How should I find Splitting Field of $x^3-2$ over $\mathbb Q$.
**My try **:
$x^3-2=(x-2^\frac{1}{3})(x^2+2^\frac{1}{3}x+2^\frac{2}{3})$
On solving I am getting the roots as $2^\frac{1}{3},\dfrac{2^\frac{1}{3}[-1+\sqrt 3 i]}{2},\dfrac{2^\frac{1}{3}[-1-\sqrt 3 i]}{2},$
But I dont know the answer is given that roots are $2^\frac{1}{3},2^\frac{1}{3}\omega ,2^\frac{1}{3}\omega ^2$ where $\omega $ is cube root of unity
Best Answer
The roots you find are correct; just observe that $$ 2^{1/3}\frac{-1+i\sqrt{3}}{2}=2^{1/3}\omega $$ and that $\omega$ is indeed a cube root of unity.
You can also proceed by steps. It's clear that $\sqrt[3]{2}$ must belong to the splitting field, so we can start adding it. Now, in $\mathbb{Q}(\sqrt[3]{2})$, we can factor the polynomial as $$ (x-\sqrt[3]{2})(x^2+\sqrt[3]{2}\,x+\sqrt[3]{4}) $$ If $\alpha$ is a root of the second factor, it must belong to the splitting field $F$, but then also $\omega=\alpha/\sqrt[3]{2}\in F$. The other root is $\sqrt[3]{4}/\alpha$.
From $$ \alpha^2+\sqrt[3]{2}\,\alpha+\sqrt[3]{4}=0 $$ we get $$ \omega^2+\omega+1=0 $$ so $\omega$ is a (non real) root of $x^3-1=(x-1)(x^2+x+1)$, hence a cubic root of unity. The other non real root is its conjugate $\bar{\omega}=\omega^{-1}=\omega^2$.
Thus $$ F=\mathbb{Q}(\sqrt[3]{2},\alpha,\sqrt[3]{4}/\alpha)= \mathbb{Q}(\sqrt[3]{2},\sqrt[3]{2}\,\omega,\sqrt[3]{2}/\omega)= \mathbb{Q}(\sqrt[3]{2},\sqrt[3]{2}\,\omega,\sqrt[3]{2}\,\omega^2)= \mathbb{Q}(\sqrt[3]{2},\omega) $$ (the last equality can be easily proved by double inclusion).