Is there a cubic irreducible polynomial over $\Bbb Q$ with all of its three roots are irrational?

Question

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 ?

Answer 1

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.

Answer 2

Let's take a nice cubic polynomial with 3 real roots (like $x(x-2)(x+2) = x^3 - 4x$), then add a small constant (like $2$) to it. What we get is $x^3 - 4x + 2$, which is a polynomial with three real roots (calculating the value of the polynomial at $-3, 0, 1, 2$ shows that it still has three real roots) and by Eisenstein it is irreducible.

One might need some tweaking to make this work (my original attempt used $x^3 - x$ instead, which doesn't work quite as nicely), but it's a simple way to make examples relatively consistently.

Answer 3

I believe the submitter has misinterpreted the question. The expression '$F \nsubseteq \Bbb R$' has nothing to do with irrationality per se; it expresses the notion that $F$ is a field contained in $\Bbb C$ but NOT in $\Bbb R$ - i.e that $F$ necessarily contains complex-valued elements whose imaginary parts are non-zero. So the question is, whether EVERY cubic has at least one non-real complex root - which plainly is not true, as other answers show.