Find the Galois group of a polynomial over a field

Question

How to determine the structure of the Galois group of $(x^4+3)(x^3-2)$ over $\mathbb{Q}(\sqrt{-3})$? (This is not a homework problem.)

Answer 1

I would suggest factoring as $$(x^2-\sqrt{-3})(x^2+\sqrt{-3})(x^3-2).$$ Since each term in the factorization is irreducible it follows that if $F$ is the splitting field of our polynomial, and $F_1,F_2,F_3$ are the splitting fields of $(x^2-\sqrt{-3}),(x^2+\sqrt{-3}),(x^3-2),$ respectively, then $$\text{Gal}(F/\mathbb{Q}(\sqrt{-3}))=\text{Gal}(F_1/\mathbb{Q}(\sqrt{-3}))\times \text{Gal}(F_2/\mathbb{Q}(\sqrt{-3}))\times \text{Gal}(F_3/\mathbb{Q}(\sqrt{-3})).$$ As $F_1,F_2$ are degree $2$ extensions it is clear that their Galois Groups are isomorphic copies of $\mathbb{Z}_2.$ Notice that $F_3=\mathbb{Q}(\sqrt{3},\sqrt[3]{2}),$ a degree $3$ extension, hence the Galois group corresponding to $F_3$ is $\mathbb{Z}_3.$ This tells us that $$\text{Gal}(F/\mathbb{Q}(\sqrt{-3}))\cong\mathbb{Z}^2_2\times\mathbb{Z}_3.$$