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Let the field $K=\mathbb{Q}(\sqrt{p_1}, \sqrt{p_2 q}, i)$ where $p_1, p_2 \equiv 1 \mod{4}$ and $q \equiv 3 \mod{4}$, kronecker(2,$p_1$)=1 and kronecker(2,$p_2$)=kronecker($p_1$,$p_2$) = kronecker($p_1$,$q$)=-1. How to prove that $r$, the 2-rank of $K$, is equal to 2. I have proved that $2 \le r \le 3$.

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I TeX'd your question - please check that I haven't changed your intended meaning. –  Ted Sep 14 '12 at 8:29

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