I am struggling with the following problem: Let $f \in F[x]$ be an irreducible quintic polynomial with splitting field $K$, where $\mathbb{Q} \subseteq F$. Supposing that $f$ has three real roots and two complex roots, prove that $Aut(K/F)$ contains a $2$-cycle.

My attempt: I suspect the $2$-cycle will be an automorphism which fixes all real roots and sends the complex roots to their conjugates. Let the real roots of $f$ be $a_1, a_2, a_3$. Construct a tower of fields $F \subset E \subset K$ where $E = F[a_1, a_2, a_3]$. There must be a minimal polynomial $g \in E[x]$ with the two complex roots of $f$ as its only roots. Call these roots $\omega$ and $\omega^*$.

Since $K$ is a Galois extension of $F$, then so too is it a Galois extension of $E$. Hence, $|Aut(K/E)| = [K:E]$. Since the extension $K$ over $E$ is nontrivial, then there must be some nontrivial $E$-automorphism that permutes the roots $\omega, \omega^*$. The only nontrivial permutation of the roots would send $\omega \mapsto \omega^*$.

Thus, we conclude two things. First, $[K:E] = 2$. Second, $\phi(\omega) = \omega^*$ is a legitimate $F$-automorphism of $K$. Therefore, $\phi$ is our $2$-cycle. Is this correct?

In general, say that an irreducible polynomial $f \in F[x]$ has $n$ pairs of complex roots and that its splitting field is $K$. Is complex conjugation of every root a legitimate $F$-automorphism of $K$?

  • $\begingroup$ I think your work is complete. As for the second part, I would say yes and it will correspond to a product of 2-cycles. $\endgroup$ – Test123 Apr 29 '14 at 3:43
  • $\begingroup$ With multiple pairs as in the second part, how can we be sure that such an automorphism is indeed fixing the base field? My original proof depended on there being only a single pair. $\endgroup$ – Kaj Hansen Apr 29 '14 at 3:50
  • $\begingroup$ As long as $F$ doesn't contain any of the complex root we will have a similar case to the one you mentioned above. Otherwise complex conjugation will definitely not fix all the elements of $F$. $\endgroup$ – Test123 Apr 29 '14 at 3:55
  • $\begingroup$ What means pair of complex roots ? You meant those polynomials have real coefficients ? $\endgroup$ – reuns Apr 7 '17 at 8:03

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