3

Let $K$ be the splitting field the polynomial $x^{4}+3$ over $\mathbb{Q}$.Find the Galois group of $K$ over $\mathbb{Q}$?

I think $[K:\mathbb{Q}]=8$ but how can we find the group? Any help would be great.

delueze
  • 545
  • 1
    Please tell us if you understand any similar problem, e.g., have you learned how to calculate the Galois group of $x^4-2$ over $\mathbf Q$? – KCd Aug 18 '15 at 01:57
  • @KCd: Yes, the Galois group of $x^4-2$ is $D_8$, but in my example I didn't get anything from my computations. Thanks for help. – delueze Aug 18 '15 at 02:03
  • 1
    But is there truly a substantial difference in the ideas for the example you know and the one you ask about? – KCd Aug 18 '15 at 02:11

1 Answers1

3

The polynomial is irreducible (because 3 is not a square) and separable (it has no repeated roots). Since splitting fields are always normal $K/\mathbb Q$ is Galois.

If $\alpha$ is a root of the equation (take some $\sqrt[4]{-3}$) then all roots are: $\alpha,\zeta_4 \alpha,\zeta_4^2 \alpha,\zeta_4^3 \alpha$ where $\zeta_4$ is a fourth root of unity and the splitting field is $\mathbb Q(\zeta_4,\alpha)$ since you can make any root out of those two numbers and those two numbers out of the roots (e.g. by dividing $\zeta_4 \alpha/\alpha$).

Clearly one automorphism is cycling the roots of unity.

Since $\zeta_4^2 = -1$, the roots can also be written $\alpha,\zeta_4 \alpha,-\alpha,-\zeta_4 \alpha$.

Another automorphism is $\alpha \mapsto -\alpha$.

Together this generates $D_8$ and $|D_8| = 8 = |K:\mathbb Q|$ so this is the Galois group. Note that this argument applies to every polynomial $x^4 + s$ with nonsquare $s > 0$. A different result may be obtained for negative squares.

Brennan.Tobias
  • 929