 Solution

For a direct check of unramified behavior, the cyclotomic polynomial ${\Phi }_{25}$ divides $X^{25}-1$. Modulo $7$, the latter has derivative $25X^{24}=4X^{24}$ and has no common root with that derivative. Thus ${\overline{\Phi }}_{25}$ is separable, so $7$ does not divide its polynomial discriminant. The field discriminant divides this polynomial discriminant, and therefore $7$ is unramified in $\mathbb{Q}({\zeta }_{25})$. The Frobenius automorphism satisfiesIndeed its residue action is seventh powering. The reduced $25$th roots of unity are distinct because $7\nmid 25$, so the residue congruence identifies the exponent uniquely modulo $25$.

The Galois group is $(\mathbb{Z}/25\mathbb{Z}{)}^{\times }$, of order $\phi (25)=20$. Since $7^2=49\equiv -1(\mathrm{mod}25)$, $7^4\equiv 1$ but $7^2\not\equiv 1$. Consequently the Frobenius automorphism has order four. An unramified decomposition group is generated by that element, and the orbit-stabilizer theorem givesThere are therefore five primes above $7$, each with residue field ${\mathbb{F}}_{7^4}$.