This is an understanding question : to search for the order $n$ of the ideal class group, why do we only need to try to split the $(p)$ where p is a prime $\leq M$ ($M$ being the Minkowski bound) ? I think it's because every ideal class contains an ideal of norm $\leq M$ and that every ideal can be split in a product of primes, and for every ideal $I$, $I^n$ is principal but i can't put it together properly.
Thank you !
Any ideal class may be represented by a ideal $I$ of norm $\leq M$.
In particular, any prime ideal $P$ dividing $I$ has norm $\leq M$. Hence the ideal $I$ is the product of prime ideals with norm $\leq M$.
Note that any prime ideal $P$ is lying above a prime number $p$. The norm of $P$ is a non trivial power of $p$, so $p\leq M$
So the classes of prime ideals with lying above a prime $p\leq M$ generate the class group.
Now if $p$ is inert, then $P=(p)$ is principal, and its class is trivial in the class group. SO, you just need to consider the non inert $p$.
This is all you can say if you work with an arbitrary number field.
If you are only interested in quadratic fields, it is true that you can stick to split or totally ramified primes, because in this case $p$ is inert, split or totally ramified.
However, there are cases where the unique prime ideal lying above a ramified prime is not principal (so a priori you need to take it into account in the class group)
For example, for $\mathbb{Q}(\sqrt{-30})$, $M\approx 6.97$, so you can only consider $p=2,3,5$. All of them are totally ramified, and you can show that the class group is the Klein group, and generated by the classes of $\mathfrak{p}_2$ and $\mathfrak{p}_3$ (see https://kconrad.math.uconn.edu/blurbs/gradnumthy/classgpex.pdf, Example 4).
So you cannot only stick to split $p\leq M$.