MHB Polynomial of degree 3. splitting field.

[caffeinemachine]

If $F$ is the field of rational numbers, find the necessary and sufficient conditions on $a$ and $b$ so that the splitting field of $p(x)=x^3+ax+b=0$ has degree exactly $3$ over $F$.

ATTEMPT:
If $p(x)$ is not irreducible in $F[x]$ then the splitting field of $p(x)$ over $F$ can have degree $2!=2$ over $F$.
Thus $p(x)$ should have no rational roots.
Now.
Claim: $p(x)$ should not have any complex root (in the field of complex numbers).
Proof: Suppose it did. Say $\alpha$ is a complex root of $p(x)$. Now since $p(x)$ is a polynomial of degree three it has at least one real root say $a$. Let $E$ be the splitting field of $p(x)$ over $F$. Then $[F(a):F]$ divides $[E:F]=3$. $[F(a):F] \neq 1$ since $p(x)$ is irreducible in $F[x]$. Thus $[F(a):F]=3$. Clearly $\alpha \not \in F(a)$. Thus $[F(a, \alpha):F] > [F(a):F]$. Since $[F(a, \alpha):F]$ divides $[E:F]$, it follows that $[E;F]> [F(a):F]=3$.

So all three roots of $p(x)$ in the field of complex numbers are real.

Now what do I do?

 

[Deveno]

have you considered looking at the discriminant? in this case $\Delta = -4a^3 - 27b^2$. now argue that $\Delta < 0$ and $\sqrt{-\Delta}$ is rational (you might want to consider the vieta substitution:

$$x = w - \frac{a}{3w}$$

to see where I'm coming from).

 

[caffeinemachine]

have you considered looking at the discriminant? in this case $\Delta = -4a^3 - 27b^2$. now argue that $\Delta < 0$ and $\sqrt{-\Delta}$ is rational (you might want to consider the vieta substitution:

$$x = w - \frac{a}{3w}$$

to see where I'm coming from).

I don't know what the discriminant is in case of cubic equation.
I didn't know that the question requires that knowledge.
But thanks for your post Denevo. I will attempt this question again after reading more about the cubic.

 

[Deveno]

use the substitution i mentioned, and then multiply through by $w^3$. you should get a quadratic in $w^3$. what are the conditions that a quadratic has two real rational roots?