Is there an imaginary number field $K$ (ie $K\not\subset\mathbb{R}$) such that $i\notin K$, such that there is a quadratic extension $L/K(i)$ such that $L/K$ is Galois with group $\mathbb{Z}/4\mathbb{Z}$?
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Suppose that such a field exists, since $L/K(i)$ is quadratic, it is the splitting field of a degree 2 polynomial , so $L/K$ is the splitting field of $Q=P(X^2+1)$. Let $\{-i,i,a,b\}$ be the roots of $Q$ we can write $P=(X-a)(X-b)$ and $s\in Gal(L/K)$. We have $(X^2+1)^s=(X^2+1)$. This implies that $s\{-i,i\}=\{-i,i\}$. We deduce that and $s\{a,b\}=\{a,b\}$. Thus the order of $s$ is 2 and such a field doesn't exist since you have an element of order 4 in $Z/4$.
Tsemo Aristide
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But $(X^2+1)^s=(X^2+1)$ so $s$ preserves ${-i,i}$. – Tsemo Aristide Oct 04 '16 at 15:43
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Yes, you edited your answer after I posted my comment, now it's fine. +1. – Ege Erdil Oct 04 '16 at 15:43
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There is a general criterion - shorter than @Tsemo Aristide (?) - which says immediately "no" http://math.stackexchange.com/a/1915223/300700 – nguyen quang do Oct 06 '16 at 05:34