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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

Choose the real embeddings ${\sigma }_1,\dots ,{\sigma }_r$ and one embedding ${\tau }_j$ from each pair of complex-conjugate embeddings. The logarithmic embedding of number field units isIt is a group homomorphism from multiplication to addition.

If $u$ lies in its kernel, every conjugate of $u$ has modulus one. Under the Minkowski embedding, every power $u^n$ therefore lies in one fixed bounded subset. Each $u^n$ belongs to ${\mathcal{O}}_K$, whose Minkowski image is discrete, so that bounded subset contains only finitely many such lattice points. Hence $u^a=u^b$ for some $0\le a<b$, andThus $u$ is a root of unity. Conversely, every root of unity clearly has all conjugates of modulus one and lies in the kernel. This proves the kernel of the logarithmic unit embedding is precisely $\mu (K)$.

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.