Galois Theory and Galois Groups Show that $\mathbb{Q}[x]/\langle x^{3}-2\rangle = [{a + b\alpha + c\alpha^{2}: a, b, c \in \mathbb{Q}, \alpha^{3} = 2}]$ is not a Galois extension of $\mathbb{Q}$. In particular, show that every automorphism of $\mathbb{Q}[x]/\langle x^{3}-2\rangle$ that fixes $\mathbb{Q}$ also fixes all of $\mathbb{Q}[x]/\langle x^{3}-2\rangle$. To do this, we'll use the following approach. Let $\varphi$ be an automorphism of $\mathbb{Q}[x]/\langle x^{3}-2\rangle$ that fixes $\mathbb{Q}$, i.e. that satisfies $\varphi(a) = a$  $\forall a \in \mathbb{Q}$. We want to show that $\varphi$ is necessarily the identity automorphism, namely the function $e$, given by $e(\beta) = \beta$ $\forall \beta\in\mathbb{Q}[x]/\langle x^{3}-2\rangle$.
Part a. We want to show that $\varphi = e$, namely that $\varphi(\beta) = \beta$ for all $\beta$ of the form $a+b\alpha+c\alpha^{2}$ where $a,b,c \in \mathbb{Q}$ and $\alpha^{3} = 2$. Show that to do this, it suffices to show that $\varphi(\alpha) = \alpha$.
$\bf{Thoughts:}$ So this is what I have so far, and I wasn't sure if I had to actually show that $\varphi(\alpha) = \alpha$. Let K = $\mathbb{Q}[x]/\langle x^{3}-2\rangle$ and Let F = $\mathbb{Q}$. Let $\varphi$ be any automorphism of K that fixes F. $\varphi$ is defined to be $\varphi: K\to K$ such that $\forall \beta$, $\beta = a+b\alpha+c\alpha^{2}$, where $a,b,c\in F$ and $\alpha^{3} = 2$. Show $\varphi(\beta)= \beta$. $\varphi(\beta) = \varphi(a+b\alpha+c\alpha^{2}) = \varphi(a) +\varphi(b\alpha)+\varphi(c\alpha^{2}) = a+b\varphi(\alpha)+c\varphi(\alpha^{2})$ (since a,b,c are in F, which is the fixed field). I'm not sure if I actually have to show that $\varphi(\alpha) = \alpha$, because then $\varphi(\beta) = \beta$, which for this part of the problem is what we're trying to show, I think. Any help would be greatly appreciated.
Part b: Show that [$\varphi(\alpha)]^{3} = 2$. Once we have this then we can conclude that $\beta^{3} = 2$ has only one solution $\beta\in K$.
$\bf{Thoughts:}$ I already showed that [$\varphi(\alpha)]^{3} = 2$ by using that fact that since it's an automorphism, so then we can bring the power of 3 into the automorphism so that it's actually: $\varphi(\alpha^{3}) = \varphi(2) = 2$ since $2 \in F$, which is the fixed field. Then this part is done.
Part c: To show that there is only one cube root of 2 in K, factor $\beta^{3} - 2 = \beta^{3} - \alpha^{3} = (\beta -\alpha)(\beta^{2}+\alpha\beta+\alpha^{2})$. Conclude that it suffices to show that $\beta^{2}+\alpha\beta+\alpha^{2}\ne 0$ for all $\beta \in K$.
$\bf{Thoughts:}$ So I wrote the factoring out, and concluded that the only way for that to happen was if $\beta -\alpha = 0$, implying that $\beta = \alpha$ $\textit{or}$ $\beta^{2}+\alpha\beta+\alpha^{2} = 0$ for all $\beta\in K$. So then this part is done.
Part d: By way of contradiction, show that if $\exists \beta \in K$ such that $\beta^{2}+\alpha\beta+\alpha^{2} = 0$, then writing $\beta$ as $\beta = a+b\alpha+c\alpha^{2}$, we necessarily have 
$$\
0 = a^{2}+2c(1+2b),\\
0 = a(1+2b)+2c^{2},\\
0 = 1+b+b^{2}+2ac$$
Show that there are no rational numbers $a$, $b$, and $c$ which satisfy these equations. (which this ends the entire proof cuz it shows that $\varphi = e$)
$\bf{Thoughts:}$ I understand what we're supposed to do for this part of the problem, but I'm not sure how to start showing that there are no rational numbers which satisfy these equations.
Thanks, in advance, for the help. Sorry those three equations aren't centered, I'm not entirely sure how to do that in Latex quite yet, and what I did try didn't work.
 A: I think it'd be way simpler to show that quadratic has no roots in $\,\Bbb Q(\sqrt[3]2)\,$ by analyzing its discriminant:
$$\Delta=\alpha^2-4\alpha^2=-3\alpha^2$$
But $\,\alpha=\sqrt[3]2\,$ ,  so $\,0>-3\alpha^2\in\Bbb R\,$ and from here the quadratic has no real roots, and this is enough as $\,K\,$ is a real field...
A: I'm of the opinion that you're working too hard. 
Notice that $K$ is a real field so $\omega \sqrt[3]{2},\omega^2 \sqrt[3]{2} \notin K$ where $\omega=e^{2\pi i/3}$. Then the only root of $x^3-2$ in $K$ is $\sqrt[3]{2}$ so any automorphism fixes $\sqrt[3]{2}$. 
A: Even though Don Antonio showed that the quadratic has no roots, I would like to point out that your three equations are easily seen to possess no solutions at all:
Firstly, since $a^2$ and $c^2$ are $\ge 0$, we find that both $c(1+2b)$ and $a(1+2b)$ are $\le 0$. But $1+b+b^2=(b+\frac{1}{2})^2+\frac{3}{4}$, thus $ac\le 0$. This implies that $a$ and $c$ are both of the same sign, and of the opposite sign, i.e. $a=c=0$, but then the third equation is not satisfied, so there is no solution at all, in real numbers.
Tell me if I miss something, thanks.
A: Part a. What this part means is that you should show that $\phi$ is the identity automorphism if and only if $\phi(\alpha ) = \alpha$. This is straightforward.
Note: The rest of the parts are focused on showing that $\phi(\alpha) = \alpha$.
Part b. Correct
Part c. Not correct. You are right about the factoring, and it's just concluding that if we want a non--identity automorphism, then we need $\phi(\alpha) = \beta \neq \alpha$. Thus, $\beta$ must satisfy $\beta^2 + \alpha \beta + \alpha^2 = 0$.
Part d. Do what they say, and compare coefficients treating it as a polynomial in $\alpha$. Remember that $\alpha ^3 = 2$. 
P.S. To center equations, use double dollar sign instead of single dollar sign.
