I want to show that $\{1,x,x^2\}$ is $\mathbb{Q}-$base of $\mathbb{Q}(x)$.
We're looking at the field extension $\mathbb{C}/\mathbb{Q}$ and $x\in \mathbb{C}$. $\mathbb{Q}(x)$ is the smallest subfield of $\mathbb{C}/\mathbb{Q}$ that contains $x$. Also, $x$ is a complex root of $X^3-2X+2\in \mathbb{Q}[X]$.
First I showed that $m = X^3-2X+2$ is the minimal polynomial of $x$. Since it's already given that $x$ is a root of $m$, I only need to show that it's irreducible.
Let $p=2$ and $m = X^3-2X+2 = aX^3+bX+c$.
Then $p$ is a prime and $p\mid b \wedge p\mid c$ but $p^2\nmid c \wedge p\nmid a$. By Eisenstein criterion this polynomial is irreducible and therefore the minimal polynomial of $x$.
That means that $[\mathbb{Q}(x) : \mathbb{Q}] = 3$, right?
So we only need to find $3$ linearly independent vectors. Two of them are obvious to me, namely $\{1,x\}$. But I don't really get why I need $x^2$ and how I can prove that it's independent from $x$ (or $1$).
My guess is that if it was linear dependent, we then have by linear combination:
$$\lambda_1 \cdot 1 + \lambda_2 \cdot x = \lambda_3 \cdot x^2 \Rightarrow -\lambda_3\cdot x^2 + \lambda_1 \cdot 1 + \lambda_2 \cdot x =0$$ Since that polynomial has degree $2$ and $x$ is a root, it's a contradiction because the minimal polynomial of $x$ has degree $3$.
Any help is appreciated.
If $1,x,x^2$ are not linearly independent then $\lambda_0,\lambda_1,\lambda_2\in\mathbb C$ exists such that: $$\lambda_0+\lambda_1x+\lambda_2x^2=0\text{ and }(\lambda_0,\lambda_1,\lambda_2)\neq(0,0,0)$$
However then $x$ is the root of a non-zero polynomial with degree less than $3$ contradicting the fact that its minimum polynomial has degree $3$.