# Candidate for showing extension is simple

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?

• You have the right element, think about what $[\mathbb{Q}(\sqrt(5)+i):\mathbb{Q}]$ could be – Sheel Stueber Jul 4 '18 at 1:53

Since you have $(\sqrt5+i)^{-1}=\frac 16(\sqrt5-i)\implies$ we have $\sqrt5-i$, and from there we easily get $\sqrt5$ and $i$...