Prime factorisation in quadratic fields

In $R=\mathbb{Z}[i]$, $(2+i)(2-i)=5$ as elements, hence $(5)=(2+i)(2-i)$ as ideals. The quotient by $(2+i)$ is ${\mathbb{F}}_5$ via $i\mapsto -2$, and the quotient by $(2-i)$ is ${\mathbb{F}}_5$ via $i\mapsto 2$: in either quotient eliminate $i$, leaving the relation $5=0$. Thus both factors are prime of residue degree one. They are distinct, since $2+i$ maps to $4\mathrm{\ne }0$ in the second quotient.

The quotient $R\mathrm{/}(3)$ is ${\mathbb{F}}_3[X]\mathrm{/}(X^2+1)$, a field of nine elements since $X^2+1$ has no root in ${\mathbb{F}}_3$. Thus $(3)$ is itself prime with residue degree two.

Finally $(1+i{)}^2=2i$ as elements, and $i$ is a unit, so $(2)=(1+i{)}^2$ as ideals. The quotient $R\mathrm{/}(1+i)$ is ${\mathbb{F}}_2$ by substituting $i=-1$, so $(1+i)$ is prime with residue degree one. These explicit ideal products and residue fields give respectively $(e,f)=(1,1),(1,1)$; $(1,2)$; and $(2,1)$. They satisfy the degree-two fundamental identity and the definitions of split, inert, and ramified.