5
$\begingroup$

Let's assume that we have a splitting field $F$ over $Q$ that is a finite extension. Let $p(x)$ be the polynomial in $Q[x]$ that has $F$ as a splitting field and is of minimal degree. Is it correct that if $Gal(F,Q)$ is isomorphic to $S_n$ then $deg \ p(x)=n$? If yes, is it true that, more generally, $deg \ p(x)$ equals $n$ where $n$ is the minimal natural number such that $Gal(F,Q)$ is contained in $S_n$? Thanks a lot in advance.

$\endgroup$
  • $\begingroup$ Yes the first question is true. $\endgroup$ – Gregory Grant Mar 9 '15 at 17:20
  • $\begingroup$ I think that $Gal(F,Q)$ can be always seen as a subgroup in $S_n$ where $n=deg(P)$ $\endgroup$ – Abellan Mar 9 '15 at 17:42
  • $\begingroup$ @Gregory: do you have a proof? $\endgroup$ – Qiaochu Yuan Mar 9 '15 at 22:50
  • 3
    $\begingroup$ @Mesih: Every extension of $\mathbb{Q}$ is separable. $\endgroup$ – Brandon Carter Mar 9 '15 at 23:13
6
$\begingroup$

Yes and yes. In one direction, if $p(x)$ is a polynomial with splitting field $F$, then $G = \text{Gal}(F/\mathbb{Q})$ acts faithfully on the roots of $p(x)$, and this gives an embedding of $G$ into $S_n$ where $n = \deg p$.

In the other direction, the data of an embedding of $G$ into $S_n$ is the data of a faithful action of $G$ on a set with $n$ elements. Decompose this action into its orbits, and let $H_1, H_2, \dots H_k$ be the corresponding stabilizers. The condition that $G$ embeds into $S_n$ is precisely the condition that the intersection of the conjugates of all of the $H_i$ is trivial. Let

$$F_i = F^{H_i}$$

be the subextensions of $F$ corresponding to the $H_i$. Now the condition that $G$ embeds into $S_n$ is precisely the condition that the $F_i$ have Galois closure $F$. Let $p_i(x)$ be the minimal polynomial of a primitive element of $F_i$, and let

$$p(x) = \prod p_i(x).$$

Then $p(x)$ is a polynomial of degree $\sum |G/H_i| = n$ with splitting field $F$.

$\endgroup$
  • $\begingroup$ Do you mean all $F_{i}$ will have Galois closure $F$, or something different? Sorry it is not clear to me from $F_{i}=F^{H_{i}}$ as the fixed subfield under $H_{i}$ and the interaction of conjugates of $H_{i}$ is trivial. $\endgroup$ – Bombyx mori Mar 9 '15 at 23:22
  • $\begingroup$ @Bombyx: by the Galois closure of a family of subextensions I mean the smallest Galois subextension containing them. This corresponds under the Galois correspond to taking the intersection of all of the conjugates of the $H_i$. $\endgroup$ – Qiaochu Yuan Mar 9 '15 at 23:26
  • $\begingroup$ I think you meant "under the Galois correspondence". I see what you meant now. Thanks! $\endgroup$ – Bombyx mori Mar 10 '15 at 1:07

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.