Find the splitting field of $x^3 + 4$ over $\mathbb{Q}$

Question

I need to find the splitting field L of $g(x) = x^3 + 4$ over $\mathbb{Q}$. How do I do it?

I also need to determine $[L : Q]$.

Answer 1

Let $\alpha = -\sqrt[3]{4}$, and let $\omega = e^{\frac{2\pi i}{3}}$ be a primitive cube root of unity. Clearly $\alpha^3 = -4$, so $\alpha$ is a root of $f(x) = x^3 + 4$. Since $1, \omega$ and $\omega^2$ are all cube roots of $1$, we have that $\alpha, \omega\alpha, \omega^2\alpha$ are all roots of $f(x)$, and so $f(x) = (x-\alpha)(x-\omega\alpha)(x-\omega^2\alpha)$.

Let $L$ be the splitting field of $f$ over $\mathbb{Q}$. Then $L$ must contain all roots of $f$, so $\alpha$ and $\alpha\omega$ are in $L$. It follows that $\omega = \alpha\omega/\alpha$ is also in $L$. Thus, $\mathbb{Q}(\alpha, \omega) \subseteq L$. But in fact, all the roots of $f$ are in $\mathbb{Q}(\alpha, \omega)$, so $f$ splits in $\mathbb{Q}(\alpha, \omega)$, so $L= \mathbb{Q}(\alpha, \omega)$. So the splitting field of $f(x)$ over $\mathbb{Q}$ is $\mathbb{Q}(-\sqrt[3]{4},\omega) = \mathbb{Q}(\sqrt[3]{4}, \omega)$, where $\omega$ is a primitive cube root of unity.

Now we need to determine the degree $[L:\mathbb Q]$. To do this, note that, by the Tower Law,

$$ [L:\mathbb Q] = [L:\mathbb{Q}(\sqrt[3]{4})][\mathbb{Q}(\sqrt[3]{4}):\mathbb{Q}] $$ Let $m(x) = x^3 - 4 \in \mathbb{Q}[x]$. This is a cubic rational polynomial with no rational roots (its only real root is $\sqrt[3]{4}$, which is irrational), so it is irreducible over $\mathbb{Q}$, so $m(x)$ is the minimum polynomial of $\sqrt[3]{4}$ over $\mathbb{Q}$, and hence $[\mathbb{Q}(\sqrt[3]{4}):\mathbb{Q}] = \deg m = 3$.

Note that $L = \mathbb{Q}(\sqrt[3]{4})(\omega)$, so $[L:\mathbb{Q}(\sqrt[3]{4})]$ equals the degree of the minimum polynomial of $\omega$ over $\mathbb{Q}(\sqrt[3]{4})$. We have that $\omega$ is a root of $x^2 + x + 1$, so $[L:\mathbb{Q}(\sqrt[3]{4})] \leq 2$. Furthermore, $\mathbb{Q}(\sqrt[3]{4})$ is contained in $\mathbb{R}$, and $L$ certainly is not (since $\omega \not \in \mathbb{R}$), and we conclude that $[L:\mathbb{Q}(\sqrt[3]{4})] \geq 2$.

Thus, we conclude that $[L:\mathbb{Q}(\sqrt[3]{4})] =2$, and hence $[L:\mathbb{Q}] = 2 \cdot 3 = 6$.