Thread: Determining a Galois Group

1. Determining a Galois Group

I am trying to compute the Galois group for the polynomial $x^4-6x^2+7$, but I'm having a bit of trouble. I know that $x^4-6x^2+7=(x^2-(3-\sqrt{2}))(x^2-(3+\sqrt{2}))$ over $\mathbb{Q}(\sqrt{2})$. The polynomial will then split over $\mathbb{Q}(\sqrt{2},\sqrt{3+\sqrt{2}},\sqrt{3-\sqrt{2}})$, which is an extension of degree 8.

Now I have the automorphisms $\tau_1: \sqrt{2}\mapsto -\sqrt{2}$, $\tau_2: \sqrt{3+\sqrt{2}}\mapsto -\sqrt{3+\sqrt{2}}$ and $\tau_3: \sqrt{3-\sqrt{2}}\mapsto -\sqrt{3-\sqrt{2}}$. I know that $\tau_1^2=\tau_2^2=\tau_3^2=e$. But I can't seem to get relations between them (and in particular, I would like to find an element of order 4 to show that this is the dihedral group). For example, I am interested in powers of $\tau_1 \tau_2$ but I'm not sure how to run the computation.

Any suggestions would be much appreciated.

2. Originally Posted by roninpro
I am trying to compute the Galois group for the polynomial $x^4-6x^2+7$, but I'm having a bit of trouble. I know that $x^4-6x^2+7=(x^2-(3-\sqrt{2}))(x^2-(3+\sqrt{2}))$ over $\mathbb{Q}(\sqrt{2})$. The polynomial will then split over $\mathbb{Q}(\sqrt{2},\sqrt{3+\sqrt{2}},\sqrt{3-\sqrt{2}})$, which is an extension of degree 8.

Now I have the automorphisms $\tau_1: \sqrt{2}\mapsto -\sqrt{2}$, $\tau_2: \sqrt{3+\sqrt{2}}\mapsto -\sqrt{3+\sqrt{2}}$ and $\tau_3: \sqrt{3-\sqrt{2}}\mapsto -\sqrt{3-\sqrt{2}}$. I know that $\tau_1^2=\tau_2^2=\tau_3^2=e$. But I can't seem to get relations between them (and in particular, I would like to find an element of order 4 to show that this is the dihedral group). For example, I am interested in powers of $\tau_1 \tau_2$ but I'm not sure how to run the computation.

Any suggestions would be much appreciated.
let $a = \sqrt{3+\sqrt{2}}$ and $b = \sqrt{3-\sqrt{2}}.$ note that $\sqrt{2} \in \mathbb{Q}(a,b).$ so the splitting field is just $\mathbb{Q}(a,b).$
as you said, the Galois group is $D_8,$ the dihedra group of order 8. just define $\sigma(a)=b, \ \sigma(b)=-a$ and $\tau(a)=a, \ \tau(b)=-b.$ then $o(\sigma)=4, \ o(\tau)=2, \ o(\tau \sigma)=2.$ thus the Galois group is $D_8.$

3. Originally Posted by roninpro
I am trying to compute the Galois group for the polynomial $x^4-6x^2+7$, but I'm having a bit of trouble. I know that $x^4-6x^2+7=(x^2-(3-\sqrt{2}))(x^2-(3+\sqrt{2}))$ over $\mathbb{Q}(\sqrt{2})$. The polynomial will then split over $\mathbb{Q}(\sqrt{2},\sqrt{3+\sqrt{2}},\sqrt{3-\sqrt{2}})$, which is an extension of degree 8.

Now I have the automorphisms $\tau_1: \sqrt{2}\mapsto -\sqrt{2}$, $\tau_2: \sqrt{3+\sqrt{2}}\mapsto -\sqrt{3+\sqrt{2}}$ and $\tau_3: \sqrt{3-\sqrt{2}}\mapsto -\sqrt{3-\sqrt{2}}$. I know that $\tau_1^2=\tau_2^2=\tau_3^2=e$. But I can't seem to get relations between them (and in particular, I would like to find an element of order 4 to show that this is the dihedral group). For example, I am interested in powers of $\tau_1 \tau_2$ but I'm not sure how to run the computation.

Any suggestions would be much appreciated.

I'm interested in knowing how did you come up with those three automorphisms, as you didn't define them

over a basis of the splitting field over the rationals...

I mean: (1) How can you actually know the three automorphisms (i.e., how can you tell they're even different, to

begin with)?, and (2) How can you possibly find out any relation between them if you don't know the maps?

Tonio

4. Originally Posted by tonio
I'm interested in knowing how did you come up with those three automorphisms, as you didn't define them

over a basis of the splitting field over the rationals...

I mean: (1) How can you actually know the three automorphisms (i.e., how can you tell they're even different, to

begin with)?, and (2) How can you possibly find out any relation between them if you don't know the maps?

Tonio
I guess I didn't write down my maps completely. For $\tau_1$, I meant that $\sqrt{2}\mapsto \sqrt{2}$, and everything else is fixed (i.e. $\sqrt{3+\sqrt{2}}\mapsto \sqrt{3+\sqrt{2}}$, and so on). For $\tau_2$, I meant that $\sqrt{3+\sqrt{2}}\mapsto -\sqrt{3+\sqrt{2}}$ and fixes everything else, etc... So the fixed field for $\tau_2, \tau_3$ would be $\mathbb{Q}(\sqrt{2})$. It then seemed reasonable to me that something like $\tau_1\tau_2$ would be an element of order 4, but I didn't know how to compute the powers.