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!

You have shown that every ideal class contains an integral ideal with norm at most $4$.
Recall how primes split in quadratic fields. Because $-47$ is $1$ mod $8$, we have 
$$(2)=\left(2, \frac{1+\sqrt{-47}}{2}\right)\left(2, \frac{1-\sqrt{-47}}{2}\right).$$
Because $-47$ is $1$ mod $3$, we have 
$$(3)=(3,1-\sqrt{-47})(3,1+\sqrt{-47}).$$
It is easy to see that these are all non-principal. Note that the norm of $1+\sqrt{-47}$ is $48=2^4\cdot 3$. Note that $(3,1+\sqrt{-47})$ divides $(1+\sqrt{-47})$, with the rest of the factors being one of the two factors of $2$ above. 
This shows that we only need the prime factors of $(2)$ to generate the class group. Note that one is the inverse of the other in the class group, so it suffices to find the order of one of the prime factors. Going through the computations shows that the fifth power is the first one that is principal, so the class group is cyclic with order $5$.
