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!

  • 1
    $\begingroup$ Have you seen the Galois correspondence theorem? If yes, then you just need to determine the Galois group of the extension $\mathbb{Q} \subset \mathbb{Q}(i,\sqrt{5})$. If not, then perhaps you need to show that such an $\alpha$ as you have does not satisfy a quadratic equation over $\mathbb{Q}$. This could get pretty nasty. $\endgroup$ Jan 6, 2016 at 9:28
  • $\begingroup$ Your reasoning is not bad, however the question is about Galois theory. $\endgroup$
    – Piquito
    Jan 6, 2016 at 9:56
  • $\begingroup$ You're missing the subfield $\mathbb{Q}(i\sqrt{5})$. $\endgroup$
    – Matt B
    Jan 6, 2016 at 11:08
  • $\begingroup$ @PrahladVaidyanathan This exercise is indeed following Galois correspondence theorem, but I'm not sure how to apply. Could you give a hint? Thanks! $\endgroup$
    – nekodesu
    Jan 6, 2016 at 21:18

1 Answer 1


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.

  • $\begingroup$ Thank you! I think I understand the flow of the proof now. Two quick questions: how do we know the bijection is isomorphic to $C_2 \times C_2$? And how do we argue that there are only 5 subgroups? $\endgroup$
    – nekodesu
    Jan 6, 2016 at 21:22
  • $\begingroup$ I meant that the Galois group of $\mathbb{Q}(i,\sqrt{5})/\mathbb{Q}$ was isomorphic to $C_2 \times C_2$ which should be easy enough to prove. As for there being 5 subgroups of $C_2 \times C_2$, you can just count them. $\endgroup$
    – Matt B
    Jan 6, 2016 at 22:09
  • $\begingroup$ We clearly have the trivial subgroup and the whole thing so that's two. Any other subgroup must have order 2 by Lagrange's theorem so is cyclic and generated by an element of order 2, and there are 3 of those: $(1,0), (0,1)$ and $(1,1)$. $\endgroup$
    – Matt B
    Jan 6, 2016 at 22:11

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