3.1 Units

One of the things we would like to know is what are the units in \(\mathcal{O}_K\) for \(K\) some number field.

Proposition 3.1.1

Let \(K\) be a number field and let \(\mathcal{O}_K^\times \) denote the group of units, then

\[ \mathcal{O}_K^\times =\{ \alpha \in \mathcal{O}_K \mid N_{K/\mathbb {Q}}(\alpha )=\pm 1\} \]

Proof ▼

If \(\alpha \) is a unit the \(\alpha ^{-1} \in \mathcal{O}_K\) and therefore \(N_{K/\mathbb {Q}}(\alpha )N_{K/\mathbb {Q}}(\alpha )^{-1}=N_{K/\mathbb {Q}}(\alpha \alpha ^{-1})=N_{K/\mathbb {Q}}(1)=1\). So \(N_{K/\mathbb {Q}}(\alpha ) \in \mathbb {Z}^\times \) and is therefore \(\pm 1\).

Now, assume \(N_{K/\mathbb {Q}}(\alpha )=\pm 1\) and let \(\sigma _i\) be the embeddings of \(K\) into \(\mathbb {C}\) with \(\sigma _1\) the identity embedding. Then

\[ \alpha \prod _{i=2}^n \sigma _i(\alpha )=\pm 1 \]

which means \(\alpha ^{-1}=\pm \prod _{i=2}^n \sigma _i(\alpha )\) but each \(\sigma _i(\alpha )\) is again an algebraic integer so \( \alpha ^{-1} \in \overline{\mathbb {Z}} \cap K=\mathcal{O}_K\) (see Remark 2.1.7).

Example 3.1.2

Let us look at \(K=\mathbb {Q}(\sqrt{2})\). In this case we have already seen that \(\mathcal{O}_K=\mathbb {Z}[\sqrt{2}]\). Now, is \(1+\sqrt{2}\) a unit? Well

\[ N_{K/\mathbb {Q}}(1+\sqrt{2})=(1+\sqrt{2})(1-\sqrt{2})=-1 \]

So yes it is.

Next we have a theorem which we will use but not prove.

Theorem 3.1.3 Dirichlet’s Unit theorem

Let \(K\) be a number field and let

\[ \mu _K=\{ x \in K^\times \mid x^n=1 \text{ for some } n \in \mathbb {Z}_{{\gt} 0}\} . \]

This is the set of roots of unity in \(K\). Let \(r_1\) denote the number of real embeddings of \(K\) and \(r_2\) the number of complex conjugate pairs of embeddings. Then

\[ \mathcal{O}_K^{\times } \cong \mu _K \times \mathbb {Z}^{r_1+r_2-1} \]

Corollary 3.1.4

If \(K=\mathbb {Q}(\sqrt{-d})\) with \(d \in \mathbb {Z}_{{\gt} 0}\) a square-free integer, then \(\mathcal{O}_K^\times \) is \(\mu _K\).

Proof ▼

In this case \(r_1=0\) and \(r_2=1\) therefore Theorem 3.1.3 gives the result.

Exercise 3.1.5

Show that if \(K=\mathbb {Q}(\sqrt{d})\) with \(d \in \mathbb {Z}\) a square-free integer, then

\[ {\mu _K=} \begin{cases} \{ \pm 1, \pm \sqrt{-1}\} & \text{if } d=-1\\ \{ 1, \zeta , \zeta ^2, \zeta ^3, \zeta ^4, \zeta ^5 \} & \text{if } d=-3\\ \{ \pm 1\} & \text{otherwise} \end{cases} \]

where \(\zeta =e^{2\pi /6}=\frac{1+\sqrt{-3}}{2}\).

Corollary 3.1.6

Let \(K=\mathbb {Q}(\sqrt{d})\) with \(d\) a positive square-free integer. Then

\[ \mathcal{O}_K^{\times } \cong \{ \pm 1\} \times \mathbb {Z}. \]

In this case there is a unique unit in \(u \in \mathcal{O}_K^{\times }\) which generates \(\mathcal{O}_K^\times / \{ \pm 1\} \) and under the standard embedding we have \(u \geq 1\). This unit \(u\) is called the fundamental unit.

