Cyclotomic Arithmetic and Reciprocity via Frobenius — Examples

Taking $m=3$ in [F1], $-{\zeta }_3$ is a primitive sixth root of unity, so $\mathbb{Q}({\zeta }_6)=\mathbb{Q}({\zeta }_3)$.

By [F2] with $n=6\equiv 2(\mathrm{m}\mathrm{o}\mathrm{d}4)$, the conductor of $\mathbb{Q}({\zeta }_6)$ is $6\mathrm{/}2=3$.

The reduced index of this field is $3$, and $2\nmid 3$, so [F3] shows that $2$ is unramified in $\mathbb{Q}({\zeta }_6)=\mathbb{Q}({\zeta }_3)$; by [F4] with $f=3$, $\mathrm{\ell }=2$, the residue degree is ${\mathrm{ord}}_3(2)=2$ and the number of primes above $2$ is $\phi (3)\mathrm{/}2=1$, so $2\mathcal{O}$ is prime of degree $2$.

Since $\phi (5)=4$, the discriminant formula gives $d_K=(-1{)}^2\cdot 5^4\mathrm{/}{5}^{4\mathrm{/}4}=5^4\mathrm{/}5=5^3=125$.

The multiplicative order of $2$ modulo $5$ is $4$: the powers of $2$ modulo $5$ are $2,4,3,1$, so ${\mathrm{ord}}_5(2)=4$.

The class $11\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}5)$ is the identity of $(\mathbb{Z}\mathrm{/}5{)}^{\times }$.

By [F3], $5{\mathcal{O}}_K=(\lambda {)}^4$ with $\lambda {\mathcal{O}}_K$ the unique prime above $5$ and residue field ${\mathbb{F}}_5$; its ramification exponent is $4=[K:\mathbb{Q}]$, so $5$ is totally ramified and its inertia group at $\lambda$ has order $4$, the full Galois group.

By [F4] and step 1.2 applied with $\mathrm{\ell }=2$, there is exactly $4\mathrm{/}4=1$ prime above $2$, of residue degree $4$; its residue field has $2^4=16$ elements, and since $2$ is unramified its inertia group is trivial by [F6], so $e=1,f=4,g=1$ with $efg=4=[K:\mathbb{Q}]$.

By [F5] and step 1.3, $11$ splits completely in $K$: there are $\phi (5)=4$ distinct primes above $11$, each of ramification exponent $1$ and residue degree $1$.

Collecting the results: ${\mathcal{O}}_K=\mathbb{Z}[{\zeta }_5]$ with $1,\zeta ,{\zeta }^2,{\zeta }^3$ an integral basis, $d_K=125$, the prime $5$ is totally ramified with $(5)=(1-\zeta {)}^4$, the prime $2$ has one prime above it of residue degree $4$ and trivial inertia, and $11$ splits into four degree-one primes.

Since $\phi (8)=4$ and the only prime divisor of $8$ is $2$, the formula gives $d_K=(-1{)}^2\cdot 8^4\mathrm{/}{2}^4=4096\mathrm{/}16=256=2^8$.

For an odd prime $\mathrm{\ell }$ the class of $\mathrm{\ell }$ modulo $8$ lies in $\{1,3,5,7\}$; by [F4] it has order $1$ exactly for the class $1$, and order $2$ for the classes $3,5,7$.

By [F2] with $\mathrm{\ell }$ odd, the primes above $\mathrm{\ell }$ number $4\mathrm{/}{\mathrm{ord}}_8(\mathrm{\ell })$ and each has residue degree ${\mathrm{ord}}_8(\mathrm{\ell })$. Hence $\mathrm{\ell }\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}8)$ gives $4\mathrm{/}1=4$ primes of residue degree $1$ (residue fields ${\mathbb{F}}_{\mathrm{\ell }}$), and by [F3] this is exactly the complete splitting condition; each of $\mathrm{\ell }\equiv 3,5,7(\mathrm{m}\mathrm{o}\mathrm{d}8)$ gives $4\mathrm{/}2=2$ primes, of residue degree $2$ (residue fields ${\mathbb{F}}_{{\mathrm{\ell }}^2}$).

