Calculating the Galois group for a splitting field

[PsychonautQQ]

Homework Statement

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.

Homework Equations

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.

 

[andrewkirk]

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.

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 $s$.

 

[PsychonautQQ]

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 $s$.

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)?

 

[fresh_42]

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

 

[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 $Q(r,s)$.

 

[PsychonautQQ]

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

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)?

 

[fresh_42]

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)?

$s=i\sqrt{-3+\sqrt{11}} \approx 0.56\, i$ 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.

 

[PsychonautQQ]

$s=i\sqrt{-3+\sqrt{11}} \approx 0.56\, i$ 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.

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?

 

[fresh_42]

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?

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.)

 

[PsychonautQQ]

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.)

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.

 

[fresh_42]

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.

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.

 

[PsychonautQQ]

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.

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.

 

[fresh_42]

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 $2^n$.)

 

[PsychonautQQ]

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 $2^n$.)

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?

 

[fresh_42]

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 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.)

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?

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.

 

[PsychonautQQ]

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.

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.

 

[PsychonautQQ]

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.

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!

 

[fresh_42]

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

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 $\sigma_3^4=\sigma_0$.

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

Not sure about this, but you certainly make me repeat some definitions and theorems I've long forgotten. Eisenstein has been such an example

 

[PsychonautQQ]

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 repetition some definitions and theorems I've long forgotten. Eisenstein has been such an example

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 let's 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?

 

[fresh_42]

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:
$\langle p_0 \rangle = \langle p_1 \rangle = \langle p_2 \rangle = \mathbb{Z}_2$
$\langle p_3 \rangle = \mathbb{Z}_4$.

 

[PsychonautQQ]

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:
$\langle p_0 \rangle = \langle p_1 \rangle = \langle p_2 \rangle = \mathbb{Z}_2$
$\langle p_3 \rangle = \mathbb{Z}_4$.

$p_0=-Id$ so it would switch the sign of both r and s, and then $p_1$ switches just the sign of s, $p_2$ switches r to s and s back to r and $p_3$ switches r to s and s to -r. So $p_3(r)=p_2(r)$ but the point is for a generic element in Q(r,s) then this equality would not hold, a generic element taking the form of a+br+cs where a,b,c are rational numbers. Is $p_3$ necessary? $p_3(a+br+cs)=a+bs-cr$ is this the same as $p_2(p_1(a+br+cs))=p_2(a+br-cs)=a+bs-cr$?

P.S. I hope my first attempt at LaTeX coding works!

 

[fresh_42]

$p_0=-Id$ so it would switch the sign of both r and s, and then $p_1$ switches just the sign of s, $p_2$ switches r to s and s back to r and $p_3$ switches r to s and s to -r. So $p_3(r)=p_2(r)$ but the point is for a generic element in Q(r,s) then this equality would not hold, a generic element taking the form of a+br+cs where a,b,c are rational numbers. Is $p_3$ necessary? $p_3(a+br+cs)=a+bs-cr$ is this the same as $p_2(p_1(a+br+cs))=p_2(a+br-cs)=a+bs-cr$?

P.S. I hope my first attempt at LaTeX coding works!

Yes, reasoning and LaTeX (except Q which would have been \mathbb{Q}. I extra installed a small program to set some keys with these repeating elements of LaTeX. Makes typing a lot easier. E.g. I set "\mathbb{}" = Alt+b).

Whether $p_3$ is necessary? No, but two of them are. Which ones is a matter of taste. But since the group is $D_4$ I thought a rotation and a reflection would make sense to fit into the usual group representation. I haven't thought about the primitive element, but as $p_3$ is of order $4$, it looks like a candidate to start with.

Edit: $G = \langle p_1,p_2 \,\vert \, p_1^2=p_2^2=(p_1p_2)^4=1 \rangle$ is o.k., too.

 

[PsychonautQQ]

Yes, reasoning and LaTeX (except Q which would have been \mathbb{Q}. I extra installed a small program to set some keys with these repeating elements of LaTeX. Makes typing a lot easier. E.g. I set "\mathbb{}" = Alt+b).

Whether $p_3$ is necessary? No, but two of them are. Which ones is a matter of taste. But since the group is $D_4$ I thought a rotation and a reflection would make sense to fit into the usual group representation. I haven't thought about the primitive element, but as $p_3$ is of order $4$, it looks like a candidate to start with.

Edit: $G = \langle p_1,p_2 \,\vert \, p_1^2=p_2^2=(p_1p_2)^4=1 \rangle$ is o.k., too.

