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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

Let $L=\mathbb{Q}({\zeta }_n)$. We use the standard cyclotomic containment lemma: every root of unity in $\mathbb{Q}({\zeta }_n)$ has order dividing $\mathrm{lcm}(2,n)$. One proof uses the conductor of a rational cyclotomic field. The conductor attached to a primitive $m$th root is $m$, except that it is $m/2$ when $m\equiv 2(\mathrm{mod}4)$; containment in $\mathbb{Q}({\zeta }_n)$ forces this conductor to divide $n$, which gives $m\mid \mathrm{lcm}(2,n)$.

If $n$ is even, $\mathrm{lcm}(2,n)=n$, and the powers of ${\zeta }_n$ already give all $n$ possible roots of unity in $L$. If $n$ is odd, $-{\zeta }_n$ has order $2n$, so $L$ contains all $2n$th roots of unity, and the containment lemma shows there are no others. Thus the roots of unity in a rational cyclotomic field number

Assume that $\zeta$ is a primitive fifth root of unity. The cyclotomic field $K=\mathbb{Q}(\zeta )$ has degree four and signature $(r_1,r_2)=(0,2)$. The Dirichlet unit theorem therefore gives free rankThe roots of unity in a rational cyclotomic field are the ten elements $\pm {\zeta }^j$, so

For explicit units, $-\zeta$ has order ten, whileHence $1+\zeta$ is a unit. It has infinite order because in the standard complex embeddingwhereas every root of unity has modulus one. Its image in the free $\mathbb{Z}$ factor is therefore nonzero and generates a finite-index subgroup of that factor. Consequentlyhas finite index in ${\mathcal{O}}_K^{\times }$, as described by the unit group of the fifth cyclotomic field.