Proof ▼

Note that in this case \(\mu _K=\{ \pm 1\} \). So the structure of the group of units follows from Theorem 3.1.3. Now, \(\mathcal{O}_{K}^{\times }/\{ \pm 1\} \cong \mathbb {Z}\) so take any unit \(v\) mapping to a generator of \(\mathcal{O}_{K}^{\times }/\{ \pm 1\} \). Then one of \(\{ \pm v, \pm v^{-1}\} \) is in the set \((1,\infty )\). So let \(u\) be this unit.

Proposition 3.1.7

Let \(K\) be a number field and let \(\alpha \in \mathcal{O}_K\). If \(N_{K/\mathbb {Q}}(\alpha )=\pm p\) for \(p\) a prime number, then \(\alpha \) is an irreducible element of \(\mathcal{O}_K\).

Proof ▼

Let \(\alpha =\beta \gamma \) we want to show that one of \(\beta ,\gamma \) is a unit. Now taking norms we have

\[ N_{K/\mathbb {Q}}(\alpha )=N_{K/\mathbb {Q}}(\beta )N_{K/\mathbb {Q}}(\gamma )=\pm p. \]

Therefore as \(p\) is prime we must have \(N_{K/\mathbb {Q}}(\beta )=\pm 1\) or \(N_{K/\mathbb {Q}}(\gamma )=\pm 1\). In either case we get the result.

Remark 3.1.8

The converse is false. For example in \(\mathbb {Z}[\sqrt{-1}]\) the element \(3\) is irreducible and has norm \(9\).

Proposition 3.1.9

Let \(K\) be a number field and \(\alpha \in \mathcal{O}_K\) be non-zero and not a unit. Then \(\alpha \) can be written as a product of irreducible elements.

Proof ▼

We prove this by induction on \(|N_{K/\mathbb {Q}}(\alpha )|\). If \(N_{K/\mathbb {Q}}(\alpha )=2\) then \(\alpha \) is irreducible so we are done. Now assume it is true for all \(\beta \) with \(|N_{K/\mathbb {Q}}(\beta )|{\lt} |N_{K/\mathbb {Q}}(\alpha )\). If \(\alpha \) is irreducible then we are done, otherwise \(\alpha =\beta \gamma \) with \(|N_{K/\mathbb {Q}}(\beta )|, |N_{K/\mathbb {Q}}(\gamma )|{\lt} |N_{K/\mathbb {Q}}(\alpha )|\). Therefore both \(\beta \) and \(\gamma \) can be factored into irreducibles and thus so can \(\alpha \).

Definition 3.1.10

We say a ring has unique factorization if whenever

\[ p_1p_2\cdots p_n=q_1q_2\cdots q_m \]

for \(p_i,q_i\) irreducible elements, then \(n=m\) and after possibly reordering we can find units \(u_i\) such that \(p_i=u_iq_i\) for all \(i\).

Example 3.1.11

The \(\mathbb {Z}\) has unique factorization as does \(\mathbb {Z}[\sqrt{-2}]\).

But this is not usually the case, for example \(\mathbb {Z}[\sqrt{-10}]\) doesn’t have unique factorization as \(10=2 \cdot 5= -\sqrt{-10}\sqrt{-10}\) all of which are irreducible as can be seen by taking norms: Note that \(N_{K/\mathbb {Q}}(2)=4,N_{K/\mathbb {Q}}(5)=25, N_{K/\mathbb {Q}}(\sqrt{-10})=10\). Now, every element in \(K=\mathbb {Q}(\sqrt{-10})\) has norm of the form \(x^2+10y^2\) and this can never be \(\pm 2, \pm 5\), so there can’t be any irreducible elements dividing \(2,5,\sqrt{-10}\) therefore they are irreducible.

In order to fix this, we will later decompose things into prime ideals and work with this, but before this we need to understand ideals better.