Degree of splitting field of $f(x) \in \mathbb Q[x]$

I'm having trouble with a homework question, and I'm wondering if I can get some tips and/or hints (I do not want someone to solve the problem for me).

The question is as follows:

Let $$f(x) \in \mathbb{Q}[x]$$ be an irreducible monic polynomial of degree 3 that does not split over $$\mathbb{R}$$. Find the degree of the splitting field of $$f(x)$$ over $$\mathbb{Q}$$.

I haven't gotten far, but this is my work:

Since $$f(x)$$ does not split over $$\mathbb{R}$$, we know that it has a complex root with a non-zero imaginary part. Denote this root by $$\alpha$$. By the complex conjugate root theorem, $$\overline{\alpha}$$ is also a root of $$f(x)$$, that is $$f(\overline{\alpha})=0$$. I then tried to check if it is true that $$\overline{\alpha} \in \mathbb Q(\alpha)$$, but I can't seem to reach a conclusion (however, my intuition says that it is not true). I also know that $$f(x)$$ must have a real root, since $$\deg(f(x))=3$$ (so it can't have another complex root by the conjugate root theorem).

Any help is appreciated!

Thanks!

• @AndreasCaranti I was sleeping, sorry... – Dietrich Burde May 10 at 12:27
• @AndreasCaranti No, it doesn't: it has two complex non-real roots...Perhaps it is the meaning of "splitting": for me, and apparently also for the OP, it means that it can be writte as a product of linear factors. – DonAntonio May 10 at 12:27

The degree of the splitting field of a polynomial of degree $$n$$ is at most $$n!$$
• And in fact it divides $n!$. – KCd May 10 at 11:34
• @lhf Something like this? Let $r \in \mathbb R$ be the real root of $f(x)$. The degree of the extension $\mathbb Q(r)/\mathbb Q$ is 3, since $f(x)$ is the minimal polynomial of $r$ over $\mathbb Q$. It is also clear that $\mathbb Q(r)$ does not include the other two roots, since these have a non-zero imaginary part. We therefore know that we need to add another element to $\mathbb Q(r)$ to make it the splitting field, and since the degree of the splitting field divides $3!=6$, we can conclude that the degree of the splitting field for $f(x)$ is equal to 6. – Trettman May 10 at 13:52
• @lhs Perfect, thanks! Seems like I got stuck trying to say something about $\mathbb Q(\alpha)$, when it's easier to study $\mathbb Q(r)$... By the way, should I answer my own question? – Trettman May 10 at 13:57