Prove or Disprove: For every irreducible cubic polynomial $f(x)\in \Bbb Q[x]$, there exist a subfield $F$ of $\Bbb C$ such that $F \nsubseteq \Bbb R$ and $F \simeq \Bbb Q[x]/\langle f(x) \rangle $
Here, $\deg f$ is odd implies $f$ has a real root $\alpha$ and by irreducibility, $\alpha$ must be irrational. Thus $$F(\alpha) \simeq \Bbb Q[x]/\langle f(x) \rangle $$ and $F(\alpha) \subset \Bbb R \subset \Bbb C$
If $\beta, \gamma $ are other roots of $f$ (in some extention) ,then $$F(\beta) \simeq \Bbb Q[x]/\langle f(x) \rangle \simeq F(\gamma) $$ If $\beta , \gamma \in \Bbb C$ then $F(\gamma) \nsubseteq \Bbb R$. So every $f$ has the desired property, so the statement is true
But the answer given is "the statement is false". So my question is :
Is there any such $f$ with all three roots are irrational ? How to disprove the statement ?
Best Answer
The answer to your title question is yes. For example, $$ x^{3}-3x+1=0 $$ has three real solutions and none of them is rational by the rational root test.