\cdots +x^{p-1} \]

is minimal. So \(m_{\zeta _p}=\Phi _p\).

More generally, here is a Lemma we will use without proof.

Lemma 2.3.1

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

Theorem 2.3.2

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 \(\ref{prop: disc of prim elt basis}\) 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 2.2.19, 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, from the computation in Example 1.2.16, 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 2.2.7 tells us the only prime we need to check is \(p\). But from Lemma 2.2.21 we know that dividing by \(p\) wont give us any new integral elements, so this must be an integral basis which give the result.

Exercise 2.3.3

Let \(p\) be a prime, \(k\) a positive integer and \(\zeta _{p^k}\) be a \(p^k\)-th root of unity and let \(\lambda _{p^k}=1-\zeta _{p^k}\). Show that

\[ \mathbb {Z}[\zeta _{p^k}]=\mathbb {Z}[\lambda _{p^k}] \]

and

\[ \Delta (\{ 1,\zeta _{p^k},\dots ,\zeta _{p^k}^{\varphi (p^k)}\} )=\Delta (\{ 1,\lambda _{p^k},\dots ,\lambda _{p^k}^{\varphi (p^k)}\} ). \]

Here \(\varphi \) is the usual Euler totient function.
Show that \(\Delta (\{ 1,\zeta _{p^k},\dots ,\zeta _{p^k}^{\varphi (p^k)}\} )\) divides \(p^{k\varphi (p^k)}\).
Let \(p\) be a prime and \(n=p^k\). Let \(S=\{ 1 \leq x \leq n \mid p\nmid x\} \) (i.e the set of elements less than \(n\) which are not divisible by \(p\)). Show that

\[ \prod _{r \in S} (1-\zeta _{p^k}^r)=p \]

and from this deduce that \(\lambda _{p^k}^{\varphi (p^k)}\) divides \(p\) in \(\mathbb {Z}[\zeta _{p^k}]\). [Hint: Consider the polynomial

\[ f(x)=\frac{x^{p^k}-1}{x^{p^{k-1}}-1}=1+x^{p^{k-1}}+x^{2p^{k-1}}+\cdots +x^{(p-1)p^{k-1}} \]

]
Using the above prove that if \(K=\mathbb {Q}(\zeta _{p^k})\) then \(\mathcal{O}_K=\mathbb {Z}[\zeta _{p^k}]=\mathbb {Z}[\lambda _{p^k}]\).

If one works harder, one can show (but we wont prove this):

Theorem 2.3.4

Let \(n\) be a positive integer and \(\zeta _n\) a root of unity. If \(K=\mathbb {Q}(\zeta _n)\) then

\[ \mathcal{O}_K=\mathbb {Z}[\zeta _n]. \]

Exercise 2.3.5

Let \(n\) be a positive integer and let \(\zeta _n\) be an \(n\)-th root of unity. Show that if \(k\) is coprime to \(n\) then

\[ 1+\zeta _n+\cdots +\zeta _n^{k-1} \]

is a unity in \(\mathbb {Z}[\zeta _n]\).

[Hint:Check that its inverse is \(\frac{1-\zeta _n}{1-\zeta _n^k}\)]

Exercise 2.3.6

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

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

where \(u \in \mathbb {Z}[\zeta _n]^{\times }\) (i.e. \(u\) is a unit).