 Solution

The quadratic field $K=\mathbb{Q}(\sqrt{-7})$ is the unique quadratic subfield of $\mathbb{Q}({\zeta }_7)$. Its nontrivial character is the quadratic characterUnder the Artin reciprocity map, Frobenius at an unramified prime $p$ restricts trivially to $K$ exactly when ${\chi }_7(p)=1$. The quadratic residues modulo $7$ are $1,2,4$, while $5$ is a nonresidue. Thusso the Artin symbol at $5$ is the nonidentity element of $\mathrm{Gal}(K/\mathbb{Q})$. Therefore $5$ has residue degree two and is inert in $K$.