Is there an elementary proof that the alternating group $A_n$, for any $n$, is the Galois group of an extension of the rationals? In fact, I am looking for a proof which does not use Hilbert's irreducibility theorem. This can be done for $S_n$ by establishing the existence of an irreducible polynomial in $\mathbb Z[X]$, of degree $n$, whose Galois group over the rationals contains a transposition and an $(n-1)$-cycle. I have been wondering whether an analogous proof could be found for $A_n$.

  • $\begingroup$ Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ – Thomas Dec 6 '14 at 16:37
  • $\begingroup$ Do you know of any non-elementary proofs of the result, so that people can decide what would be acceptable and what is not? $\endgroup$ – Andrés E. Caicedo Dec 6 '14 at 16:43
  • $\begingroup$ For instance, is your goal to avoid the use of Hilbert's irreducibility theorem? $\endgroup$ – Andrés E. Caicedo Dec 6 '14 at 16:45
  • $\begingroup$ It is a well-known problem: essentially, it boils down to finding an irreducible polynomial with a square discriminant, see en.wikipedia.org/wiki/Inverse_Galois_problem $\endgroup$ – Jack D'Aurizio Dec 7 '14 at 21:29
  • 1
    $\begingroup$ @Jack: that only shows that the Galois group is contained in $A_n$. $\endgroup$ – Qiaochu Yuan Dec 8 '14 at 2:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.