Let $K=\Bbb{Q}(\sqrt{5})$ and consider the ring of integers $\Bbb{Z}\left[\frac{1+\sqrt{5}}{2}\right] = \mathcal{O}_K\subset K$.
I want to explicitly determine the prime ideals of $\mathcal{O}_K$ over $11\Bbb Z$. By considering the minimal polynomial $x^2-x-1$ of $(1+\sqrt{5})/2$ modulo $11$ I find: $$x^2-x-1\equiv (x+7)(x+3)\pmod{11}.$$ This tells me that the ideals over $11\Bbb Z$ are $\langle 11, x + 7 \rangle, \langle 11, x + 3 \rangle$, with $x = \frac{1 + \sqrt{5}}{2}$.
I.e. the ideals are $$\left\langle 11, \frac{7 + \sqrt{5}}{2} \right\rangle \text{ and } \left\langle 11, \frac{15 + \sqrt{5}}{2} \right\rangle.$$
I know that this $K$ has class number $1$, i.e. $\mathcal{O}_K$ is a PID. With this in mind, I would like to know what single generator generates these two ideals over $11\Bbb Z$.
It is clear that $\left\langle 11, \frac{7 + \sqrt{5}}{2} \right\rangle = \left\langle 11, \frac{15 + \sqrt{5}}{2} \right\rangle$. But I can't see why the second one would be generated by $\frac{15+\sqrt{5}}{2}$ if this is so.
Furthermore, since these are principal ideals, I should be able to find a generator $\alpha$ of this ideal such that $\left\langle \frac{7 + \sqrt{5}}{2} \right\rangle \langle \alpha \rangle = \langle 11 \rangle$. This leads me to believe that the second one is: $$\left\langle 11, \frac{15 + \sqrt{5}}{2} \right\rangle = \left\langle \frac{-7 + \sqrt{5}}{2} \right\rangle.$$
(After seeing this, one can see that I could just subtract $11$ from $\frac{15+\sqrt{5}}{2}$ in the second ideal, but hindsight is $\sim20/4$)
Question: Is there a faster procedure to find the principal prime ideals lying over some $p\Bbb Z$ explicitly? (Say in the non-ramified, non-inert case, and for a quadratic number field, this one or any other.)
Yes, there is, and that requires being a little less concerned about half $1 + \sqrt d$.
Given the golden ratio $$\phi = \frac{1}{2} + \frac{\sqrt 5}{2},$$ we see that $(4 - \phi)(3 + \phi) = 11$. But that confuses you humans, which is why I like it.
Because of your human limitations, it is better to ignore the "half-integers" and the infinitely many units, and instead commit to memory the following formula: if $\gcd(p, d) = 1$ and you can solve $x^2 \equiv d \pmod p$, then $\langle p \rangle = \langle p, x - \sqrt d \rangle \langle p, x + \sqrt d \rangle$.
So in your example, you have $p = 11$ and $d = 5$. Then $4^2 = 16 \equiv 5 \pmod{11}$, and so $\langle 11 \rangle = \langle 11, 4 - \sqrt 5 \rangle \langle 11, 4 + \sqrt 5 \rangle$. But since $(4 - \sqrt 5)(4 + \sqrt 5) = 11$, the $11$'s in the ideal generators turn out to be redundant, and hence $\langle 11 \rangle = \langle 4 - \sqrt 5 \rangle \langle 4 + \sqrt 5 \rangle$.