= Solution
Let $K=\mathbb Q(\alpha)$ with $\alpha$ an <algebraic integer>, let $m\in\mathbb Z[X]$ be its monic <minimal polynomial of an algebraic element>, and write $A=\mathbb Z[\alpha]$. The <Kummer-Dedekind theorem> applies to a rational prime $p$ not dividing the finite index $[\mathcal O_K:A]$. If
$$
\overline m=\prod_{j=1}^r g_j^{e_j}\quad\text{in }\mathbb F_p[X]
$$
with distinct monic <irreducible polynomials> $g_j$, choose monic integer lifts $\widetilde g_j$. Then the distinct <prime ideals> above $p$ are
$$
\mathfrak p_j=(p,\widetilde g_j(\alpha)),\qquad
\boxed{p\mathcal O_K=\prod_{j=1}^r\mathfrak p_j^{e_j},\quad
[\mathcal O_K/\mathfrak p_j:\mathbb F_p]=\deg g_j}.
$$
In particular, $p$ is unramified precisely when $\overline m$ has no repeated irreducible factor. The exclusion of primes dividing the index is essential.
Here is the <quotient-ring proof of the Kummer-Dedekind theorem>. The finite group $\mathcal O_K/A$ has order coprime to $p$. Localizing at the integers coprime to $p$ therefore gives $A\otimes\mathbb Z_{(p)}=\mathcal O_K\otimes\mathbb Z_{(p)}$, and reduction yields
$$
\mathcal O_K/p\mathcal O_K\simeq A/pA\simeq\mathbb F_p[X]/(\overline m).
$$
The second isomorphism follows from $A\simeq\mathbb Z[X]/(m)$. The <Chinese remainder theorem> gives
$$
\mathcal O_K/p\mathcal O_K\simeq\prod_j\mathbb F_p[X]/(g_j^{e_j}).
$$
Each factor is a <local ring> with unique <maximal ideal> generated by $g_j$ and <residue field> $\mathbb F_p[X]/(g_j)$. Pulling its <maximal ideal> back to $\mathcal O_K$ gives exactly $\mathfrak p_j$; these exhaust the primes above $p$. This proves the asserted <residue degrees> and their independence of the chosen lifts.
It remains to prove the exponents rather than simply identify the primes. Unique <prime ideal factorization> in the <Dedekind domain> $\mathcal O_K$ writes $(p)=\prod_j\mathfrak p_j^{b_j}$. At $\mathfrak p_j$, its <discrete valuation ring> has $p=u\pi^{b_j}$, with $u$ a unit. Thus the <maximal ideal> of $\mathcal O_{K,\mathfrak p_j}/p\mathcal O_{K,\mathfrak p_j}$ has <nilpotence index of an ideal> $b_j$: its $b_j$th power is zero and its $(b_j-1)$st is nonzero. In $\mathbb F_p[X]/(g_j^{e_j})$ this index is $e_j$, since $g_j^{e_j}=0$ but $g_j^{e_j-1}\ne0$. The displayed product isomorphism identifies these local factors, hence $b_j=e_j$. This completes the theorem's proof. The same local argument works over a base <number field> whenever $\mathcal O_K[\alpha]$ equals the full relative <ring of integers> after localizing at the chosen base prime.
For the required <cyclotomic field>, the allowed <ring of integers> formula makes the index one. Its <cyclotomic polynomial> is
$$
\Phi_{80}(X)=\frac{X^{40}+1}{X^8+1}=X^{32}-X^{24}+X^{16}-X^8+1.
$$
For $p\notin\{2,5\}$, the polynomial $X^{80}-1$ has derivative $80X^{79}$ coprime to it modulo $p$. Its divisor $\overline\Phi_{80}$ is therefore squarefree, so the <Kummer-Dedekind theorem> gives no ramification.
The <cyclotomic reduction at a conductor prime> is explicit at the remaining two primes:
$$
\overline\Phi_{80}=(X^4+X^3+X^2+X+1)^8\quad\text{in }\mathbb F_2[X],
$$
$$
\overline\Phi_{80}=(X^8+1)^4=(X^4+2)^4(X^4+3)^4\quad\text{in }\mathbb F_5[X].
$$
The quartic over $\mathbb F_2$ has no linear factor. Its only possible irreducible quadratic divisor would be $X^2+X+1$, whose roots have order three, whereas its own roots have order five. Hence it is irreducible. The two quartics over $\mathbb F_5$ are also irreducible: their roots have multiplicative order sixteen because their fourth powers are $2$ or $3$, both of order four; the <finite-field Frobenius automorphism> has orbit length four on an element of order sixteen, since $5^4\equiv1\pmod{16}$ and neither $5$ nor $5^2$ is $1$ modulo sixteen.
Consequently the <prime ideal factorization> above $2$ has one prime with ramification index eight and <residue degree> four. Above $5$ it has two primes, each with ramification index four and <residue degree> four. In particular,
$$
\boxed{\text{The ramified rational primes are exactly }2\text{ and }5.}
$$
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