Let $p$ be a negative prime number such that $p \equiv 5\pmod 8$.
Let $K = \mathbb{Q}(\sqrt{p})$ and denote its ideal class group by $Cl_K$. I aim to prove that $Cl_K[2] := \{a \in Cl_K \mid 2a = 0\}$ is trivial.
In this context, $p$ and $2$ are unramified in $\mathbb{Q}(\sqrt{p})/\mathbb{Q}$ because one can verify that the corresponding local field extension is unramified.
Given this information on ramification, how can we conclude that $Cl_K[2] = 1$?
Remarks:
1.A similar question is discussed at https://mathoverflow.net/questions/141150/2-class-group-of-a-quadratic-imaginary-extension, but it lacks detailed answers to this specific question.
2.The concept of the "2-Hilbert class field" may be key to this question. If it can be shown that the 2-Hilbert class field of $K$ is $K$ itself, then we have our conclusion.
3.This formula follows from 'genus theory', but what I'm seeking is to understand this formula in terms of class field theory or ramification theory.