Splitting field with Galios group A4

1. Nov 7, 2012

Jim Kata

the polynomial x^4+8x+12=0 has the Galois group A4. I have all its roots, but can't figure out its splitting field. The roots are
$$\alpha_1=\sqrt{2}(\sqrt{\cos{(\pi/9)}}+i\sqrt{\cos{(2\pi/9)}}+i\sqrt{\cos{(4\pi/9)}})$$

$$\alpha_2=\sqrt{2}(\sqrt{\cos{(\pi/9)}} - i\sqrt{\cos{(2\pi/9)}}-i\sqrt{\cos{(4\pi/9)}})$$

$$\alpha_3=\sqrt{2}(-\sqrt{\cos{(\pi/9)}} + i\sqrt{\cos{(2\pi/9)}}-i\sqrt{\cos{(4\pi/9)}})$$

$$\alpha_4=\sqrt{2}(-\sqrt{\cos{(\pi/9)}} - i\sqrt{\cos{(2\pi/9)}}+i\sqrt{\cos{(4\pi/9)}})$$

Last edited: Nov 7, 2012
2. Nov 8, 2012

mathwonk

what do you mean? its the field generated by those roots. do you want a smaller set of generators? (presumably two of those would do.)

3. Nov 10, 2012

mathwonk

hellooo?

4. Nov 12, 2012

Jim Kata

Sorry I didn't respond earlier, but I don't think it is as simple as just attaching two roots since the degree of the field in that case wouldn't be twelve.

5. Nov 12, 2012

Hurkyl

Staff Emeritus
Why not?

What can you say about, e.g., the minimal polynomial of a2 in the field Q(a1)?

It can't be x4 + 8x + 12, because this polynomial is reducible over Q(a1)! (it has a root)

Last edited: Nov 12, 2012
6. Nov 12, 2012

Jim Kata

I think I get what you're saying let me mess with this idea for a bit and get back to you.

7. Nov 12, 2012

Jim Kata

Here is where I am at.

$$x^4+8x+12=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4)=(x^3+\alpha_1x^2+{\alpha_1}^2x+({\alpha_1}^3+8))(x-\alpha_1)=0$$

So I define $$F \equiv \mathbb{Q}(\alpha_1)$$ So it is obvious that the coefficients of that cubic are in the field $$F \equiv \mathbb{Q}(\alpha_1)$$ and that the roots of that cubic are $$\alpha_2, \alpha_3, \alpha_4$$ but it is not obvious to me that $$\alpha_3$$ can be written in terms of $$\alpha_2$$ and the field $$\mathbb{Q}(\alpha_1)$$ making the the Galois group of the cubic $$\mathbb{Z_3}$$

8. Nov 12, 2012

Jim Kata

I guess I just have to assume that's the case since if it wasn't the Galois group would be S4 not A4. In that case I get a splitting field:
$$\mathbb{Q}(\alpha_1,\alpha_2)= A +B\alpha_1 + C{\alpha_1}^2 + D{\alpha_1}^3 + E\alpha_2 + F\alpha_1\alpha_2 +G{\alpha_1}^2\alpha_2 +H{\alpha_1}^3\alpha_2 + I{\alpha_2}^2 + J\alpha_1{\alpha_2}^2 + K{\alpha_1}^2{\alpha_2}^2 +L{\alpha_1}^3{\alpha_2}^2$$ Is this correct?