minimal field extension of Q($\sqrt[3] {2}$) I need to describe the minimal field extension $\mathbb Q(\sqrt[3] {2})$ of the rational numbers $\mathbb Q$ that contain $\sqrt[3] {2}$.
$\mathbb Q(\sqrt[3] {2}) =\{a+b\sqrt[3] {2}+c(\sqrt[3] {2})^2|a,b,c \in \mathbb{Q}\}$.
I tried to use the rationalization of $x^3 + y^3 + z^3 - 3xyz$ ?
 A: Could you elaborate on what you mean with describing? I don't really understand what you mean by minimal field extension or what you mean by rationalization. Also you pretty much gave the answer yourself when you described the field.
You could first describe the minimal polynomial (Hint: it is $x^3 -2$)
The degree of this extension is 3 (use a theorem or show that the above polynomial is irreducible, (or some other argument)).
There are not many groups of order 3 so the galois group shouldn't be hard to figure out.
EDIT: As pointed out below in a comment, this extension is not normal so no galois group.
A: There are two issues here:


*

*Why $R=\{a+b\sqrt[3] {2}+c(\sqrt[3] {2})^2:a,b,c \in \mathbb{Q}\}$ is indeed a field.

*Why this is the minimal extension containing $\sqrt[3] {2}$.
Here are some answers:


*

*It is easy to prove that $R$ is a ring. To prove that it is a field, consider $\alpha \in R$ with $\alpha\ne0$ and consider the map $x \mapsto \alpha x$. This is a $\mathbb{Q}$-linear injective map and so is surjective because $R$ is a finite-dimensional $\mathbb{Q}$-vector space. So there is an $x$ such that $\alpha x=1$. This is your inverse. If you want to find this inverse explicitly, you have to solve a linear system.

*It is the minimal extension because it has degree 3 and 3 is prime. Any intermediate extension would have degree a divisor of 3.
