Prime decomposition in Q(zeta_8)

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 $efg=4=[K:\mathbb{Q}]$.