Let $f(x)\in\mathbb{Q}[x]$ be an irreducible polynomial of degree $n$, whose zeros are all real. If $K$ is a splitting field of $f(x)$, it is true that $\mathrm{Gal}(K/\mathbb{Q})$ can be embedded into $A_{n}$? If $f(x)$ has complex zeros, then its Galois group has a complex conjugation map which corresponds to a transposition in $S_{n}$. Hence I thought that Galois group of polynomial with real zeros will have no odd permutation, which implies that its Galois group is a subgroup of $A_{n}$.

I hope that this is true since this may give a new proof of inverse Galois problem for $A_{n}$. We can prove that $p(x) = A(x-1)\cdots (x-n)+1$ is irreducible and only has real zeros for some large $A$, so it would have a Galois group which is a subgroup of $A_{n}$ (if the above statement is true) and it seems that the Galois group will be $A_{n}$ for suitable choice of $A$, although I don't know whether this argument is true or not.

  • $\begingroup$ The complex conjugation is a transposition of the zeros of $f(x)$ if and only if $f$ has exactly $n-2$ real zeros. $\endgroup$ – Jyrki Lahtonen Aug 24 '18 at 9:43

I guess in your notation you have that $n = \deg f$. If this is the case, then your assumption isn't right. For example take the polynomial:

$$p(x) = x^4 - 5x^3 + 3x^2 + 3x - 1$$

which is irreducible over $\mathbb{Q}$ and also has $4$ real roots. On the other side PARI/GP gives that the Galois group of $p(x)$ over $\mathbb{Q}$ is $S_4$ and from here it's obvious that it can't be embedded into $A_4$.

We can also calculate the Galois group of $p(x)$ over $\mathbb{Q}$ by using the method described in here. It's not hard to find out the cubic resolvent, which is:

$$R_3(x) = x^3 - 3x^2 - 11x + 4$$

This is an irreducible polynomial in $\mathbb{Q}[x]$. Combining it with the fact that $\operatorname{disc} f = 8789 = 11 \cdot 17 \cdot 47$, a non-square we conclude that the Galosi group of $p(x)$ over $\mathbb{Q}$ is isomorphic to $S_4$

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    $\begingroup$ Thank you very much! I hope this to be true... $\endgroup$ – Seewoo Lee Aug 21 '18 at 0:02
  • $\begingroup$ @SeewooLee I added some more informations, that avoids a use of some computer assistance. On the other side $S_4$ is isomorphic to a subgroup of $A_6$, so it might be possible that the the Galois group can be seen as a subgroup of $A_n$, for some $n \in \mathbb{N}$ $\endgroup$ – Stefan4024 Aug 21 '18 at 0:10
  • $\begingroup$ For the latter one, did you mean $S_{3}$? $R_{3}$ has degree 3.. $\endgroup$ – Seewoo Lee Aug 21 '18 at 0:15
  • $\begingroup$ @SeewooLee I'm not sure what you mean. $R_3$ is the cubic resolvent of the quartic polynomial $p(x)$. It's used to determine the Galois group of $p(x)$ over $\mathbb{Q}$, as the roots of $R_3(x)$ are in the splitting field of $p(x)$ over $\mathbb{Q}$ $\endgroup$ – Stefan4024 Aug 21 '18 at 0:17
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    $\begingroup$ Oh I'm sorry, I think you mean that Galois group of $R_{3}$ is $S_{4}$. $\endgroup$ – Seewoo Lee Aug 21 '18 at 0:44

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