Calculating the Galois group for a splitting field

1. Nov 18, 2016

PsychonautQQ

1. The problem statement, all variables and given/known data
Let f(x) = x^4 - 6x^2 - 2. Let K be a splitting field for this polynomial over Q, show that Gal(K:Q) is non-abelian of order 8.

2. Relevant equations

3. The attempt at a solution
So I calculated the roots of this polynomial, one root was r = (3+(11^1/2))^1/2, and the others were -r, ir and -ir where i=(-1)^1/2. Thus a Q(i,r) is a splitting field of f(x) over Q.

Now time to calculate the Galois group. The minimal polynomial of r over Q is f(x). I know this because it is a root of f(x) and f(x) is irreducible over Q by Eisenstein criteria with p=2. Thus there is an element of Gal(K:Q) that takes r to -r, and another that takes r to ir, and another that takes r to -ir, because it must take r to the other roots of it's minimal polynomial in Q.

The minimal polynomial of i over Q(r) is x^2+1, and so there is are elements of the Galois group that take i to i and i to -i.

We also know that [Q(r,i):Q] = [Q(r,i):Q(r)] * [Q(r):Q]. Since the minimal polynomial of i over Q(r) is x^2 + 1 we know that [Q(r,i):Q(r)] = 2 and thus [Q(r,i):Q] = 8.
Since Char(Q)=0 and Q(r,i) is a finite splitting field of this polynomial over Q, [Q(r,i):Q] is a galois extension and so Gal(Q(r,i):Q) = [Q(r,i):Q] = 8.

So any element in the Galois group can take r to one of four different places and i to 1 of 2 different places, and there is an element for each of these possibilities, and so these are the 8 elements in the Galois group. (is this logic shaky?)

Now, to show it's non abelian,. Let b the map that permutes the roots of the minimal polynomial of i over Q(r) and c be the map that takes r to ir, (which are both roots of its minimal polynomial over Q). Then bc(r) = b(ir) = -ir and cb(r) = c(r) = ir. Thus cb(r) does not equal bc(r) so this group is not abelian.

2. Nov 18, 2016

andrewkirk

That's not what I get.
I get $\pm r$ and $\pm s$ where $r=\sqrt{3+\sqrt{11}}$ and $s=i\sqrt{-3+\sqrt{11}}$. My root $r$ is the same as yours but the other root is not.

I suggest checking the working that led to the roots in the OP.

If my roots are correct, that will change the reasoning later in the OP. In particular, f(x) will be the minimal polynomial for both $r$ and $s$.

Last edited: Nov 18, 2016
3. Nov 19, 2016

PsychonautQQ

You are correct in that my roots were incorrect. So now the splitting field for f(x) will be Q(r,s) rather than Q(r,i), are we sure that the real part of s is not in Q(r)?

4. Nov 19, 2016

Staff: Mentor

I bet it is. Have you looked at $s$?

5. Nov 19, 2016

andrewkirk

The general element of $Q(r,s)$ will be
$$\sum_{j=0}^\infty \sum_{k=0}^\infty a_{jk}r^js^k$$
with $a_{jk}\in\mathbb Q$ and only finitely many of the $a_{jk}$ being nonzero.

I suspect, but have not proven, that the $r^js^k$ are all linearly independent, when $Q(r,s)$ is viewed as a vector space over $\mathbb Q$.
EDIT: No that's not right since it appears to be the case that $r^2+s^2=6$.

But the answer to your question is much simpler than any of that, as @fresh_42 has hinted. Perhaps you were meaning to ask whether $s/i$ is in $Q(r)$, to which I suspect the answer is No, and I suspect that neither is it in $Q(r,s)$.

Last edited: Nov 19, 2016
6. Nov 21, 2016

PsychonautQQ

if the real part of s is in Q(r) then wouldn't all the roots be contained in Q(r,i)? therefore Q(r,i) would be a splitting field, no? Which seems weird because Q(r,s) is also a splitting field, but I guess the minimal polynomial of s over Q(r) would have a degree of 2 and so it would make sense. Maybe Q(r,i) = Q(r,s)?

7. Nov 21, 2016

Staff: Mentor

$s=i\sqrt{-3+\sqrt{11}} \approx 0.56\, i$ and the real part of it is zero which is definitely in $\mathbb{Q}(r)$.
@andrewkirk 's hint $r^2+s^2=6$ looks like a good point to start at, because $r$ and $s$ have to be part of $K$ somehow.

8. Nov 22, 2016

PsychonautQQ

Ah,, so s is not in Q(r,I) then. So then the splitting field is actually Q(r,s) then, where s has degree 2 over Q(r) and r has degree 4 over Q, correct?

9. Nov 22, 2016

Staff: Mentor

Until now we have $x^4-6x^2-2=(x+r)(x-r)(x+s)(x-s)$ with $r:=\sqrt{3+\sqrt{11}}\, , \,s=i\sqrt{-3+\sqrt{11}}$. Why should there be another degree than four? But we don't have to bother. You already know what automorphisms $\sigma$, we're looking for, do on these roots $\{\pm r\, , \,\pm s\}$. This means we can write them down by their values $\sigma(r)\, , \,\sigma(s)$ and built the multiplication table of the Galois group to see which one it is. (I have a suspicion but didn't calculate it.)

10. Nov 22, 2016

PsychonautQQ

I was thinking that the minimal polynomial of s in Q(r) would be less than degree four because you could factor out (x+r)(x-r) from x^4-6x^2-2 and thus the minimal polynomial of s in Q(r) would be of degree 2. (x+s)(x-s) = x^2-3+11^(1/2) which is in Q(r). Is this thinking not correct?

If is it, then the galois group would have elements that take r to itself, -r, s, and -s, whilst there is another element that takes s to -s.

