Show that over any field $K$, such that $\mathbb Q \subset K$ the polynomial $x^3-3x+1$ is either irreducible or splits into linear factors Show that over any field $K$, such that $\mathbb Q \subset K$ the polynomial $x^3-3x+1$ is either irreducible or splits into linear factors
I think if we prove that: if $K$ contain one root it contains all three, we prove the statement, the problem is I don't know Galois Theory and this problem seems very difficult to me, I need help.
Thanks 
 A: Suppose $a\in\mathbb{R}$ is a root of your polynomial and compute
$$(a^2-2)^3-3(a^2-2)+1=a^6-6a^4+12a^2-8-3a^2+6+1=a^6-6a^4+9a^2-1=(a^3-3a)^2-1=(a^3-3a+1)(a^3-3a-1)=0$$
therefore also $a^2-2$ is a root. So, $\mathbb{Q}(a)$ contains two of the three complex root, hence it has to contain also the third one.
It follows that for every field $K\supset\mathbb{Q}$, the polynomial is irreducible if and only if $a\not\in K$, otherwise it splits into linear factors.
A: The discriminant of the cubic above is $81$ which is a square in $\Bbb{Q}$ and so the Galois group of the polynomial above is $A_3$. Now suppose we are given that $K$ contains a root $a$ of $f$. Write $L$ for the splitting field of $f(x)$; we have that $K \cap \Bbb{Q}(a)$ is not empty.Then $K \cap \Bbb{Q}(a)$ is an intermediate field between $L$ and $\Bbb{Q}$ and hence is either the rationals or is equal to $L$ since $A_3$ has only two subgroups. It cannot be equal to the rationals for $K$ contains $a$ that is a root of $f$. Hence  $K \cap \Bbb{Q}(a)$ must be equal to $L$ and then we have $\Bbb{Q}(a) = L$ and consequently $K$ contains all three roots of $f$. 
A: With a little calculus is not hard to show this polynomial has three different real roots.
If $\,\alpha\,$ is one of the roots, then 
$$x^3-3x+1=(x-\alpha)\left(x^2+\alpha \,x+(\alpha^2-3)\right)$$
The quadratic's discriminant is $\,\Delta:=12-3\alpha^2=3(2-\alpha)(2+\alpha)\,$
Now, again using a little a calculus, we can see the roots are in the following intervals:
$$(-2,-1)\to\alpha_1\;\;\;\;\;\;\;(-1,1)\to\alpha_2\;\;\;\;\;\;(1,2)\to\alpha_3$$
and since all the roots are in $\,(-2,2)\,$ we get that $\,\Delta>0\,$ no matter what root $\,\alpha\,$ we chose from the  
three existing ones.
Thus, since the roots of the quadratic are 
$$\frac{-\alpha\pm\sqrt\Delta}{2}\in\Bbb Q(\alpha)$$
We can see that $\,\Bbb Q(\alpha)\,$ contains them all and, thus, this is also true for any field containing $\,\alpha\,$.
Of course, if the polynomial has no roots in $\,\Bbb K\supset\Bbb Q\,$ then it must remain irreducible there (why?)
A: Yes you are on the right track. If you show that given one root you can express the other two, you are done (given one root $y_1$, the other two roots are in $\Bbb Q(y_1)$).
This is equivalent to showing that the decomposition field of $P$ over $\Bbb Q$ has degree $3$ and its Galois group is cyclic of order $3$. Usually the Galois group of a degree $3$ polynomial is $S_3$, and it collapses to $A_3$ (cyclic of order $3$) when the discriminant is a square.
Call $y_1,y_2,y_3$ the roots of $x^3-3x+1$ in $K$.
You can compute $ \Delta = (y_1-y_2)^2(y_1-y_3)^2(y_2-y_3)^2$. It is symmetric when permuting the roots, so $\Delta \in \Bbb Q$ (it is the discriminant of the polynomial).
Then, compute $\delta' = (y_1-y_2)(y_1-y_3) \in \Bbb Q(y_1)$. This one is easy, for example you can write $y_1^2-(y_2+y_3)y_1+y_2y_3 = y_1^2 - (-y_1)y_1 + (-1)/y_1 = 2y_1^2 - 1/y_1$.
If $\Delta = \delta^2$ with $\delta \in \Bbb Q$, then $(y_2-y_3)^2 = (\delta/\delta')^2$, and thus $(y_2-y_3) = \pm \delta/\delta' \in \Bbb Q(y_1)$. Combining this with $y_2+y_3 = -y_1 \in \Bbb Q (y_1)$, you obtain the expressions of $y_2$ and $y_3$ in $\Bbb Q(y_1)$.