Summary: $d_K=2^8$, every odd prime is unramified, and its splitting type in $K$ depends only on $\mathrm{\ell }<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>8$: four degree-one primes for $\mathrm{\ell }\equiv 1$, two degree-two primes for $\mathrm{\ell }\equiv 3,5,7$; in all cases $efg=4=[K:\mathbb{Q}]$.

The data $\phi (12)=4$, ${12}^4=20736$ and $2^4\cdot 3^2=16\cdot 9=144$ give $d_K=(-1{)}^2\cdot 20736\mathrm{/}144=144=2^4\cdot 3^2$.

For $\mathrm{\ell }=2$ write $12=2^2\cdot 3$, so $a=2$, $m=3$: $e=\phi (4)=2$, $d={\mathrm{ord}}_3(2)=2$ (as $2^2\equiv 1$ and $2\equiv ̸1(\mathrm{m}\mathrm{o}\mathrm{d}3)$) and $g=\phi (3)\mathrm{/}2=1$; hence $2{\mathcal{O}}_K=P^2$ for the unique prime above $2$, of residue degree $2$.

For $\mathrm{\ell }=3$ write $12=3\cdot 4$, so $a=1$, $m=4$: $e=\phi (3)=2$, $d={\mathrm{ord}}_4(3)=2$ (as $3^2\equiv 1$ and $3\equiv ̸1(\mathrm{m}\mathrm{o}\mathrm{d}4)$) and $g=\phi (4)\mathrm{/}2=1$; hence $3{\mathcal{O}}_K=P^{\mathrm{\prime }2}$ for the unique prime above $3$, of residue degree $2$.

For a prime $\mathrm{\ell }\nmid 12$ the class of $\mathrm{\ell }$ modulo $12$ is one of $1,5,7,11$. If $\mathrm{\ell }\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}12)$ then ${\mathrm{ord}}_{12}(\mathrm{\ell })=1$, so by [F3] there are $\phi (12)=4$ primes of degree $1$, and by [F4] $\mathrm{\ell }$ splits completely; if $\mathrm{\ell }\equiv 5,7,11(\mathrm{m}\mathrm{o}\mathrm{d}12)$ then the order is $2$ by [F5], so there are $4\mathrm{/}2=2$ primes, each of residue degree $2$.

Collecting steps 1.1 through 2.1: $d_K=144$; the primes $2$ and $3$ are ramified with a single prime each, of $e=f=2$; and every $\mathrm{\ell }\nmid 12$ splits into four degree-one primes for the class $1$, or two degree-two primes for the classes $5,7,11$ modulo $12$.

Substituting the Legendre values, $\tau =\zeta -{\zeta }^2$.

Since ${\zeta }^3=1$ we have ${\zeta }^2={\zeta }^{-1}$, and by [L5] also ${\zeta }^2=\bar{\zeta }$; thus $\tau =\zeta -{\zeta }^{-1}=\exp (2\pi i\mathrm{/}3)-\exp (-2\pi i\mathrm{/}3)=2i\sin (2\pi \mathrm{/}3)=2i\sin (\pi \mathrm{/}3)$, using Euler's formula and the reflection identity.

From step 2.1, ${\tau }^2=(2i\sin (\pi \mathrm{/}3){)}^2=-4{\sin }^2(\pi \mathrm{/}3)$, while [L2] gives ${\tau }^2=-3$; hence ${\sin }^2(\pi \mathrm{/}3)=3\mathrm{/}4$, and $\sin (\pi \mathrm{/}3)>0$ because $0<\pi \mathrm{/}3<\pi$, so $\sin (\pi \mathrm{/}3)=\sqrt{3}\mathrm{/}2$.

Substituting back, $\tau =2i\cdot (\sqrt{3}\mathrm{/}2)=i\sqrt{3}$, and ${\tau }^2=(i\sqrt{3}{)}^2=-3$, in agreement with the general square formula.

By [L1], $\tau =\zeta -{\zeta }^2-{\zeta }^3+{\zeta }^4$.

