Why is the $\mathbb{Q}(i, \sqrt{3}, \sqrt[3]{5}) $splitting of the polynomial $(x^3 - 5)(x^2 + 1)$ and not just $(x^3 - 5)$?

Question

I know that $\zeta = \sqrt[3]{5}$ satisfies the polynomial $x^3-5 \in \mathbb{Q}[x]$, and splits completely in $\mathbb{Q}(i, \alpha, \zeta)$, where $\alpha = \sqrt{3}$. I found the roots $x_1 = \zeta$, $x_2 = \zeta/2 + \zeta i \alpha /2$ and $x_3 = \zeta/2 - \zeta i \alpha /2$

And I can write $\zeta = x_1$, $\alpha = (2x_1/x_2 - 1)^4/3$ and $i = (x_3-x_2)/(x_1\alpha)$, and since the polynomial $(x^3-5)$ already splits completely over $\mathbb{Q}(i, \alpha, \zeta)$, why do I need the extra term $(x^2+1)$?

Answer 1

The splitting field of $x^3-5$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[3]{5}, i\sqrt{3})$. In particular, $i,\sqrt{3} \not\in \mathbb{Q}(\sqrt[3]{5}, i\sqrt{3}),$ which becomes apparent if you realize $\mathbb{Q}(\sqrt[3]{5}, i\sqrt{3})$ as a $\mathbb{Q}(\sqrt[3]{5})$-vector space with basis $\{1, i\sqrt{3}\}$ (along with the provable fact $\sqrt{3} \not\in \mathbb{Q}(\sqrt[3]{5})$)

Answer 2

To find the splitting field of $x^3-5$ on $\mathbb{Q}$ you have to extend $\mathbb{Q}$ with the root of the polynomial. So the first extension is $\mathbb{Q}(\sqrt[3]{5})$ then the other two roots of the polinomial, $\frac{-1+i\sqrt{3}}{2}\sqrt[3]{5}, \frac{-1-i\sqrt{3}}{2}\sqrt[3]{5}$, aren't in this extension, infact there isn't complex number in $\mathbb{Q}(\sqrt[3]{5})$. So now let's extend with another roots, like $\frac{-1+i\sqrt{3}}{2}\sqrt[3]{5}$, and we obtein $\mathbb{Q}(\sqrt[3]{5}, \frac{-1+i\sqrt{3}}{2}\sqrt[3]{5})$ but $\frac{\frac{-1+i\sqrt{3}}{2}\sqrt[3]{5},}{\frac{\sqrt[3]{5}}{2}}=-1+i\sqrt{3}$, so $i\sqrt{3} \in \mathbb{Q}(\sqrt[3]{5}, i\sqrt[3]{5}+\frac{1}{2}\sqrt[3]{5},)$. Then we can conclude that $\mathbb{Q}(\sqrt[3]{5}, \frac{-1+i\sqrt{3}}{2}\sqrt[3]{5})=\mathbb{Q}(\sqrt[3]{5}, i\sqrt{3})$ and this is the the splitting field of $x^3-5$. But $i, \sqrt{3} \not \in \mathbb{Q}(\sqrt[3]{5}, i\sqrt{3})$.

If we consider the splitting field of $(x^3-5)(x^2+1)$ we nead to extend $\mathbb{Q}(\sqrt[3]{5}, i\sqrt{3})$ with a root of $x^2+1$ and we obtain $\mathbb{Q}(\sqrt[3]{5}, i\sqrt{3}, i)=\mathbb{Q}(\sqrt[3]{5}, \sqrt{3}, i)$. So how it's easly to check the splitting field of $(x^3-5)(x^2+1)$ contains the splitting field of $(x^3-5)$.

Answer 3

The splitting field of $x^3-5$ has degree at most $3!=6$.

$\mathbb{Q}(i, \sqrt{3}, \sqrt[3]{5})$ contains $\mathbb{Q}(i, \sqrt{3})$, which has degree $4$, and $\mathbb{Q}(\sqrt[3]{5})$, which has degree $3$. Therefore, $\mathbb{Q}(i, \sqrt{3}, \sqrt[3]{5})$ has degree a multiple of $12$, and so cannot be at most $6$.