Dirichlet's 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 $K$.

Write $u=\zeta \prod {\epsilon }_i^{b_i}$ in a fixed fundamental-unit basis. The logarithmic embedding is injective on the exponent lattice and lands in the weighted sum-zero hyperplane. Norm equivalence gives a logarithmic coordinate of size at least a constant times $\Vert b{\Vert }_{\infty }$; the weighted sum-zero relation then forces a negative coordinate of comparable size. There are only finitely many torsion choices. This exponential smallness, together with $\alpha u+\beta v=1$ and a linear form in logarithms of algebraic numbers for the nearby value $\beta v$, bounds the unit exponents effectively.

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 field norm minus one and is the least unit greater than one in the field embedding sending $\sqrt{2}$ to its positive value. It is a fundamental unit. The Dirichlet unit theorem therefore gives

Writing ${\epsilon }^n=a_n+b_n\sqrt{2}$, direct multiplication givesFor $n=0,1,2,3,4$, the pairs $(a_n,b_n)$ are $(1,0),(1,1),(3,2),(7,5),(17,12)$, andIn particular, $\eta ={\epsilon }^2=3+2\sqrt{2}$ has field norm one. Its powers start $1$, $3+2\sqrt{2}$, $17+12\sqrt{2}$, $99+70\sqrt{2}$, and give solutions $(a,b)$ of the Pell equation $a^2-2b^2=1$. This turns finding one unit into generating infinitely many exact integer solutions.

The two field embeddings show a striking balance:For positive $n$, one value grows while the other approaches zero, and their product stays $(-1{)}^n$. Taking logarithms of their absolute values converts this multiplication into equally spaced points, the simplest example of the logarithmic embedding of number field units.^[ref]

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.