Put $t:=\zeta +{\zeta }^4=\zeta +{\zeta }^{-1}$. Then $t=2\cos (2\pi \mathrm{/}5)>0$ by [L5] and [L6].

Put $s:={\zeta }^2+{\zeta }^3={\zeta }^2+{\zeta }^{-2}$. From ${\mathrm{\Phi }}_5(\zeta )=0$ we get $1+t+s=0$, so $s=-1-t$; therefore $\tau =t-s=t-(-1-t)=1+2t$.

Moreover $t^2=(\zeta +{\zeta }^{-1}{)}^2={\zeta }^2+2+{\zeta }^{-2}=s+2=1-t$, so $t^2+t-1=0$ and $t=\frac{-1\pm \sqrt{5}}{2}$; since $t>0$ we have $t=\frac{\sqrt{5}-1}{2}$.

Substituting into step 2.1, $\tau =1+2t=1+(\sqrt{5}-1)=\sqrt{5}$, and this is consistent with the general theorem, which gives ${\tau }^2=5$.

The stabilizer of $\tau$ under ${\sigma }_b$ is the set of $b$ with $(b\mathrm{/}5)=1$, because ${\sigma }_b(\tau )=(b\mathrm{/}5)\tau$ and $\tau =\sqrt{5}\mathrm{\ne }0$; as the nonzero squares modulo $5$ are $1$ and $4$, the stabilizer is $\{1,4\}$.

Substituting $p=7$ into $p^{\ast }=(-1{)}^{(p-1)\mathrm{/}2}p$ gives $p^{\ast }=(-1{)}^3\cdot 7=-7$.

By [F1] with $p=7$, the unique degree-two intermediate field of $\mathbb{Q}({\zeta }_7)\mathrm{/}\mathbb{Q}$ is $\mathbb{Q}({\tau }_7)=\mathbb{Q}(\sqrt{p^{\ast }})=\mathbb{Q}(\sqrt{-7})$, where ${\tau }_7$ is the Gauss sum attached to the chosen primitive root ${\zeta }_7$.

By [F2] with $p=7$, ${\tau }^2=-7$, so $\tau =\pm \sqrt{-7}$ and the generator $\mathbb{Q}(\tau )$ is indeed $\mathbb{Q}(\sqrt{-7})$, in agreement with the identification of step 2.1.

For $p=5$ the quadratic subfield is $\mathbb{Q}(\sqrt{p^{\ast }})=\mathbb{Q}(\sqrt{5})=\mathbb{Q}({\tau }_5)$, with ${\tau }_5=\epsilon \sqrt{5}\mathrm{\ne }0$ for some rational sign $\epsilon \in \{1,-1\}$.

$\left(\frac{3}{5}\right)=-1$ and $\left(\frac{5}{3}\right)=\left(\frac{2}{3}\right)=-1$.

By [F1] with $p=5$, $q=3$, the arithmetic Frobenius satisfies ${\mathrm{Frob}}_3({\tau }_5)=\left(\frac{5}{3}\right){\tau }_5=-{\tau }_5$. Since it fixes $\epsilon \in \mathbb{Q}$, step 1.1 gives $\epsilon {\mathrm{Frob}}_3(\sqrt{5})=-\epsilon \sqrt{5}$, hence ${\mathrm{Frob}}_3(\sqrt{5})=-\sqrt{5}$. The restriction identity gives $\left(\frac{5}{3}\right)=\left(\frac{3}{5}\right)=-1$.

Since ${\tau }_5\mathrm{\ne }0$, one has ${\mathrm{Frob}}_3({\tau }_5)=-{\tau }_5\mathrm{\ne }{\tau }_5$, so the arithmetic Frobenius of $3$ acts nontrivially on $\mathbb{Q}(\sqrt{5})$: it is the nontrivial element of $\mathrm{Gal}(\mathbb{Q}(\sqrt{5})\mathrm{/}\mathbb{Q})$, and the two Legendre symbols both equal $-1$, in agreement with the reciprocity law $(3\mathrm{/}5)(5\mathrm{/}3)=(-1{)}^{2\cdot 1}=1$.