 Inert prime

A nonzero prime ideal $\mathfrak{p}$ of a number field $K$ is inert in a finite extension $L/K$ when $\mathfrak{p}{\mathcal{O}}_L$ remains a prime ideal. There is then one prime above it, with ramification index one and residue degree $[L:K]$.

For example, $3$ is inert in $\mathbb{Z}[i]$: its quotient is ${\mathbb{F}}_3[X]/(X^2+1)$, a field because $X^2+1$ has no root in ${\mathbb{F}}_3$. By contrast $5=(2+i)(2-i)$ splits. For quadratic extensions, irreducibility of a suitable defining polynomial modulo an unramified prime gives an elementary test for this behavior.