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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!

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  • $\begingroup$ @AndreasCaranti I was sleeping, sorry... $\endgroup$ – Dietrich Burde May 10 at 12:27
  • $\begingroup$ @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. $\endgroup$ – DonAntonio May 10 at 12:27
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Hint: Use this general statement, which can be proved by induction:

The degree of the splitting field of a polynomial of degree $n$ is at most $n!$

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    $\begingroup$ And in fact it divides $n!$. $\endgroup$ – KCd May 10 at 11:34
  • $\begingroup$ @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. $\endgroup$ – Trettman May 10 at 13:52
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    $\begingroup$ @Trettman, yes, exactly. $\endgroup$ – lhf May 10 at 13:55
  • $\begingroup$ @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? $\endgroup$ – Trettman May 10 at 13:57
  • $\begingroup$ @Trettman, sure, go ahead. $\endgroup$ – lhf May 10 at 13:57

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