Gaussian and eisenstein frobenius

The quadratic integral-basis theorem gives ${\mathcal{O}}_{\mathbb{Q}(i)}=\mathbb{Z}[i]$ and ${\mathcal{O}}_{\mathbb{Q}({\zeta }_3)}=\mathbb{Z}[{\zeta }_3]$, since ${\zeta }_3=(-1+\sqrt{-3})\mathrm{/}2$. The polynomials are $T^2+1$ and $T^2+T+1$, with discriminants -4 and -3. At the stated nonexceptional primes they have good reduction.

For odd p, roots of $T^2+1$ are elements of order four in ${\mathbb{F}}_p^{\times }$. Cyclicity says they exist exactly when $4\mid p-1$. For $p\mathrm{\ne }3$, roots of $T^2+T+1$ are elements of order three, since $(T-1)(T^2+T+1)=T^3-1$ and T=1 is not a root unless p=3. Cyclicity gives roots exactly when $3\mid p-1$. A quadratic without a root is irreducible. Thus the cycle types are two fixed points or a transposition, giving split or inert cases in the quadratic table.

In good reduction the two roots are distinct. The Frobenius congruence sends each chosen root to its p-th power modulo P; that power is itself a root, so injectivity on the two root reductions forces equality in the number field. This gives both claimed formulas.

In the Gaussian ring, $(1+i{)}^2=2i$ and $\mathbb{Z}[i]\mathrm{/}(1+i)={\mathbb{F}}_2$ by substituting i=-1. Thus (1+i) is prime and $(2)=(1+i{)}^2$ as ideals. In the Eisenstein ring, $(1-{\zeta }_3{)}^2=-3{\zeta }_3$ and the quotient by $(1-{\zeta }_3)$ is ${\mathbb{F}}_3$ by substituting ${\zeta }_3=1$. Hence $(3)=(1-{\zeta }_3{)}^2$ as ideals. Both exceptional primes ramify.