11. Nov 22, 2016

Staff: Mentor

Yes, that's true. Either way $\mathbb{Q}\subseteq \mathbb{Q}(r) \subseteq \mathbb{Q}(r,s)$ or $\mathbb{Q}\subseteq \mathbb{Q}(s) \subseteq \mathbb{Q}(r,s)$ is a $4$&$2$ step. And you will probably find two non-commutative automorphisms by a smart guess.
But I think the entire Galois group is interesting, too. And coming so far, it would be a pity not to determine it.

12. Nov 23, 2016

PsychonautQQ

I must now find all subgroups of this Galois group and thus all subfields of K. To do this i'll first find the Galois group. The Galois group must have elements that take r to the other roots of it's minimal polynomial in Q, thus there is an automorphism p such that p(r) = r,-r,s, or -s and fixes everything else. I'm confused though because I do not understand the order of this element, say p(r) = s. Then p(p(r)) = p(s) = ??. I want to say that p fixes and s and thus p(p(r)) = p(r) = s but this is rubbish as this would not be an element of a well defined group. is there a link to how p acts on s that I am not seeing? I was thinking maybe s = -ir but then I realized that is not true.

So basically I'm thinking this galois group is going to be some sort of semi direct product, either (Z_2 x Z_2) * Z_2 where * is a SDP or Z_4 * Z_2, and I think the key to figuring out which of these it is is understand the elements of Gal(Q(r):Q) which I'm having trouble with.

13. Nov 23, 2016

Staff: Mentor

I wouldn't bother the subgroups at this stage. Simply look at the automorphisms:
$\sigma_0(r)=r\, , \,\sigma_0(s)=s$
$\sigma_1(r)=r\, , \,\sigma_1(s)=-s$
... etc...
and then $- \sigma_i$.

The (normal) subgroups should become visible with the multiplication table. With the setting $\pm \sigma_i$ you only need to name four automorphisms, which makes it easier. (There are only two non Abelian groups with eight elements, since direct products of cyclic groups aren't and there are only subgroups of orders $2^n$.)

14. Nov 23, 2016

PsychonautQQ

I suppose I am confused because I feel like there should be an element that takes r-->s and I don't understand the order of this element. I believe this element of the Galois group should exist because the minimal polynomial of r over Q has four roots, one of which is s, so shouldn't there be an automorphism in this Galois group that takes r to s?

15. Nov 23, 2016

Staff: Mentor

I suppose this, too. But if we take $r \mapsto s$, then what is the image of $s$? To get a bijection only $\pm r$ is available. (This gives the $\sigma_i$ I didn't mention. $\pm 1$ can't be used as images, because all elements of $\mathbb{Q}$ have to stay fixed.)
And the order is either $2$, if you simply change $r$ and $s$, and $4$, if you one is negative, because in this case, we need another two applications to eliminate the sign.

16. Nov 23, 2016

PsychonautQQ

Exactly, I'm confused by the fact that there must be an automorphism r-->s and then wondering what the image of s would be. I don't think you listed an automorphism that takes r-->s in the nice list you made me.

17. Nov 23, 2016

PsychonautQQ

Can you rephrase the second thing you said? specifically "if you one is negative", If this is a typo I bet any competent person could figure out what you meant from the context but I am the PsychonautQQ so ahem yeah

edit: woot 600 posts, 100:1 post to like ratio, I must be really contributing high level thoughts!

18. Nov 23, 2016

Staff: Mentor

Well, if we define $\sigma_0=id$ and $\sigma_1$ by $\sigma_1(r)=r$ and $\sigma_1(s)=-s$ then $\sigma_1^2 = \sigma_0$.
If we define $\sigma_2$ by $\sigma_2(r)=s$ and $\sigma_2(s)=r$ then, too, $\sigma_2^2 = \sigma_0$.
But if we define $\sigma_3$ by $\sigma_3(r)=s$ and $\sigma_3(s)=-r$ then $\sigma_3^2=- \sigma_0$ and $\sigma_3^4=\sigma_0$.
Not sure about this, but you certainly make me repeat some definitions and theorems I've long forgotten. Eisenstein has been such an example

Last edited: Nov 23, 2016
19. Nov 23, 2016

PsychonautQQ

Wow i am a fool, I was wondering if r-->s then what is the image of s, and of course the answer was I have yet to define it, but I need to define it in such a way that the element is well defined in the group. Thank you for your patience.

Now I need to find ALL the subgroups and thus ALL the subfields.

So lets say p_0(r)=-r and fixes s, p_1(s)=-s and r fixes r, while p_2(r)=s and p_2(s)=r while p_3(r)=s and p_3(s)=-r. So the galois group is generated by these four elements but with some overlap on p_2 and p_3 because they both act on r the same. But we will have a subgroup generated be each element, so there will be subgroups <p_0>,<p_1>,<p_2>,<p_3>, where the first three have order 2 and the last one has order 4. Then We could have a subgroup generated by basically any combination of these generates, right?

20. Nov 23, 2016

Staff: Mentor

I would drop $p_0$ because $p_0=-p_1$ and define $p_0 = -id = -1$ instead.
And you are right, $p_1^2=p_2^2=1$ while $p_3^4=1$ but $p_3^2=-1$.
Then the Galois group becomes $G = \{\pm p_0,\pm p_1,\pm p_2,\pm p_3\}$ where the first six are reflections and the fourth is a rotation by $\pi /2$. Now can you generate $G$ by just a reflection and a rotation, say $p_1$ and $p_3$?

Not sure whether any combination gives you different subgroups, probably not. But yes, you have some cyclic subgroups:
$\langle p_0 \rangle = \langle p_1 \rangle = \langle p_2 \rangle = \mathbb{Z}_2$
$\langle p_3 \rangle = \mathbb{Z}_4$.