I am thinking about the splitting of 11 in $Q(\zeta_5)$. This totally splits, and I was told that since the minimal polynomial of $\zeta$ completely factorizes mod 11 as $$ x^4 + x^3 + x^2 + x + 1 = (x-3)(x-4)(x-5)(x-9)\quad \mod 11, $$ the four prime factors are $$ P_1 = (11, \zeta-3), P_2 = (11, \zeta-4), \cdots. $$
I just realized that since the class number of $Q(\zeta_5)$ is one, each ideal is principal.
My question is what is a generator of, say, $P_1$. Thanks.
I played with this using Maple, and got the following answer. Recall that $\zeta_5 = \frac{\sqrt{5}-1}{4} + \frac{i}{2}\sqrt{\frac{5+\sqrt{5}}{2}}$. We have $$ 11 = (1+2\zeta)(1+2\zeta^2)(1+2\zeta^3)(1+2\zeta^4). $$ In particular, $(1+2\zeta)(1+2\zeta^4)=|1+2\zeta|^2 = 4+\sqrt{5}$ and $(1+2\zeta^2)(1+2\zeta^3)=|1+2\zeta^2|^2 = 4-\sqrt{5}$.
To relate this with the other factorization $$(11)=(11, \zeta-3)\cdot (11, \zeta-4)\cdot (11, \zeta-5)\cdot (11, \zeta-9), $$ we can check that for example $\zeta-3$ is a factor of one of the $1+2\zeta^i$ mod 11. This amounts to that $3$ is a root mod 11, and that term would be $1+2\zeta^3$. Then we have the following correspondence, and I also list the division relations, computed using Maple. \begin{align*} (1+2\zeta) &= (11, \zeta-5),\quad \zeta-5=(1+2\zeta)(3+3\zeta+2\zeta^2+4\zeta^3),\\ (1+2\zeta^2) &= (11, \zeta-4),\quad \zeta-4=(1+2\zeta^2)(\zeta-2\zeta^3),\\ (1+2\zeta^3) &= (11, \zeta-3),\quad \zeta-3=(1+2\zeta^3)(-1-\zeta-2\zeta^2),\\ (1+2\zeta^4) &= (11, \zeta-9),\quad \zeta-9=(1+2\zeta^4)(-1-5\zeta+2\zeta^2-2\zeta^3). \end{align*}