Algebraic extensions on $\mathbb{Q}$ Let be $\alpha = \sqrt[3]2$ $\in \mathbb{R}$ and $\beta = \sqrt[3]2 (-\frac{1}{2} + \frac{\sqrt3}{2}i)$ $\in \mathbb{C}$. Show that
a) $\mathbb{Q}[\alpha] \cong \mathbb{Q}[\beta]$.
b) $|Aut \ \ \mathbb{Q}[\alpha]| = 1$.
c) $Gal[x^3 - 2, \mathbb{Q}] = \mathbb{Q} [\alpha, \beta] = \mathbb{Q} [\alpha, \overline{\beta}] = \mathbb{Q} [\beta, \overline{\beta}]$.
d) If $L = Gal[x^3 - 2, \mathbb{Q}]$, so $|Aut \ \ L| = 6$ 
(Hint: observe that if $\sigma \in Aut \ \ L$ and $u^3 = 2$, so $\sigma(u)^3 = 2$).
I would like to know if my attempt is correct and how to do the letter $d$. Thanks in advance!
My attempt:
a)
$\alpha^3 = (\sqrt[3]2)^3$
$\alpha^3 = 2$
$\alpha^3 - 2 = 0$,
so $\alpha$ is algebraic on $\mathbb{Q}$ and $p(x) = x^3 - 2 \in Ker(v_{\alpha})$, where $v_{\alpha}$ is the evaluation map. Observe that $x^3 - 2$ is irredutible on $\mathbb{Q}$ by Eisenstein's criterion for $p = 2$, therefore $p(x)$ is minimal of $\alpha$ on $\mathbb{Q}$.
$x^3 - 2 = 0$
$x^3 = 2$
$x^3 = 2 (\cos \ \ 0 + i \sin \ \ 0)$,
so the roots of $p(x)$ are $x_k = \sqrt[3]2 (\cos \ \ (\frac{2k\pi}{3}) + i \sin \ \ (\frac{2k\pi}{3}) )$, $k = 0,1$ and $2$, this is $x_0 = \alpha$, $x_1 = \beta$ and $x_2 = \overline{\beta}$, so $\beta$ is algebraic on $\mathbb{Q}$ and $p(x) = x^3 - 2 \in Ker(v_{\beta})$, where $v_{\beta}$ is the evaluation map. Observe that $x^3 - 2$ is irredutible on $\mathbb{Q}$ by Eisenstein's criterion for $p = 2$, therefore $p(x)$ is minimal of $\beta$ on $\mathbb{Q}$.
$p(x)$ is the minimal of $\alpha$ and $\beta$ on $\mathbb{Q}$, so $Ker(v_{\alpha}) = Ker(v_{\beta})$, because $p(x)$ is the generator of $Ker(v_{\alpha})$ and $Ker(v_{\beta})$, so $\mathbb{Q}[x] / Ker(v_{\alpha}) = \mathbb{Q}[x] / Ker(v_{\beta})$ and, by the Theorem of Isomorphism, we have $\mathbb{Q}[\alpha] \cong \mathbb{Q}[x] / Ker(v_{\alpha})$ and $\mathbb{Q}[x] / Ker(v_{\beta}) \cong \mathbb{Q}[\beta]$.
$\mathbb{Q}[\alpha] \cong \mathbb{Q}[x] / Ker(v_{\alpha})$, $\mathbb{Q}[x] / Ker(v_{\alpha}) = \mathbb{Q}[x] / Ker(v_{\beta})$, $\mathbb{Q}[x] / Ker(v_{\beta}) \cong \mathbb{Q}[\beta]$ $\Longrightarrow$ $\mathbb{Q}[\alpha] \cong \mathbb{Q}[\beta]$.
$\square$
b) 
In this part, I used the hint.
Le be $\sigma \in Aut \ \ \mathbb{Q}[\alpha]$ and $\alpha^3 = 2$, so $\sigma(\alpha)^3 = \sigma(\alpha)*\sigma(\alpha)*\sigma(\alpha) = \sigma(\alpha * \alpha * \alpha) = \sigma(\alpha^3) = \sigma(2) = 2$, this is $\sigma(\alpha)^3 = 2$, therefore $\sigma(\alpha) = \alpha$ and, with this, we have $\sigma(\alpha^n) = \alpha^n$ for $n \in \mathbb{N}$.
Let be $q(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n \in \mathbb{Q}[X]$, so $(\sigma \circ q)(\alpha) = \sigma(a_0 + a_1 \alpha + a_2 \alpha^2 + \cdots + a_n \alpha^n) = \sigma(a_0) + \sigma(a_1 \alpha) + a_2 \alpha^2 + \cdots + a_n \alpha^n = \sigma(a_0) + \sigma(a_1) \sigma(\alpha) + \sigma(a_2) \sigma(\alpha^2) + \cdots + \sigma(a_n) \sigma(\alpha^n) = a_0 + a_1 \alpha + a_2 \alpha^2 + \cdots + a_n \alpha^n = q(\alpha)$, thus $\sigma = id_{\mathbb{Q}[\alpha][X]} \square$.
c)
$Gal (x^3 - 2) = \mathbb{Q}[\alpha, \beta]$, because $\alpha$ and $\beta$ are the roots of $x^3 - 2$ and $\alpha$ and $\beta$ are linearly independent.
