According to Wikipedia, if $f(x)$ is an irreducible polynomial of prime degree $p$ over $\mathbb{Q}$ and it has two nonreal roots, then the Galois group of $f$ is the full symmetric group $S_p$. For instance, a standard example is often $x^3 - 2$ over $\mathbb{Q}$ whose Galois group is $S_3$, i.e. the symmetric group of permutations on $3$ letters. Why then is the general case true?

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    $\begingroup$ Perhaps this works: If it has two nonreal roots then there's an element of the Galois group (which is a subgroup of S_p) that switches them. This is a transposition. Finally, since the Galois group should have order divisible by p, it should have an element of order p. Subgroups of S_n which contain a 2-cycle and an n-cycle are all of S_n. $\endgroup$ – Dylan Yott Jan 15 '15 at 0:38
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    $\begingroup$ @DylanYott Yes that works! (I guess by "it has two nonreal roots" the OP means that it has exactly two nonreal roots.) $\endgroup$ – Derek Holt Jan 15 '15 at 10:28

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