How do I find this polynomials splitting field over $\mathbb{Q}$?

Question

I’m trying to find the splitting field of the polynomial p(x) = $x^8 + 11x^4 + 24$. Here is my attempt: The obvious solutions to the polynomial are $x = \pm \sqrt[4]{-3}$, and $x = \pm \sqrt[4]{-8}$. Letting $\zeta_{4}^k = e^{\frac{2k\pi i}{4}}$, we have $\zeta_{4}(\pm\sqrt[4]{-3}) = \pm i\sqrt[4]{-3}$, and $\zeta_{4}(\pm\sqrt[4]{-8}) = \pm i\sqrt[4]{-8}$. So the splitting field for the individual factors are $\mathbb{Q}(\sqrt[4]{-3}, i)$, and $\mathbb{Q}(\sqrt[4]{-8}, i)$. Does this mean the splitting field of $x^8 + 11x^4 + 24$ is $\mathbb{Q}(\sqrt[4]{-3},\sqrt[4]{-8}, i)$? If not, why so?

Answer 1

As you seem to have noted, the polynomial factors as $(x^4 + 3)(x^4 + 8)$. However, note that the factors are more than the ones you've written. (Although there's some ambiguity in writing $\sqrt[4]{-3}$ to begin with. Presumably, you're fixing one complex fourth root of $-3$. Then the roots of the first factor will be $\sqrt[4]{-3}, \iota \sqrt[4]{-3}, -\sqrt[4]{-3}, \iota \sqrt[4]{-3}$, where $\iota^2 = -1$.)

Similarly, one can write the roots for the other polynomial. Note that the splitting field is then $$\Bbb Q(\sqrt[4]{-3}, \iota \sqrt[4]{-3}, \sqrt[4]{-8}, \iota \sqrt[4]{-8}).$$ As you've (implicitly) noted, we have $\Bbb Q(\sqrt[4]{-3}, \iota \sqrt[4]{-3}) = \Bbb Q(\sqrt[4]{-3}, \iota)$ and similarly for the other roots. Thus, we finally get the extension as $$\Bbb Q(\sqrt[4]{-3}, \iota , \sqrt[4]{-8}),$$ as written.

---

As a personal preference, I always avoid writing $\sqrt[4]{-3}$ due to ambiguity. Instead, I would say at the beginning that I'm fixing a root $\alpha$ at the beginning satisfying $\alpha^4 = -3$.
If you wish to be explicit, you can write $$\alpha = \sqrt[4]{3} \exp\left(\frac{\iota \pi}{4}\right).$$ The above is okay to write since we do have a canonical choice of $\sqrt[4]{3} \in \Bbb R$.