On class groups of algebraic number fields I've been making a light study of the relationships between Diophantine equations and their relations to class groups of algebraic number fields. For the most part (aside from degree two) there is no direct relationship however an interesting aside question comes up (assuming my math isn't just plain inaccurate)
Consider the set of all ideals in an algebraic number field $K/\mathbb{Q}$
For each prime $p \in \mathbb{Z}$ let us choose exactly one prime above $\mathfrak{P} | (p) $ in $O_K$
Consider then $I'_K$ to be the subset of the fractional ideal group $I_K$ generated entirely by this subset of all the ideals in $K$
So if this is not wrong thinking my question is what would the group $I'_K/P'_K$ be ?
 A: (Working with David Loeffler's interpretation of the question).
Let $K/\mathbf{Q}$ be a cyclic cubic extension with class group $C = (\mathbf{Z}/2 \mathbf{Z})^2$, for example
$$K = \mathbf{Q}[x]/(x^3 -163 x + 163)
\subset \mathbf{Q}(\zeta_{163})$$
of discriminant $163^2$. There is a natural action of $\mathbf{Z}/3\mathbf{Z} = \mathrm{Gal}(K/\mathbf{Q})$ on $C$, and the action is non-trivial. There is a unique such action, and it acts transitively on the non-zero elements of $C$.
Since $K/\mathbf{Q}$ is Galois, either $p = 163 = \mathfrak{p}^3$ with $\mathfrak{p} = (x)$ principal, or $p$ is inert and thus principal, or $p$ splits completely. In the latter case, we have $p = \mathfrak{p}_1 \mathfrak{p}_2 \mathfrak{p}_3$. In this case, the action of $\mathrm{Gal}(K/\mathbf{Q})$ permutes the $\mathfrak{p}_i$. Hence they are all trivial in the class group, or they represent each of the three non-trivial elements of the class group.
In particular, if one is allowed to perversely choose any particular $\mathfrak{p}_i$ for any $p$, the corresponding subgroup of $C$ they generate can be forced to be the entire group or to be any of the three non-trivial subgroups of order $2$.
