<s>[INST] Show that $\mathbb Q(\sqrt[3]{2})[Y]/\langle Y^2+Y+1\rangle$ is a splitting field of $X^3-2$ Show that $\mathbb Q(\sqrt[3]{2})[Y]/\langle Y^2+Y+1\rangle$ is a splitting field of $f(X) = X^3-2$ where $Y$ is an indeterminate over $\mathbb Q$
 [/INST] Since $Y^2 + Y + 1$ is irreducible, the elements of $\mathbb{Q}(\sqrt[3]{2})(Y)/ \langle Y^2 + Y + 1 \rangle$
are of the form $ a+b \theta$, where $\theta$ is a root of $Y^2 + Y + 1$, and $a,b \in \mathbb{Q}(\sqrt[3]{2})$. The roots of $Y^2 + Y + 1$ are
$$ -\frac{1}{2}\pm i\frac{\sqrt{3}}{2}.$$
Choose the one with the plus sign, hence the elements of $\mathbb{Q}(\sqrt[3]{2})(Y)/ \langle Y^2 + Y + 1 \rangle$
are of the form $ a+b \theta$, where $\theta=-\frac{1}{2}+ i\frac{\sqrt{3}}{2}$, and $a,b \in \mathbb{Q}(\sqrt[3]{2})$
Now, the roots of $X^3-2=0$ are $\sqrt[3]{2}$, $\zeta \sqrt[3]{2}$ and $\zeta^2 \sqrt[3]{2}$ where $\zeta$ is a the primitive third root of $1$, hence its splitting field is $\mathbb{Q}(\sqrt[3]{2}, \zeta)$. Since 
$$ \zeta = e^{\frac{2\pi i}{3}}=\cos \frac{2\pi i}{3} + i \sin \frac{2\pi i}{3} = -\frac{1}{2}+i \frac{\sqrt{3}}{2},$$
then $\mathbb{Q}(\sqrt[3]{2}, \zeta) =\mathbb{Q}(\sqrt[3]{2},-\frac{1}{2}+i \frac{\sqrt{3}}{2})$, that is, the elements of the form $a+b \theta$, where $\theta=-\frac{1}{2}+ i\frac{\sqrt{3}}{2}$, and $a,b \in \mathbb{Q}(\sqrt[3]{2})$.
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