Class Group of $\mathbb{Q}(\sqrt{-47})$ Calculate the group of $\mathbb{Q}(\sqrt{-47})$.
I have this: The Minkowski bound is $4,36$ approximately.
Thanks!

Here is another attempt. In case I made any mistakes, let me know and I will either try and fix it, or delete my answer.

We have Minkowski bound $\frac{2 \sqrt{47}}{\pi}<\frac{2}{3}\cdot 7=\frac{14}{3}\approx 4.66$. So let us look at the primes $2$ and $3$:
$-47\equiv 1$ mod $8\quad\Rightarrow\quad 2$ is split, i.e. $(2)=P\overline P$ for some prime ideals $P,\overline P$.
NB: In fact we have $P=(2,\delta)$ and $\overline P=(2,\overline \delta)$ with $\delta=\frac{1+\sqrt{-47}}{2}$ and $\overline\delta=\frac{1-\sqrt{-47}}{2}$. But this is going to be irrelevant in the rest of the proof.
$-47\equiv 1$ mod $3\quad\Rightarrow\quad 3$ is split, i.e. $(3)=Q \overline Q$ for some prime ideals $Q,\overline Q$.
So the class group has at most 5 elements with representatives $(1),P,\overline P, Q, \overline Q$.
Note that $P$ is not principal, because $N(\frac{a+b\sqrt{-47}}{2})=\frac{a^2+47b^2}{4}=2$ does not have an integer solution (because $8$ is not a square). So $P$ does not have order $1$.
Suppose $P$ has order $2$. Then $P^2$ is a principal ideal with $N(P^2)=N(P)^2=2^2=4$. The only elements with norm $4$ are $\pm2$. But $P^2$ cannot be $(2)$, because $2$ is split.
Suppose $P$ has order $3$. Then $P^3$ is a principal ideal with $N(P^3)=N(P)^3=2^3=8$. But $N(\frac{a+b\sqrt{-47}}{2})=\frac{a^2+47b^2}{4}=8$ does not have an integer solution (because $32$ is not a square).
Suppose $P$ has order $4$. Then $P^4$ is a principal ideal with $N(P^4)=16$. The only elements with norm $16$ are $\pm4$. But $P^4$ cannot be $(4)$, because $(4)=(2)(2)=P\overline P P\overline P$ is the unique factorisation, and $P\ne \overline P$.
Suppose $P$ has order $5$. Then $P^5$ is a principal ideal with $N(P^5)=32$. And, indeed, the element $\frac{9+\sqrt{-47}}{2}$ has norm $32$.
So $P^5=(\frac{9+\sqrt{-47}}{2})$. Hence the class group is cyclic of order $5$.
