Roots and Fixed Points Let $\alpha$ be a root of $f(x) = x^3 − 3x + 1$. Can $\mathbb{Q}(\alpha)$ contain a root of $g(x) = x^3 − 4x + 1$? Here is my idea: 
$\mathbb{Q}(\alpha) = \{a + b\alpha + c\alpha^2 : a,b,c \in \mathbb{Q}\}$. Let $\beta \in \mathbb{C}$ be a root of $g(x)$. Since $g(x) = f(x) - x$, we see that that $f(\beta) = \beta$, and so all roots of $g(x)$ are fixed points of $f(x)$. Similarly if $\kappa$ is a fixed point of $f(x)$, then $g(\kappa) = 0$. Hence $\gamma \in \mathbb{C}$ is a root of $g(x) \iff$ $\gamma$ is a fixed point of $f(x)$. So is there such $\gamma \in \mathbb{Q}(\alpha)$? Suppose there is a $\gamma = a + b\alpha + c\alpha^2$. Then $f(\gamma) = \gamma$. Plugging this in an collecting terms of $\alpha^0, \alpha, \alpha^2$, I got these equations:
$a^3 - 6ab - 9bc^2 + c^3 - a = 0$
$3a^2b - 4b + 18ab - 3ac^2 - b^3 - 3b^2c - 27bc^2 - 6c^3 = 0$
$3a^2c + 3ab^2 - 4c + 9ac^2 + 3b^3 + 9b^2 - 3bc^2 + 9c^3 = 0$
(I assumed the $\{1, \alpha, \alpha^2\}$ are linearly independent; it seems they should be. but not 100 % sure)
I dont know what to do after this. Is a simpler, more elegant way to do this problem? If so please help.
Update (After the second Comment):
The discriminant of $f(x)$ is 81 which is square and so $Gal(\mathbb{Q}(\alpha)/\mathbb{Q}) \cong C_{3}$ and $[\mathbb{Q}(\alpha): \mathbb{Q}] = 3$; $\mathbb{Q}(\alpha)$ is the splitting field of f(x). Similarly, the discriminant of g(x) is 229, which is not square free and so $Gal(L_{g}/\mathbb{Q}) \cong S_{3}$ and $[L_{g}: \mathbb{Q}] = 3$. If g(x) has a root in $\mathbb{Q}(\alpha)$, then g(x) splits completely in $\mathbb{Q}(\alpha)$. Since $L_g/\mathbb{Q}$ and $\mathbb{Q}(\alpha)/\mathbb{Q}$ are splitting fields of g(x), $L_{g}/\mathbb{Q} \cong \mathbb{Q}(\alpha)/\mathbb{Q}$.Would this be a contradiction, because the degrees are different? Any comments or suggestions will be appreciated.
 A: An idea that's often useful to deal with these types of problems is to use the local-global principle. If you can exhibit some prime $ p $ modulo which either one of these polynomials has a root with unit multiplicity while the other doesn't, you've established that $ \mathbf Q(\alpha) $ doesn't contain a root of $ X^3 - 4X + 1 $.
For instance, $ X^3 - 4X + 1 \equiv (X+1)(X^2+X+1) \pmod{2} $, so that it has a root of unit multiplicity modulo $ 2 $. This root lifts to a root in $ \mathbf Q_2 $ by Hensel's lemma, so that there's an embedding $ \mathbf Q(\beta) \to \mathbf Q_2 $ with $ \beta $ a root of $ X^3 - 4X + 1 $ over $ \mathbf Q $. On the other hand, $ X^3 - 3X + 1 $ is irreducible modulo $ 2 $, which means it can't have any roots in $ \mathbf Q_2 $. If we had $ \beta \in \mathbf Q(\alpha) $, then by a degree argument we would have $ \mathbf Q(\alpha) = \mathbf Q(\beta) $, which is impossible in light of the fact that one of them embeds into the dyadic numbers whereas the other does not. Another way of stating the same argument is that the prime $ 2 $ has a different splitting behavior in the extensions $ \mathbf Q(\alpha)/\mathbf Q $ and $ \mathbf Q(\beta)/\mathbf Q $, so they can't be equal.
There are other ways of approaching the problem using a locality argument (looking at ramified primes in both number fields, for instance), but this one is the easiest in this particular case, I think.
