MHB What is the Galois group of a prime degree polynomial with two nonreal roots?

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I realized that I haven't yet given MHB members a Galois problem to solve. ;) So here is this week's POTW:

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Let $f(x)$ be an irreducible prime degree polynomial with rational coefficients, such that only two of its roots are nonreal complex numbers. Determine the Galois group of $f(x)$ over $\Bbb Q$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

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This week's problem was answered correctly by Deveno. Here is his solution.

Spoiler

Let $\alpha$ be any root of $f$, then $\Bbb Q(\alpha)$ has dimension $p$ over $\Bbb Q$. If $E$ is the splitting field of $f$, then:

$|G| = |\text{Gal}(f)| = |\text{Aut}(E/\Bbb Q)| = [E:\Bbb Q] = [E:\Bbb Q(\alpha)][\Bbb Q(\alpha):\Bbb Q]$, so that $p||G|$.

Now we can regard $G = \text{Gal}(f)$ as a subgroup of $S_p$ (by its actions on the $p$ roots of $f$), and since $p||G|$, $G$ contains an element of order $p$ (Cauchy's theorem), which must be a $p$-cycle.

On the other hand, since $f$ has exactly two non-real roots (which must be complex conjugates), complex conjugation is an element of $G = \text{Aut}(E/\Bbb Q)$ which acts on the roots as a transposition.

Since a $p$-cycle and a transposition generate $S_p$, this must be the Galois group of $f$.