= Solution
Let $L/K$ be an extension of <number fields>, let $\alpha\in\mathcal O_L$ be a <primitive element of a field extension>, and let $f\in\mathcal O_K[X]$ be its monic <minimal polynomial of an algebraic element>. Put $R=\mathcal O_K[\alpha]$ and $S=\mathcal O_L$. The <Kummer-Dedekind theorem> applies at a nonzero <prime ideal> $\mathfrak p$ of $\mathcal O_K$ when $R_{\mathfrak p}=S_{\mathfrak p}$: equivalently, $\mathfrak p$ is outside the support of the finite-index quotient $S/R$. Write the factorization over the <residue field> $k=\mathcal O_K/\mathfrak p$ as
$$
\overline f=\prod_{j=1}^r\overline g_j^{\,m_j},
$$
with distinct monic <irreducible polynomials> $\overline g_j$, and choose monic lifts $g_j\in\mathcal O_K[X]$. Then
$$
\boxed{\mathfrak pS=\prod_{j=1}^r\mathfrak P_j^{\,m_j},\qquad
\mathfrak P_j=(\mathfrak p,g_j(\alpha)),\qquad
f(\mathfrak P_j/\mathfrak p)=\deg\overline g_j.}
$$
Here $f(\mathfrak P_j/\mathfrak p)$ denotes the <residue-field degree>, and the <ramification index of a prime ideal> is $m_j$.
To prove this, the local index condition gives
$$
S/\mathfrak pS\simeq R/\mathfrak pR
\simeq k[X]/(\overline f)
\simeq\prod_{j=1}^r k[X]/(\overline g_j^{\,m_j}).
$$
The final isomorphism is the <Chinese remainder theorem>. Each factor is a local ring with unique maximal ideal generated by $\overline g_j$, so its inverse image in $S$ is exactly $\mathfrak P_j$, and its <residue field> is $k[X]/(\overline g_j)$. This proves the list of primes and the <residue-field degrees>. For the exponents, localize at $\mathfrak P_j$. The <ring of integers of a number field> is a <Dedekind domain>, so $S_{\mathfrak P_j}$ is a <discrete valuation ring>. If $\mathfrak pS_{\mathfrak P_j}=\mathfrak P_j^{e_j}S_{\mathfrak P_j}$, the quotient has $e_j$ successive layers isomorphic to $S/\mathfrak P_j$, and hence $k$-dimension $e_j\deg\overline g_j$. The corresponding polynomial factor has $k$-dimension $m_j\deg\overline g_j$. Thus $e_j=m_j$, completing the proof rather than merely identifying the radicals.
For the calculation, set $K=\mathbb Q(\sqrt5)$ and $t=(\sqrt5-1)/2$. The <ring of integers of a quadratic field> gives $\mathcal O_K=\mathbb Z[t]$, with $t^2+t-1=0$. The <algebraic norm> of $4+\sqrt5$ is $11$, so its <principal ideal> has <ideal norm> $11$. Its quotient ring has prime order and is therefore the <finite field> $\mathbb F_{11}$; consequently \b[$(4+\sqrt5)$ is a prime ideal]. In that <residue field>, $\sqrt5=-4=7$ and $t=3$.
Let $L=\mathbb Q(\zeta_5)$. Since $t=\zeta_5+\zeta_5^{-1}$ lies in $\mathbb Z[\zeta_5]$, the permitted identity $\mathcal O_L=\mathbb Z[\zeta_5]$ gives $\mathcal O_L=\mathcal O_K[\zeta_5]$. Thus the local index condition holds at every prime. The relative <minimal polynomial of an algebraic element> is $X^2-tX+1$, whose reduction is
$$
X^2-3X+1=(X-5)(X-9)\quad\text{in }\mathbb F_{11}[X].
$$
The roots are distinct. Applying the <Kummer-Dedekind theorem> gives
$$
\boxed{(4+\sqrt5)\mathcal O_L
=(4+\sqrt5,\zeta_5-5)(4+\sqrt5,\zeta_5-9)
=(11,\zeta_5-5)(11,\zeta_5-9).}
$$
For the last equality, evaluation at $5$ or $9$ in $\mathbb F_{11}$ gives $\zeta_5+\zeta_5^{-1}=3$ and therefore $4+\sqrt5=5+2t=0$. Both descriptions are the same kernels of these evaluation maps. Each prime has <ideal norm> $11$, <residue-field degree> one and <ramification index of a prime ideal> one.
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