Why is the extension of Q with $ \sqrt2 $ equal to $ \{ a + b\sqrt2 | a,b \in Q \} $
But the extension of $Q $ with $ \sqrt[3]2 $ not equal to $ {\{ a + b\sqrt[3]2 | a,b \in Q \}} $ ?
Should I look to the fact that $ \sqrt2 $ is a solution of $x^2 -2$ and that $ \sqrt[3]2 $ is a solution of $x^3 -2$?
I know that has something to do with it but I don't get how exactly.
For any $\alpha\in\mathbb{C}$ the field $\mathbb{Q}(\alpha)$ is the smallest subfield of $\mathbb{C}$ which contains $\mathbb{Q}$ and $\alpha$. Actually, any subfield contains $\mathbb{Q}$, so it just has to contain $\alpha$.
Now, let $F:=\{a+b\sqrt{2}: a,b\in\mathbb{Q}\}$. You can check that this is indeed a field which contains $\sqrt{2}$. Hence $\mathbb{Q}(\sqrt{2})\subseteq F$. On the other hand, $\mathbb{Q}(\sqrt{2})$ itself contains $\sqrt{2}$, so since it is closed under addition and multiplication it must contain all elements of $F$. So $F=\mathbb{Q}(\sqrt{2})$.
Now we try to find $\mathbb{Q}(\sqrt[3]{2})$. It is tempting to think this should be similar to the case with $\sqrt{2}$. But note that the set $\{a+b\sqrt[3]{2}:a,b\in\mathbb{Q}\}$ isn't a field, it isn't closed under multiplication. So it has to be something else.
So I'll give the general theorem: suppose $\alpha\in\mathbb{C}$ is algebraic over $\mathbb{Q}$ with degree $n$. Then:
$\mathbb{Q}(\alpha)=\{b_0+b_1\alpha+...+b_{n-1}\alpha^{n-1}: b_0,...,b_{n-1}\in\mathbb{Q}\}$.
So returning to $\sqrt[3]{2}$, the minimal polynomial of this element is $x^3-2$, it has degree $3$. Hence $\mathbb{Q}(\sqrt[3]{2})=\{a+b\sqrt[3]{2}+c\sqrt[3]{4}: a,b,c\in\mathbb{Q}\}$.