Given $K$ the splitting field of $x^4 - x^2 - 1$ over $\mathbb{Q}$, we want to determine all intermediate fields $L$, $\mathbb{Q} \subset L \subset K$ and determine which are Galois.

I've shown that this polynomial is irreducible over $\mathbb{Q}$ and that $K$ is $\mathbb{Q}\left(\sqrt{\frac{1 + \sqrt{5}}{2}}, \sqrt{\frac{\sqrt{5}-1} {2}}i\right)$. I also know that the Galois group of $f$ is isomorphic to $D_4$, so all of the intermediate fields have to be isomorphic to subgroups of $D_4$. Consequently the Galois extensions will be the fields $L$ that correspond to normal subgroups of $D_4$. My issue is that I'm struggling to determine the intermediate fields in a sensible manner. Any help would be much appreciated.

  • $\begingroup$ I am not sure that I agree with your $K$; I am pretty sure this is a real extension? Set $y = x^2$ then you get $y^2 - y - 1 = 0$ giving $y = \frac{1\pm\sqrt{5}}{2}$ which will then, upon taking the square root will give the roots... there is no $i$ $\endgroup$ Commented Apr 12, 2016 at 20:42
  • $\begingroup$ I know that $\mathbb{Q}(\sqrt{\frac{1+\sqrt{5}}{2}})$ and $\mathbb{Q}(\sqrt{\frac{\sqrt{5}-1}{2}i})$ both are, as is $\mathbb{Q}(\sqrt{5})$. But there are 8 intermediate fields and this is only 3. $\endgroup$ Commented Apr 12, 2016 at 20:45
  • $\begingroup$ the extension should be $K = \mathbb{Q}\left(\sqrt{\frac{1+\sqrt{5}}{2}}\right)$. Can you see what the intermediate fields are here? Also, why are you saying that the Galois group of each intermediate field will necessarily be $D_4$? The intermediate fields will have smaller degree and therefore a smaller Galois group. $\endgroup$ Commented Apr 12, 2016 at 20:48
  • $\begingroup$ When you square root your $y$ though to get the values of $x$, you'll introduce an $i$ since $\sqrt{5} > 1$. $\endgroup$ Commented Apr 12, 2016 at 20:49
  • $\begingroup$ Ah, right sorry... when you take the negative root you get an $i$. That's my fault. What intermediate fields have you determined? $\endgroup$ Commented Apr 12, 2016 at 20:51

1 Answer 1


Let $\alpha=\sqrt{(1+\sqrt{5})/2}$ and $\beta=\sqrt{(-1+\sqrt{5})/2}i$. So the roots of $f(x)=x^4-x^2-1$ are $\pm \alpha$ and $\pm \beta$.

You might want to start a building project. Build a few towers: $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{5}) \subseteq \mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\alpha,i)$. You also have a tower $\mathbb{Q} \subseteq \mathbb{Q}(i) \subseteq \mathbb{Q}(i,\sqrt{5}) \subseteq \mathbb{Q}(\alpha,i)$. You can also build a tower by first adjoining $\sqrt{5}i$.

Recall that if $D_4 = \{1,r,r^2,r^3,s,rs,r^2s,r^3s\}$, then the subgroups are $D_4$, $\{1,r,r^2,r^3\}$, $\{1,r^2,s,r^2s\}$, $\{1,r^2,rs,r^3s\}$, $\{1,r^2\}$, $\{1,s\}$, $\{1,rs\}$, $\{1,r^2s\}$, $\{1,r^3s\}$, and $\{1\}$.

Notice that you can identify several of these already ($\mathbb{Q}(i,\sqrt{5})$ has the Klein 4-group as a Galois group. The degree of $\mathbb{Q}(\alpha,i)$ over $\mathbb{Q}(i,\sqrt{5})$ is 2 so this subfield is fixed by one of the subgroups of order 2. It's a splitting field of $(x^2+1)(x^2-5)$ so it is Galois itself, thus it's fixed by $\{1,r^2\}$ (the only normal subgroup of order 2).

You should be able to fill in most everything else by using the lattice correspondence and intersecting/joining subfields.

I hope this helps you get started. :)

EDIT: Now that you've had time to fight with your example for a while. I should note that this exact splitting field is dealt with in Keith Conrad's blurb/handout Galois correspondence examples. Specifically it's pages 7-10 (example #5).

  • $\begingroup$ Yes. I'm upside down!! Thanks @JyrkiLahtonen ! $\endgroup$
    – Bill Cook
    Commented Apr 13, 2016 at 20:22

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