fundamental group of a quadratiс extension of the ring of integers I want to calculate $\pi_{1}^{et} (\mathbb{Z}[\sqrt{-5}])$ but I don't quite see how to do that. This ring is the ring of integers of $\mathbb{Q}(\sqrt{-5})$. If $\mathbb{Z}[\sqrt{-5}] \to A$ is finite, etale and connected, then $A$ is the normalisation of $\mathbb{Z}[\sqrt{-5}]$ in $Frac (A)$. Thus $\pi_{1}^{et} (\mathbb{Z}[\sqrt{-5}])$ is the Galois group of the maximal extension of $\mathbb{Q}(\sqrt{-5})$ which is unramified at all primes of $\mathbb{Z}[\sqrt{-5}]$.

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*I don't quite understand how primes in this ring look like. For example, $2, 3$ are primes and $5$ is not. Is there a way to describe the primes in a general quadratic ring of integers?

*Even assuming that I know all the primes I'm not sure I know how to solve it. Is there a general idea behind calculating $\pi_{1}^{et} (\mathcal{O}_{K})$?

 A: First, $O=\mathbb{Z}[\sqrt{-5}]$ has three types of primes.

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*ramified primes ($\mathfrak{p} \subset O$ whose residue characteristic $p$ is such that $O/pO$ is not reduced).

*inert primes ($\mathfrak{p}=pO$ for certain prime integers $p$).

*split primes ($\mathfrak{p}$ not equal to its conjugate such that $pO=\mathfrak{p}\overline{\mathfrak{p}}$).

All the primes above a prime integers $p$ fall into the same case (if it’s 1) or 2), there is only one anyway).
Now, $(\sqrt{-5})^2=5O$ so $5$ is ramified. An easy computation shows that $O/2O\cong \mathbb{Z}[x]/(2,(x+1)^2)$ is not reduced either so $2$ is ramified. On the other hand, $\sqrt{-5}$ vanishes $X^2+5$ which is separable mod any $p \notin \{2,5\}$ so they’re all unramified.
We also compute the absolute discriminant of $K=\mathbb{Q}(\sqrt{-5})$ is $20$.
Now, given an unramified prime $p$, how can we tell whether $p$ is inert or split? Well, it depends on what the ring $O/pO=\mathbb{F}_p[x]/(x^2+5)$ is. This ring is a $2$-dimensional reduced $\mathbb{F}_p$-algebra, so it can be either $\mathbb{F}_p^2$ (iff $-5$ is a square mod $p$, iff by quadratic reciprocity $(-1)^{(p-1)/2}(p/5)=1$ (where $(p/5)$ is the Legendre symbol) – then $p$ is split – or the degree $2$ extension of  $p$ – then $p$ is inert.
In that last case, the prime ideal is easy, it’s $pO$. In the first case, the prime ideals are generated by $p$ and $\sqrt{-5}-t$ where $t \in \mathbb{Z}$ is such that $p|t^2+5$.
For your second question: we are trying to find the maximal unramified extension $E$ of $K=\mathbb{Q}(\sqrt{-5})$.
A natural starting point is the Hilbert class field $H$. Note that $K$ is totally imaginary, so everything happens at the finite places, hence $Gal(H/K)$ is the abelianization of $Gal(E/K)$.
The Minkowski bound is $\sqrt{20}\frac{4}{\pi}\frac{2}{4}=\frac{4\sqrt{5}}{\pi} < 3$, so the ideal above $2$ generates the class group and its square is principal, so $H/K$ is of degree $2$.
In general, I don’t think it’s easy to compute the étale $\pi_1$ of the ring of integers of a number field. But here, things might be small enough for us to manage!
Now, let $F/K$ be any finite extension of degree $d$, unramified everywhere. Then its relative discriminant is the unit ideal, and the absolute discriminant of $F$ is $20^d$ (by composition).
But by Minkowski, $20^{d/2} \geq \frac{(2d)^{2d}}{(2d)!}\left(\frac{\pi}{4}\right)^d$. By Wolfram (I hope), this means $d <10$.
(Note that this kind of bound happens only because $20$ is small enough. If it had been a larger determinant, then $LHS/RHS$ could have gone to infinity as $d$ went large. See also: towers of class fields, for reasons that should become clearer below).
It follows in particular that $[E:K] < 10$, and $Gal(H/K)$ (cyclic of order $2$) is the abelianization of $Gal(E/K)$.
By finite group theory, since $Gal(E/K)$ has order less than $10$, it can be either $\mathbb{Z}/2\mathbb{Z}$ or $S_3$.
Either way, $E/H$ is abelian unramified everywhere and is either trivial or $\mathbb{Z}/3\mathbb{Z}$, so all that remains to check is whether the class group of $H$ is trivial.
The Minkowski bound is $20 \times (4/\pi)^2 \times 24/256=15 \times 2^9 / (\pi^2 \times 256)=30/\pi^2 < 4$.
So it is enough to check whether every prime ideal of $H$ of norm $2$ or $3$ is principal. (If the answer is ‘yes’, then $E=H$, otherwise $Gal(E/K)=S_3$).
Now, let $\mathfrak{p} \subset O_H$ be of norm $2$ or $3$, ie the residue field has cardinality $2$ or $3$. Then $H/K$ has trivial residue extension at $\mathfrak{p}$. Since it’s a quadratic extension, $\mathfrak{p}$ cannot be inert over $K$, and it cannot be ramified either, so that $\mathfrak{q}=\mathfrak{p} \cap O$ splits in $H$ (and thus has norm $2$ or $3$).
But by “class field theory” (who would have guessed?), a prime ideal splits totally in the class field iff it is principal in the base field.
But if $\mathfrak{q}$ is principal in $O$, then its generator has norm $\pm 2$ or $\pm 3$, which is seen to be impossible (the norm is $x+y \sqrt{-5} \in \mathbb{Z}[\sqrt{-5}] \longmapsto x^2+5y^2$).
So $H$ has no prime ideals of norm $2$ or $3$, so $O_H$ is a PID and thus $E=H$. QED.
