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I have $\mathbb Q (i,\sqrt 5)$ and i need to find $a \in \mathbb Q (i,\sqrt 5)$ that $\mathbb Q (i,\sqrt 5) = \mathbb Q (a)$, i have been playing with $\sqrt 5 + i$ but got nowhere, can anyone give a hint on how to look for candidates?

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    $\begingroup$ You have the right element, think about what $[\mathbb{Q}(\sqrt(5)+i):\mathbb{Q}]$ could be $\endgroup$ – Sheel Stueber Jul 4 '18 at 1:53
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Since you have $(\sqrt5+i)^{-1}=\frac 16(\sqrt5-i)\implies $ we have $\sqrt5-i$, and from there we easily get $\sqrt5$ and $i$...

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