Basis of $\mathbb{Q}[\sqrt[3]{2}]$

Question

How do I prove that $1, \sqrt[3]{2}, (\sqrt[3]{2})^2$ is a basis of $\mathbb{Q}[\sqrt[3]{2}] = \{ a + b \sqrt[3]{2} + c (\sqrt[3]{2})^2\: a,b,c \in \mathbb{Q} \}$. It's one of these cases where the thing to be proved looks so obvious to me that I don't know how to prove it.

Answer 1

The minimal polynomial of $\sqrt[3]{2}$ is $\mu(X)=X^3-2$. Therefore the elements you mention are a basis.

Answer 2

The only thing you might have to prove is that for $a,b,c\in\mathbb Q$ $$a + b\sqrt[3]2 + c(\sqrt[3]2)^2 = 0 \implies a=b=c=0$$ Since the spanning property is obvious and this proves independence.

Answer 3

An elementary, high-school-level, approach would be to note that elements in $\Bbb Q [\sqrt[3] 2]$ are, in general, of the form $a_0 + a_1 \sqrt[3] 2 + a_2 (\sqrt[3] 2)^2 + \dots + a_n (\sqrt[3] 2)^n$. Now, if $k=3q+r$, then $(\sqrt[3] 2)^k = 2^q \space (\sqrt[3] 2)^r$, with $r \in \{0, 1, 2\}$, so you collect all the terms that contain powers divisible by $3$ into a single rational number, all the terms that contain powers $\equiv 1 \pmod 3$ into a rational number times $\sqrt[3] 2$, and all the terms that contain powers $\equiv 2 \pmod 3$ into a rational number times $(\sqrt[3] 2)^2$.

Answer 4

The family you exhibit is clearly a generating family by Alex M's argument.

Furthermore the dimension of the vector space is $3$ because the minimal polynomial associated is $X^3-2$ (this polynomial clearly vanishes on the root AND is irreducible over $\mathbb{Q}$ by Eisenstein) hence we have got a generating family of cardinal $3$ for a vectorial space of dimension $3$ this cannot but be a base.

Answer 5

Here is a basic approach that doesn't rely on future results freeing up any worries about circular arguments.

Consider the following Theorem (Theorem 13.4 in Dummit & Foote)

Let $p(x) \in F[x]$ be an irreducible polynomial of degree $n$ and $K = F[x]/(p(x))$. Let $\theta = x \pmod{p(x)} \in K$. Then the elements $1,\theta, \dots, \theta^{n-1}$ are a basis for $K$ as a vector space over $F$.

Now taking $\theta = \sqrt[3]{2}$, we can easily deduce the original result by noting that $\mathbf{Q}[\sqrt[3]{2}] = \mathbf{Q}[x] / (x^3 - 2)$.