Show that the only subfields of $\mathbb{Q}(i, \sqrt{5})$ is $\mathbb{Q}, \mathbb{Q}(i),\mathbb{Q}(\sqrt{5}), \mathbb{Q}(i \sqrt{5})$ and itself? I'm reading Stewart's Galois Theory and encountered this exercise in Chapter 8. I want to show this by contradiction: Assuming there exists a proper subfield $\mathbb{Q}(\alpha)$ of $\mathbb{Q}(i, \sqrt{5})$, then $\alpha = a + bi + c\sqrt{5} + di\sqrt{5}$ for $a,b,c,d \in \mathbb{Q}$ but $\alpha$ cannot be expressed as $a + bi$ or $a + b\sqrt{5}$ solely. Then $\mathbb{Q}(\alpha)$ has to be the $\mathbb{Q}(i, \sqrt{5})$, contradicting it being a proper subfield? I think my argument isn't strong enough so could anyone give me a hint of how to show it more effectively? Also, since this is a chapter where we used a lot of field extension skills, I wonder if there is a way of seeing these field and subfields as towers and field extensions and prove the desired result. Thanks a ton! 
 A: I'll give you two different ways to do this as I don't know if you know the main theorem yet but even if you don't you can revisit this once you do learn it.
Method 1:
The simplest method is to use the Fundamental Theorem/ Galois Correspondence Theorem which says the intermediate fields are in bijection with subgroups of the Galois group, which is isomorphic to $C_2 \times C_2$ in this case.
This has $5$ subgroups so we have $5$ intermediate fields which are $\mathbb{Q}$, $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt{5})$, $\mathbb{Q}(i\sqrt{5})$ and $\mathbb{Q}(i, \sqrt{5})$ which are easily spotted.
Method 2:
If you haven't got that far yet then we can use the tower law instead. Note $[\mathbb{Q}(i,\sqrt{5}):\mathbb{Q}]=4$ so any nontrivial subfield will have degree $2$.
It's a standard result that any quadratic field has the form $\mathbb{Q}(\sqrt{D})$ for some squarefree integer $D$. Using your basis $a+bi+c\sqrt{5} +di\sqrt{5}$, we can see $i=\sqrt{-1}$, $\sqrt{5}$ and $i\sqrt{5}=\sqrt{-5}$ all lie in $\mathbb{Q}(i,\sqrt{5})$ so all give quadratic subfields.
Now suppose $\mathbb{Q}(\sqrt{D})$ was also a subfield. Then$\sqrt{D} \in \mathbb{Q}(i,\sqrt{5})$. This means $\sqrt{D} = a+bi+c\sqrt{5} +di\sqrt{5}$ for some $a,b,c,d$.
Squaring both sides we get $D= (a^2 - b^2 +5c^2 - 5d^2) + (2ab+10cd)i + (2ac-2bd)\sqrt{5} +(2ad+2bc)i\sqrt{5}$.
We are then left with solving the simulataneous equations:
$\begin{eqnarray*}
D &=& a^2 - b^2 +5c^2 - 5d^2, \\
0 &=& 2ab+10cd, \\
0 &=& 2ac-2bd, \\
0 &=& 2ad+2bc,
\end{eqnarray*}$
which then gives you solutions only for $D=-1,5,-5$ (remembering that we only consider squarefree $D$), but this is quite tedious.
