One of my qualifying exam practice questions asks the degree of the field extension of the splitting field $K$ of $x^3-3$ over $\mathbb{Q}(\sqrt{-3})$. Since the roots of $x^3-3$ are all just the cube root of the times the cubic roots of unity, I think the extension is of degree three, since $\mathbb{Q}(\sqrt{-3})$ already contains the cubic roots of unity. But then when I start thinking of the size of the basis of $K$ over $\mathbb{Q}(\sqrt{-3})$, I start thinking of the square roots times the cube roots, and I get the degree of the field extension as much more than three, and now I'm all confused.
[Math] Splitting field of $x^3-3$ over $\mathbb{Q}(\sqrt{-3})$
abstract-algebra
Best Answer
Note that $\rho=(-1+\sqrt{3}i)/2$ is a generator of the cube roots of unity, and that $$x^3-3=(x-3^{1/3})(x-3^{1/3}\rho)(x-3^{1/3}\rho^2).$$ And $\rho$ lies in $Q(\sqrt{-3})$ So the coefficients of the factored form of $x^3-3$ all lie in the latter field with the real number $3^{1/3}$ adjoined to $Q(\sqrt{-3})$. So it looks like the degree is 3, considered over $Q(\sqrt{-3})$