# When is the extension $\Bbb{Q}(\sqrt{\sqrt{5} +a})/\Bbb{Q}$ Normal?

For what values of $a \in \Bbb{Q}$ is the extension $\Bbb{Q}(\sqrt{\sqrt{5} +a})/\Bbb{Q}$ Normal? I know that a finite extension $L/K$ such as this is normal iff it is the splitting field of some $f \in K[x]$.

My Attempt:
I think that this is the extension generated by polynomial $x^4-2ax^2 + (a^2-5)$. If we have a field $L$ s.t. $\sqrt{\sqrt{5} +a} \in L$ then $(\sqrt{\sqrt{5} +a})^2=\sqrt{5} +a \in L$.
So $\sqrt{5} \in L$ and so $\pm \sqrt{\pm \sqrt{5} +a} \in L$.
So $\Bbb{Q}(\sqrt{\sqrt{5} +a})$ is the splitting field for polynomial $x^4-2ax^2 + (a^2-5) \in K[x]$ so $\Bbb{Q}(\sqrt{\sqrt{5} +a})/\Bbb{Q}$ is always a normal exension. Is this correct??

• Where did the first "So" in the line before the last one came from? Why, or how, do you deduce from your work in the first lines that that is a normal extension? How do you think you've proved it contains all the roots of $\;x^4-2ax^2+(a^2-5)\;$ ?? – DonAntonio Jun 9 '17 at 22:06
• Take a look at the conditions given in this question math.stackexchange.com/q/649466/254075 – sharding4 Jun 9 '17 at 22:07
• @DonAntonio I think that the roots of the polynomial $x^4 - 2ax^2 +(a^2-5)$ are precisely $\pm \sqrt{\pm \sqrt{5} + a}$. So I've shown that if we have one of the roots, then it splits, so it splits in $\Bbb{Q}(\sqrt{\sqrt{5}+a})$ and if it splits in a field $L$ then $L$ must clearly contain $\Bbb{Q}(\sqrt{\sqrt{5}+a})$ and we have the splitting field – SEWillB Jun 9 '17 at 22:12
• @SEWillB First, I think you're right about the roots. Second, why didn't you stress this clearly in your post? I mean, that those four numbers are the roots of the quartic? Third, I think you're, finally, correct...but some order must be put in the way of writing things. Nice. +1 – DonAntonio Jun 9 '17 at 22:16
• I thought this made sense, but I don't actually think it's correct and I can't see why! – SEWillB Jun 9 '17 at 22:18

## 2 Answers

$\Bbb{Q}(\sqrt{\sqrt{5} +a})/\Bbb{Q}$ is a normal extension iff the extension is Galois iff the order of the Galois group is 4. That means the Galois group is either isomorphic to $C_2\times C_2$ or to $C_4$. Using well known criteria for Galois groups of quartic polynomials shows the first case occurs when $a^2-5$ is a square and the second case when $5(a^2-5)$ is a square.

$a=3$ is the first example of $C_4$ and $a=5$ the first example of $C_2\times C_2$. Those 2 values and $a=85$ are the only cases of $a<100$ with Galois group $C_2\times C_2$ or $C_4$.

• This means that my attempt is false, is there any chance you could point out my error? – SEWillB Jun 9 '17 at 22:35
• @SEWillB I think the problem is the conclusion $\sqrt{\pm \sqrt{5} +a} \in L$ You don't know that the square root is in $L$. – sharding4 Jun 9 '17 at 22:38
• I see it now! I can't take that square root! – SEWillB Jun 9 '17 at 22:39
• @SEWillB that's where the criterion $a^2-5$ is a square comes in handy! – sharding4 Jun 9 '17 at 22:40

The minimal polynomial can be computed with $b=\sqrt{\sqrt{5}+a}$ and $$b^2=\sqrt{5}+a$$ so $$b^4-2ab^2+a^2-5=0$$ The roots are indeed $\pm\sqrt{a\pm\sqrt{5}}$. Suppose $0<a<\sqrt{5}$: then two of the roots are complex not real. Therefore the minimal polynomial doesn't split in the given extension, which is a subfield of $\mathbb{R}$ because $\sqrt{5}+a>0$.