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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 $\mathbb Q$

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    You can use `\langle Y^2 + Y + 1 \rangle` to get $\langle Y^2 + Y + 1 \rangle$. Also, what have you tried? Do you know how to write down $k[Y]/\langle Y^2 + Y + 1 \rangle$ in the form $k[\alpha]$? – 2017-01-10
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    The \langle \rangle fairy is sad for being beaten to this one, but overjoyed that the \langle \rangle is being spread by other devotees! – 2017-01-10

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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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    I think the roots of $Y^2+Y+1 should be -\frac{1}{2} \pm i \frac{\sqrt{3}{2}$. Isn't it? – 2017-01-11
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    You're right. Fixed. – 2017-01-11
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we know that roots of $f(x)=x^3-2$ is $\{\sqrt[3]2,\zeta_3 \sqrt[3]2,\zeta_3^2\sqrt[3]2 \}$ then $\mathbb Q(\sqrt[3]2,\zeta_3)$ is splitting field of $f(x)$, In other hand we have $$\mathbb Q(\sqrt[3]2,\zeta_3)= \mathbb (Q(\sqrt[3]2))(\zeta_3)\cong \mathbb Q(\sqrt[3]2)(Y)/(Y^2+Y+1)$$ with $Y^2+Y+1$ irreducible polunomial of $\zeta_3$.

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Let us write $a$ for $\sqrt[3]{2}$, that's easier. Now with coefficients in $\Bbb Q$ the polynomial $x^3-2$ is irreducible, but if we allow coefficients in $\Bbb Q(a)$ we can split off the root $a$, using the identity $x^3-a^3=(x-a)(x^2+ax +a^2)$. The quadratic polynomial is irreducible in $\Bbb Q(a)$ but we can solve it using the classic formula for finding the roots of quadratic equations giving: $$ \frac{-a \pm \sqrt{a^2 - 4a^2}}{2} = a(\frac{-1 \pm \sqrt{-3}}{2})= (a\zeta, a\zeta^2) $$So we can find all the roots back in the extension $\Bbb Q(\sqrt[3]{2}, \zeta)$. The notation $\Bbb Q[X]/X^2+X+1$ is only another way to write $\Bbb Q(\zeta)$.