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Let $L$ be the splitting field of $X^4+2X^2-4$ over $\mathbb{Q}(i)$
Show that $Gal(L:\mathbb{Q})\cong \mathbb{Z}_2\times \mathbb{Z}_2$ .
I was wondering why my approach is incorrect:
Consider the cubic resolvent of $X^4+2X^2-4$ $,CR = U^3+4U^2+20U$. Hence $CR$ has roots $U_1=0$, $U_2=-2-4i$ and $...