TITLE: 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
QUESTION [7 upvotes]: 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

REPLY [6 votes]: 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.

REPLY [2 votes]: 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 $f$.