Wow, I didn't realize this, your saying $p_1*p_2=p3$? And the group is $D_4$ is easy to find the subgroups for. Thanks a ton man you the bomb!

 

[fresh_42]

your saying $p_1 \cdot p_2=p_3$?

No.

$p_3(a+br+cs)=a+bs-cr$ is this the same as $p_2(p_1(a+br+cs))=p_2(a+br−cs)=a+bs-cr$?

You did it. And it is correct. O.k. the other way around $p_3=p_2\cdot p_1$ but this a minor issue (and a matter of how to define the multiplication order of automorphisms), although it's the desired example for non-commutativity.

 

[andrewkirk]

I don't get that $p_0$ as defined in #19 is $p_1$, since $p_0(s)=s\neq -s=-p_1(s)$. However I think $p_0=p_2p_1p_2$ under the definitions of #19, so it is correct that $p_0$ is redundant.

Also, we wouldn't want to redefine $p_0$ as $-id$ because that is not an automorphism, as then we'd have $p_0(1)\times p_0(1)=(-1)\times(-1)=1\neq -1=p_0(1)=p_0(1\times 1)$.

Instead let's just make $p_0$ the identity automorphism, which needs to be included in our group. I think the three non-identity generators $p_1,p_2,p_3$ are enough to generate our group.

I find diagrams very helpful with this sort of thing. I'll see if I can dig up an old latex file on homology chain diagrams and use the codes to make a diagram of this.

 

[fresh_42]

I don't get that $p_0$ as defined in #19 is $p_1$, since $p_0(s)=s\neq -s=-p_1(s)$. However I think $p_0=p_2p_1p_2$ under the definitions of #19, so it is correct that $p_0$ is redundant.

$p_0=-p_1$

Also, we wouldn't want to redefine $p_0$ as $-id$ because that is not an automorphism, as then we'd have $p_0(1)\times p_0(1)=(-1)\times(-1)=1\neq -1=p_0(1)=p_0(1\times 1)$.

I see what you mean, but $p_3^2(r,s)=(-r,-s)$ and we need eight elements. So you are right, $-id$ is wrong here. It's only $-1$ on $r$ and $s$. $\mathbb{Q}$ is fixed. Thanks for correction.

I find diagrams very helpful with this sort of thing. I'll see if I can dig up an old latex file on homology chain diagrams and use the codes to make a diagram of this.

Here is a cycle graph (found on Wiki)

https://en.wikipedia.org/wiki/List_of_small_groups#List_of_small_non-abelian_groups

 

[andrewkirk]

$p_0=-p_1$

Quite right. I mucked up that comment. Please disregard it.

The latex code for the diagram didn't work here because it needed the add-in package xypic, which is not available here. So here's one from LibreOffice draw instead. It's easy to see in this diagram how $p_1,-p_1$ and $p_2$ are reflections while $p_3$ is a rotation.

Damn this inserting a picture is fiddly. It keeps on disappearing! [note to self for future reference: drag and drop worked (sometimes), but only with .jpg, not with .bmp, .png or .gif]

I gave up in the end and used imgur. Here it is:http://imgur.com/KyjdxWJ

I'll have to ask Math Amateur for tips on how to get non-web images onto here. He seems to manage it reliably with his textbook images.

http://imgur.com/KyjdxWJ http://imgur.com/a/xzvXm

 

[PsychonautQQ]

Sorry for bumping this post, but I'm having trouble figuring out how to show that s is not an element of Q(r+s). Anyone have advice?

 

[andrewkirk]

I'm having trouble figuring out how to show that s is not an element of Q(r+s)

I can't see how that relates to the problem at hand, which is to show that the Galois group is non-commutative and of order 8.

Reviewing the posts so far, it looks like six distinct elements have been found which, using $e$ for the identity, and $p_1,p_2,p_3$ as defined in post 20 and depicted in the diagram of post 27, are:

1. $e:r\mapsto r,s\mapsto s$
2. $p_1:r\mapsto r,s\mapsto -s$
3. $p_2:r\mapsto s,s\mapsto r$
4. $p_3:r\mapsto s,s\mapsto -r$
5. $p_3*p_2:r\mapsto -r,s\mapsto s$ (using the convention that $p_3*p_2(x)\triangleq p_3(p_2(x))$, ie the second operand is applied to the field element $x$ first)
   
6. $p_3{}^2(=-e):r\mapsto -r,s\mapsto -s$

We need to (a) identify two more distinct elements, (b) show that there are no further elements, and (c) show that the group is non-commutative.

Have you made any progress on any of these? What avenues have you investigated?