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Let $K$ be a degree $n \geq 3$ extension over $\mathbb{Q}$, and let $\mathcal{O}_K$ be its ring of integers. We say a rational prime $p$ splits completely in $\mathcal{O}_K$ if the principal ideal $(p) = \mathfrak{p}_1 \cdot \cdots \cdot \mathfrak{p}_n$ for pairwise distinct degree one prime ideals $\mathfrak{p}_i$, $1 \leq i \leq n$ (indeed, we do not want $p$ to ramify in $\mathcal{O}_K$). Let $C_K$ be the ideal class group of $\mathcal{O}_K$. For an ideal $I \subset \mathcal{O}_K$, let $[I]$ denote the corresponding ideal class in the class group.

Let $g_1, \cdots, g_{n-1}$ be arbitrary (not necessarily distinct) elements of $C_K$. Does there necessarily exist a rational prime $p$ which splits completely in $\mathcal{O}_K$ such that the ideal class equations $g_i = [\mathfrak{p}_i]$ holds for $1 \leq i \leq n-1$? Note that there are at most $n-1$ degrees of freedom since the product of the $\mathfrak{p}_i$'s is principal.

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No. If $K$ is Galois, since the Galois group act transitively on the $\mathfrak{p_i}$, their class must be the same up to automorphism of the class group. Thus you can take as a counterexample any Galois extension $K$ of degree at least $3$ with nontrivial class group, $g_1$ the identity and $g_2$ any nonidentity element.

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