Ramification primes of a reduced cyclotomic conductor

Given: A reduced index $f\ge 1$, a primitive $f$-th root of unity ${\zeta }_f$, the field $K=\mathbb{Q}({\zeta }_f)$, a rational prime $\mathrm{\ell }$, and the factorisation $f={\mathrm{\ell }}^{a}m$ with $\gcd (\mathrm{\ell },m)=1$ (so $a$ is the $\mathrm{\ell }$-adic valuation of $f$ and $\mathrm{\ell }\nmid f$ exactly when $a=0$).

By the prime factorisation theorem for reduced indices, $\mathrm{\ell }{\mathcal{O}}_K=(P_1\cdots P_g{)}^e,e=\phi ({\mathrm{\ell }}^a),$ with pairwise distinct primes $P_1,\dots ,P_g$; in particular every prime above $\mathrm{\ell }$ has exponent exactly $e$ in $\mathrm{\ell }{\mathcal{O}}_K$, so $\mathrm{\ell }$ is unramified in $K$ if and only if $e=1$ (Prime factorisation in a cyclotomic field).

For a prime $\mathrm{\ell }$ and $a\ge 1$, $\phi ({\mathrm{\ell }}^a)={\mathrm{\ell }}^a-{\mathrm{\ell }}^{a-1}={\mathrm{\ell }}^{a-1}(\mathrm{\ell }-1)$, and $\phi (1)=1$; hence for $a\ge 1$ one has $\phi ({\mathrm{\ell }}^a)=1$ exactly when $\mathrm{\ell }=2$ and $a=1$ (For a prime $p$ and $k\ge 1$, $\phi (p^k)=p^k-p^{k-1}$).

Since $f$ is reduced, $f$ is odd or $4\mid f$; consequently, if $2\mid f$ then $4\mid f$, so the exponent $a=v_2(f)$ is not $1$ — it is either $0$ or at least $2$. [definition of reduced index, arithmetic]