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There is no infinite component in the modulus $(5)$. The ordinary ideal class group of $\mathbb{Q}$ is trivial, and its unit group is $\{\pm 1\}$. The ray class exact sequence therefore givesConcretely, a fractional ideal prime to $5$ has a rational generator $r=a/b$ with $5\nmid ab$. Reducing $r$ modulo $5$ is well-defined up to multiplication by the units $\pm 1$. Its ray class is trivial precisely when one of the two generators $r,-r$ is congruent to one. Thus the quotient has the two classes $\{1,4\}$ and $\{2,3\}$. No positivity restriction is imposed on a generator.

To identify the corresponding field, put $E=\mathbb{Q}({\zeta }_5+{\zeta }_5^{-1})=\mathbb{Q}(\sqrt{5})$. The maximal real subfield of the fifth cyclotomic field is fixed by complex conjugation, the element $-1$ of $(\mathbb{Z}/5\mathbb{Z}{)}^{\times }$. For a prime $\ell \ne 5$, the Frobenius automorphism of $\mathbb{Q}({\zeta }_5)/\mathbb{Q}$ is ${\zeta }_5\mapsto {\zeta }_5^{\ell }$. Its restriction to $E$ thus corresponds exactly to the residue class of $\ell$ modulo $\{\pm 1\}$. Multiplicativity extends this to the Artin reciprocity map on all fractional ideals prime to $5$, whose kernel is exactly $P_{\mathbb{Q},1}((5))$. Its field discriminant $5$ also confirms that $E$ is unramified outside $5$; moreover $E$ is a totally real number field. The uniqueness in the Main theorem of class field theory now gives

Including the real place changes the ray class group: a principal generator must then be positive as well as congruent to one. The residue and signature group of a modulus givesHere the isomorphism sends $(u,\epsilon )$ to $\epsilon u$. Consequently ${\mathbb{Q}}_{(5)\infty }=\mathbb{Q}({\zeta }_5)$, of degree four. The finite modulus gives the real quadratic field; the modulus with the real place gives the full cyclotomic field. The printed finite-modulus assertion needs no correction.

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