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Put $L=\mathbb{Q}(\omega )$ and $F=\mathbb{Q}(\omega +{\omega }^{-1})$. The maximal real subfield of the fifth cyclotomic field is $F=\mathbb{Q}(\sqrt{5})$. The roots of unity in a rational cyclotomic field show that the roots of unity in $L$ are exactly $\{\pm {\omega }^a\}$.

Let $u\in {\mathcal{O}}_L^{\times }$. For every embedding $\sigma :L\to \mathbb{C}$,Thus the kernel of the logarithmic unit embedding implies that $u/\bar{u}$ is a root of unity. Since conjugation acts trivially modulo the prime element $1-\omega$, we have $u/\bar{u}\equiv 1(\mathrm{mod}1-\omega )$. Among the tenth roots of unity this excludes the elements $-{\omega }^a$, so $u/\bar{u}\in ⟨\omega ⟩$. Squaring is a bijection on this group of order five, so choose a root of unity $\eta$ with $\eta /\bar{\eta }=u/\bar{u}$. Then $v=u/\eta$ is fixed by conjugation and is therefore a unit of $F$.

The units of the quadratic field Q square root of five are $\pm {\varphi }^b$, where $\varphi =(1+\sqrt{5})/2$. Absorbing the sign and $\eta$ into $\pm {\omega }^a$ proves

The Dirichlet unit theorem states that for a number field of signature $(r_1,r_2)$,

For $K=\mathbb{Q}(\sqrt{5})$,The field is real quadratic, so the unit rank is one and its only roots of unity are $\pm 1$. To see that $\varphi$ is fundamental, suppose a positive unit satisfies $1<u<\varphi$. If its norm is $1$, its integral trace $u+u^{-1}$ lies strictly between $2$ and $\sqrt{5}$; if its norm is $-1$, its trace $u-u^{-1}$ lies strictly between $0$ and $1$. Neither interval contains an integer. Reducing any positive unit by a suitable power of $\varphi$ now proves the units of the quadratic field Q square root of five formula

If ${\mathcal{O}}_L^{\times }/{\mathcal{O}}_K^{\times }$ is finite, the two unit groups have the same rank, namely one. Put $d=[L:K]$ and let $(r_1,r_2)$ be the signature of $L$. ThenTherefore $r_1=4-2d$ and $r_2=2d-2$. For a proper extension, nonnegativity forceswith signature $(0,2)$.

This degree occurs: take $L=K(i)$. It is a totally imaginary quadratic extension of $K$, so the unit ranks agree and the quotient is finite. It is nontrivial because $i\in {\mathcal{O}}_L^{\times }$ but $i\notin {\mathcal{O}}_K^{\times }$ (and its coset has order two). This is the finite relative unit quotient over a real quadratic field example.

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.