I just need a detailed explanation of how to go about finding the intermediate fields and galois group of $x^4-x^2-6$. This is not a homework question, I am just confused on how to go about computing these two things. Also, is there a general method of approach when trying to find intermediate fields? Thank you.
Asked
Active
Viewed 136 times
1
-
2$x^4-x^2-6=(x^2-3)(x^2+2)$, so the splitting field (over ${\mathbb Q}$) is $K={\mathbb Q}(\sqrt{3},\sqrt{-2})$. Iterated quadratic extensions of this kind are relatively easy. The Galois group is $C_2 \times C_2$ with two generators of order two interchanging $\pm\sqrt{3}$ and $\pm\sqrt{-2}$, respectively. The three proper subgroups of order $2$ each fix one of the three quadratic subextensions. – Derek Holt May 01 '14 at 12:59
1 Answers
1
Write $f(x)=x^4-x^2-6=(x^2-3)(x^2+2)$ to see that $K=\mathbb Q(\sqrt{3},i\sqrt{2})$ is the splitting field of $f$ over $\mathbb Q$. Then (considering permutations of the roots), $$G(K/\mathbb Q)=\{1,\sigma,\tau,\sigma\tau\}\cong \mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$$ where $\sigma$ is the $\mathbb Q(i\sqrt 2)$-automorphism given by $\sqrt 3\mapsto -\sqrt 3$, and $\tau$ is the $\mathbb Q(\sqrt 3)$-automorphism given by $i\sqrt 2\mapsto -i\sqrt 2$. So we can see that $\mathbb Q(\sqrt 3)$ and $\mathbb Q(i\sqrt 2)$ are two of the intermediate fields. What is the remaining one?