Dirichlet unit theorem

If a number field has $r$ real embeddings and $s$ conjugate pairs of complex embeddings, thenwhere $\mu (K)$ is the finite group of roots of unity in $K$.

In a real quadratic number field, a fundamental unit $ϵ>1$ generates the infinite cyclic factor of its unit group, so every unit is $\pm {ϵ}^n$. Existence follows from the Dirichlet unit theorem; choosing the positive generator of the rank-one logarithmic lattice gives this normalization.

If a squarefree positive $d$ has a prime divisor $p\equiv 3(\mathrm{mod}4)$, a unit of norm $-1$ in $\mathbb{Q}(\sqrt{d})$ would give $a^2-d{b}^2=-4$ for integers $a,b$. Modulo $p$ this makes $-1$ a quadratic residue, which is impossible. The obstruction is sufficient, not a full criterion for negative-norm units.

For $K=\mathbb{Q}(\sqrt{5})$,Writing $\varphi =(1+\sqrt{5})/2$, the full unit group is

The unit $\epsilon =1+\sqrt{2}$ has norm minus one and is the least unit greater than one in the positive real embedding. Dirichlet unit theorem therefore gives

Reduction modulo $(3+\sqrt{2})$ identifies $\mathbb{Z}[\sqrt{2}]/(3+\sqrt{2})$ with ${\mathbb{F}}_7$, sends $\sqrt{2}$ to $4$, and sends $\epsilon =1+\sqrt{2}$ to $5$. Since $5$ has order six and $5^3=-1$ in ${\mathbb{F}}_7^{\times }$, the kernel on units is the infinite cyclic group

If $K$ is real quadratic and $L/K$ is a number-field extension with ${\mathcal{O}}_L^{\times }/{\mathcal{O}}_K^{\times }$ finite, then the unit ranks agree. If $d=[L:K]$ and $L$ has signature $(r_1,r_2)$, this forcesFor a proper extension the only possibility is $d=2$ and signature $(0,2)$. For $K=\mathbb{Q}(\sqrt{5})$, the example $L=K(i)$ has this signature and a nontrivial finite quotient because $i$ is a new root of unity.

Choose the real embeddings ${\sigma }_i$ and one embedding ${\tau }_j$ from each complex-conjugate pair. The weighted logarithmic embedding isFor a unit, the sum of these coordinates is $\log |N_{K/\mathbb{Q}}(u)|=0$.

The kernel of $\mathrm{Log}:{\mathcal{O}}_K^{\times }\to {\mathbb{R}}^{r+s}$ is exactly $\mu (K)$. If every conjugate of a unit has modulus one, all its powers lie in a bounded subset of the discrete Minkowski lattice ${\mathcal{O}}_K$, so two powers coincide and the unit is a root of unity.

The logarithms of a fundamental system of $r+s-1$ units form a basis of the hyperplaneand their integer span is a full lattice in $H$.

Every principal ideal $I$ of a number field has a generator $\alpha$ satisfyingat every archimedean embedding. Multiply an arbitrary generator by a unit so that its logarithmic unit-lattice coordinates lie in a fixed bounded fundamental parallelepiped.

For $K=\mathbb{Q}(\sqrt{10})$,The fundamental unit $3+\sqrt{10}$ has norm minus one.

For $K=\mathbb{Q}(\sqrt{-3})$ and $\omega =(-1+\sqrt{-3})/2$,

For a primitive fifth root of unity $\zeta$,

Let $L=\mathbb{Q}({\zeta }_5)$ and let complex conjugation be denoted by a bar. For $u\in {\mathcal{O}}_L^{\times }$, every conjugate of $u/\bar{u}$ has modulus one, so the kernel of the logarithmic unit embedding shows that $u/\bar{u}$ is a root of unity. Reduction modulo $1-{\zeta }_5$ gives $u/\bar{u}\equiv 1$, so this root belongs to ${\mu }_5$. Squaring permutes ${\mu }_5$, hence multiplication by a suitable root of unity makes $u$ real. The real unit belongs to $\mathbb{Q}(\sqrt{5})$ and the units of the quadratic field Q square root of five finish the calculation.