5 Kummer’s Lemma

In this section we prove Theorem 4.8. The proof follows [ SD01 ] . We begin with some lemmas that will be needed in the proof.

Lemma 5.1

Let \(K\) be a number fields and \(\alpha \in K\). Then there exists a \(n \in \mathbb {Z}\backslash \{ 0\} \) such that \(n \alpha \) is an algebraic integer.

Theorem 5.2

Let \(K/F\) be a Galois extension of number fields whose Galois group \(\operatorname{Gal}(K/F)\) is cyclic with generator \(\sigma \). If \(\alpha \in K\) is such that \(N_{K/F}(\alpha )=1\), then

\[ \alpha =\beta / \sigma (\beta ) \]

for some \(\beta \in \mathcal{O}_K\).

Proof ▼

Pick some \(\gamma \in K\) and consider

\[ \beta =\gamma \alpha + \sigma (\gamma )\alpha \sigma (\alpha )+\cdots +\sigma ^{n-1}(\gamma )\alpha \sigma (\alpha )\cdots \sigma ^{n-1}(\alpha ), \]

where \([K/F]=n\). Since the norm \(\alpha \) is \(1\) then we have \(\alpha \sigma (\alpha )\cdots \sigma ^{n-1}(\alpha )=1\) from which one verifies that \(\alpha \sigma (\beta )=\beta \). This also uses that since we know that Galois group is cyclic, all the embeddings are given by \(\sigma ^i\). Note that \( \alpha + \sigma \alpha \sigma (\alpha )+\cdots +\sigma ^{n-1}\alpha \sigma (\alpha )\cdots \sigma ^{n-1}(\alpha )\) is a linear combination of the embeddings \(\sigma ^i\). Now, using 2.4 we can check that they must be linearly independent. Therefore, it is possible to find a \(\gamma \) such that \(\beta \neq 0\). Putting everything together we have found \(\alpha =\beta /\sigma (\beta )\). Using 5.1 we see that its possible to make sure \(\beta \in \mathcal{O}_K\) as required.

Lemma 5.3

Let \(K/F\) be a Galois extension of \(F=\mathbb {Q}(\zeta _p)\) with Galois group \(\operatorname{Gal}(K/F)\) cyclic with generator \(\sigma \). Then there exists a unit \(\eta \in \mathcal{O}_K\) such that \(N_{K/F}(\eta )=1\) but does not have the form \(\epsilon /\sigma (\epsilon )\) for any unit \(\epsilon \in \mathcal{O}_K\).

Proof ▼

Let

\[ U_1= \mathcal{O}_K^\times / \mathcal{O}_F^\times \qquad \text{and} \qquad U=U_1/U_{1,tors}. \]

By Dirichlet’s unit theorem we know \(U\) is a free abelian group of rank \(m=(p-1)^2/2\) generators ¹ . For any \(f =\sum _n a_n x^n \in \mathbb {Z}[x]\) and \(\alpha \in K\) we let \(\alpha ^f\) denote \(\prod _n \sigma ^n(\alpha )^{a_n}\). For any \(\theta _i \in U\), let \(S(\theta _1, \dots ,\theta _r)\) denote the set of \(\sigma ^i(\theta _j)\) for \(i \in \{ 0,\dots ,p-2\} \) and \(j \in \{ 1,\dots ,r\} \).

Claim 1:

There exist \(\theta _1,\dots ,\theta _m\) such that the elements of \(S(\theta _1, \dots ,\theta _m)\) are multiplicatively independent, therefore \(S(\theta _1, \dots ,\theta _m)\) generates a subgroup of finite index in \(U\).

Claim 2:

If \(\theta _i\) are chosen so that the index of \(S(\theta _1, \dots ,\theta _m)\) in \(U\) is minimal, then no \(\theta _i\) has the form \(\theta ^{1-\sigma }\) for \(\theta \in U\).

Claim 3:

If \(a_i \in \mathbb {Z}\) are not all divisible by \(p\), then \(\prod _i \theta _i^{a_i}\) is not of the form \(\theta ^{1-\sigma }\) for \(\theta \in U\).

Using these claims we can finish the proof as follows: We fist assume that \(\zeta _p= \xi ^{1-\sigma }\) with \(\xi \in \mathcal{O}_K^\times \) as otherwise we can take \(\eta =\zeta _p\). By construction it follows that \(\xi ^p=N_{K/F}(\xi )\) is a unit in \(F\) and \(\xi \) is not in \(F\). Picking representatives \(\eta _i \in \mathcal{O}_K^\times \) for the \(\theta _i\) we can consider the group they generate, denoted \(S(\eta _i)\). There is a map

\[ \phi : S(\eta _i) \to \mathcal{O}_F^\times /\mu _F \]

induced by \(N_{K/F}\), where \(\mu _F\) denote the \(p\)-th roots of unity in \(F\). There are now three possibilities:

\(\xi ^p\) is a primitive \(p\)-th root of unity. In this case we can pick \(\eta \) as follows:
No non-trivial power of \(\xi ^p\) is in the image of \(\phi \)
Neither (a) or (b) hold.

Proof ▼

Lemma 5.4

Let \(F=\mathbb {Q}(\zeta _p)\) with \(\zeta _p\) a primitive \(p\)-th root of unity. If \(\xi \in \mathcal{O}_F\) and coprime to \(\lambda _p:=1-\zeta _p\) and if

\[ \xi \equiv \alpha _0^p \mod \lambda _p^p \]

for some \(\alpha _0 \in \mathcal{O}_{F_\mathfrak {p}}^\times \) (the ring of integers of the completion of \(F\) at \(\lambda _p\)), then the ideal \((\lambda _p)\) is unramified in \(K/F\) where \(K=F(\sqrt[p]{\xi })\).

Before giving the proofs of the above, we will first show how one can use this to prove the following lemma (from which Kummer’s lemma is immediate):

Lemma 5.5

Let \(u \in F\) with \(F=\mathbb {Q}(\zeta _p)\) be a unit such that

\[ u \equiv a^p \mod \lambda _p^p \]

for some \(a \in \mathcal{O}_F\). The either \(u = \epsilon ^p\) for some \(\epsilon \in \mathcal{O}_F^\times \) or \(p\) divides the class number of \(F\).

Proof ▼

Assume \(u\) is not a \(p\)-th power. Then let \(K=F(\sqrt[p]{u})\) and \(\eta \in K\) be as in 5.3. Then \(K/F\) is Galois and cyclic of degree \(p\). Not by Hilbert 90 (5.2) we can find \(\beta \in \mathcal{O}_K\) such that

\[ \eta = \beta /\sigma (\beta ) \]

(but note that by our assumption \(\beta \) is not a unit). Note that the ideal \((\eta )\) is invariant under \(\operatorname{Gal}(K/F)\) by construction and by 5.4 it cannot be ramified, therefore \((\beta )\) is the extension of scalars of some ideal \(\mathfrak {b}\) in \(\mathcal{O}_F\). Now if \(\mathfrak {b}\) is principal generated by some \(\gamma \) then \(\beta =v \gamma \) for some \(v \in \mathcal{O}_K^\times \). But this means that

\[ \eta = \beta /\sigma (\beta )= v/(\sigma (v)) \cdot \gamma / \sigma (\gamma )=v/\sigma (v) \]

contradicting our assumption coming from 5.3. On the other hand, \(\mathfrak {b}^p= N_{K/F}(\beta )\) is principal, meaning \(p\) divides the class group of \(F\).