How to find the dimension of this field extension? 
Let $p(x)=x^3-x+1$ in $\mathbb{Q[x]}$ and $\alpha$ algebraic in $\mathbb{Q}$, such that $p(\alpha)=0$. If $\beta = 1+\alpha-2\alpha^2 \in \mathbb{Q}(\alpha)$. It's true that $[\mathbb{Q}(\beta): \mathbb{Q}]=3$?

I've tried see if $p(\beta)=0$, but it is not true.
I did not find a polynomial in $\mathbb{Q[x]}$ such that $\beta$ is a root.
 A: The real question here is if
$p(x) = x^3 - x + 1 \in \Bbb Q[x] \tag 1$
is irreducible over $\Bbb Q$?  This may in fact be obvious to some; nevertheless I proffer the following simple demonstration, largely an exercise in elementary number theory:
If $p(x)$ were reducible, being a monic cubic polynomial it would have a linear factor $x - \rho$ over $\Bbb Q$, hence $\rho \in \Bbb Q$ would be a root of $p(x)$:
$p(\rho) = 0; \tag 2$
we set
$\rho = \dfrac{r}{s}, \; r, s \in \Bbb Z, \; s \ge 1, \tag 3$
with
$\gcd(r, s) = 1; \tag 4$
then (1), (2) and (3) together yield
$\dfrac{r^3}{s^3} - \dfrac{r}{s}+ 1 = 0, \tag 5$
which in turn yields
$r^3 = s^2r - s^3 \Longrightarrow s \mid r^3; \tag 6$
thus for any prime $q$ dividing $s$,
$q \mid s \Longrightarrow q \mid r^3 \Longrightarrow q \mid r \Rightarrow \Leftarrow \gcd(r, s) = 1; \tag 7$
this shows there is in fact no prime $q \mid s$, hence
$s = 1; \tag 8$
therefore,
$(r - 1)r(r + 1) = r^3 - r = -1; \tag 9$
which is impossible since $-1$ cannot be the product of three consecutive integers, as is easy to see; that $p(x)$ is irreducible over $\Bbb Q$ has thus been proved.
Since $p(x) \in \Bbb Q[x]$ is irreducible, it follows then that
$[\Bbb Q(\alpha):\Bbb Q] = 3; \tag{10}$
this allows us to write
$[\Bbb Q(\alpha):\Bbb Q(\beta)] [\Bbb Q(\beta):\Bbb Q] = [\Bbb Q(\alpha):\Bbb Q] = 3, \tag{11}$
which in turn implies that
$[\Bbb Q(\beta):\Bbb Q] = 1 \; \text{or} \; 3; \tag{12}$
but
$[\Bbb Q(\beta):\Bbb Q] = 1 \Longrightarrow \beta \in \Bbb Q \Longrightarrow 2x^2 - x  + \beta - 1 \in \Bbb Q[x],  \tag{13}$
and since
$\beta = -2\alpha^2 + \alpha + 1, \tag{14}$
or
$2\alpha^2 - \alpha + \beta - 1 = 0, \tag{15}$
we find
$[\Bbb Q(\alpha):\Bbb Q] \le 2\Rightarrow \Leftarrow [\Bbb Q(\alpha):\Bbb Q] = 3; \tag{16}$
we conclude therefore that
$[\Bbb Q(\beta):\Bbb Q] = 3. \tag{14}$
