Splitting field with Galios group A4

[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
$$\alpha_1=\sqrt{2}(\sqrt{\cos{(\pi/9)}}+i\sqrt{\cos{(2\pi/9)}}+i\sqrt{\cos{(4\pi/9)}})$$

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

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

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

 

[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.)

 

[mathwonk]

hellooo?

 

[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.

 

[Hurkyl]

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.

Why not?

Spoiler

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

Spoiler

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

 

[Jim Kata]

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

 

[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 $$\mathbb{Z_3}$$

 

[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?