We show that $\mathbb{Q} [\alpha, \beta] = \mathbb{Q} [\alpha, \overline{\beta}] = \mathbb{Q} [\beta, \overline{\beta}]$. For this, it is enough to show that $\overline{\beta} \in \mathbb{Q} [\alpha, \beta]$, $\beta \in \mathbb{Q} [\alpha, \overline{\beta}]$ and $\alpha \in \mathbb{Q} [\beta, \overline{\beta}]$, which is clear, because $\overline{\beta} = - \alpha - \beta$, $\beta = - \alpha - \overline{\beta}$ and $\alpha = \overline{\beta} - \beta$, respectively. $\square$
d)
I don't know how to prove this, but I think that I need use the hint.
EDIT:
My attempt for letter $d$ using the fact $Gal (x^3 - 2) = \mathbb{Q}[\alpha, \beta] = dim_{\mathbb{Q}} \mathbb{Q}[\alpha,\beta] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q}]$.
Observe that $[\mathbb{Q}(\alpha):\mathbb{Q}] = 3$, because $x^3 - 2$ is minimal of $\alpha$ on $\mathbb{Q}$ and $[\mathbb{Q}(\beta):\mathbb{Q}] = 3$, because $x^3 - 2$ is minimal of $\beta$ on $\mathbb{Q}$, so
$[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q(\beta)}] * [\mathbb{Q}(\beta):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q(\beta)}] * 3 $, so $[\mathbb{Q}(\alpha,\beta):\mathbb{Q(\beta)}] \leq 3$
$[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q(\alpha)}] * [\mathbb{Q}(\alpha):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q(\alpha)}] * 3 $, so $[\mathbb{Q}(\alpha,\beta):\mathbb{Q(\alpha)}] \leq 3$
From here, I don't know what to do.
 A: Your solution is basically correct, but overcomplicated.
(a) If $\alpha$ is a root of the irreducible monic polynomial $f$ in $\mathbb{Q}[x]$, then
$$
\mathbb{Q}[\alpha]\cong\mathbb{Q}/\langle f\rangle
$$
because $f$ is the minimal polynomial for $\alpha$.
Therefore, if $\beta$ is another root of $f$, $\mathbb{Q}[\alpha]\cong\mathbb{Q}[\beta]$.
In your case $\beta^3=2$, so $\beta$ is a root of $x^3-2$ as well as $\alpha$. By Eisenstein, $x^3-2$ is irreducible.
(b) If $\sigma\in\operatorname{Aut}(\mathbb{Q}[\alpha])$, then $\sigma(\alpha)^3=\sigma(\alpha^3)=\sigma(2)=2$. Since $\alpha$ is real, $\sigma(\alpha)=\sqrt[3]{2}=\alpha$. Therefore $\sigma$ is the identity, because $\sigma$ is completely determined by its action on $\alpha$.
(c) The splitting field for $x^3-2$ is $\mathbb{Q}[\alpha,\beta,\bar\beta]$. In order to show
$$
\mathbb{Q}[\alpha,\beta,\bar\beta]=
\mathbb{Q}[\alpha,\beta]=
\mathbb{Q}[\alpha,\bar\beta]=
\mathbb{Q}[\beta,\bar\beta]
$$
we have to show that if $L$ is an extension of $\mathbb{Q}$ that contains two roots of $x^3-2$, then it contains also the third. But, if $r_1,r_2\in L$ are distinct roots of $x^3-2$, we have
$$
x^3-2=(x-r_1)(x-r_2)(x-r_3)
$$
and the quotient $x-r_3\in L[x]$, so $r_3\in L$.
(d) The Galois group of $x^3-2$ is the group of automorphisms of $\mathbb{Q}[\alpha,\beta]$. However, if we set $\omega=\beta/\alpha$, we have that $\omega$ is a primitive cubic root of $1$ and $\mathbb{Q}[\alpha,\beta]=\mathbb{Q}[\alpha,\omega]$.
Note that the minimal polynomial over $\mathbb{Q}$ of $\omega$ is $x^2+x+1$, which is also irreducible on $\mathbb{Q}[\alpha]$. So
$$
\bigl[\mathbb{Q}[\alpha,\beta]:\mathbb{Q}\bigr]=
\bigl[\mathbb{Q}[\alpha,\omega]:\mathbb{Q}\bigr]=
\bigl[\mathbb{Q}[\alpha,\omega]:\mathbb{Q}[\alpha]\bigr]
\bigl[\mathbb{Q}[\alpha]:\mathbb{Q}\bigr]=2\cdot3=6
$$
