The smallest field containing $\sqrt[3]{2}$ Someone could explain how to build the smallest field containing to $\sqrt[3]{2}$.
 A: Consider the set $ F $ of expressions of the form
$$
\left\{ a + b \cdot \sqrt[3]{2} + c \cdot \left( \sqrt[3]{2} \right)^{2} \,\Big|\, (a,b,c) \in \mathbb{Q}^{3} \right\}.
$$
This is the smallest subfield of $ \mathbb{R} $ containing $ \mathbb{Q} $ and $ \sqrt[3]{2} $.
Clearly, $ \mathbb{Q} \cup \{ \sqrt[3]{2} \} \subseteq F $. Next, observe that $ F $ is closed under addition and multiplication. The only non-trivial thing that needs to be verified is that every non-zero element of $ F $ has a multiplicative inverse in $ F $. After having established that $ F $ is a field, notice that it contains precisely all numbers that should be in any subfield of $ \mathbb{R} $ containing $ \mathbb{Q} $ and $ \sqrt[3]{2} $. This proves that $ F $ is the smallest subfield of $ \mathbb{R} $ with such a property.
Notice that $ F $ is a vector space of dimension $ 3 $ over $ \mathbb{Q} $, which agrees nicely with the fact that the degree of the minimal polynomial of $ \sqrt[3]{2} $ over $ \mathbb{Q} $ is $ 3 $ (the minimal polynomial is $ X^{3} - 2 $).
A: Haskell Curry gave an internal definition, but we can also define this field externally:
$$F=\bigcap\{K\mid\Bbb Q\subseteq K, K\text{ is a field}, \sqrt[3]2\in K\}$$
Namely, $F$ is the intersection of all the fields which extend $\Bbb Q$ and contain $\sqrt[3]2$. We know this collection is not empty because $\Bbb R$ is in this collection.
Now we need to show that $F$ is a field, and that it extends $\Bbb Q$ and $\sqrt[3]2\in F$. The last two things are simple. Every rational number, as well $\sqrt[3]2$ are in $K$ for every $K$ in the collection we intersected over. Therefore $\sqrt[3]2\in F$ and $\Bbb Q\subseteq F$.
To see that $F$ is a field we see that if $x,y\in F$ then $x,y\in K$ for all $K$ in the collection, and therefore $x+y,x\cdot y\in K$ (for all $K$) and therefore $F$ is closed under addition and multiplication, and the same argument shows that additive inverse and multiplicative inverse for non-zero elements also exist in $F$.
Therefore $F$ is a field and it extends the rationals and contain $\sqrt[3]2$. To see that it is minimal observe if $K$ is in the collection then $F\subseteq K$ by definition of the intersection. Therefore the minimality is assured.

Note that this construction does not specifically depend on $\Bbb Q$, and we may replace it with an arbitrary field of our liking as the "base" field. If $\sqrt[3]2$ is already in that base field then the construction will not add new elements, for obvious reasons.
A: If the base field is $F=Z_2$ then we already have the cube root of 2, which is 0. If $F=Z_3$
then $F$ already contains one cube root of 2,  i.e. cube root of -1, namely -1;   if we want the other cube roots of 2, then the field $F(2^{1/3})=F[2^{1/3}]=\{a+b\cdot 2^{1/3}\}$
has 9 elements.
