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.

  • $\begingroup$ 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
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    $\begingroup$ @Mesih: Every extension of $\mathbb{Q}$ is separable. $\endgroup$ – Brandon Carter Mar 9 '15 at 23:13

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 $F$.

  • $\begingroup$ 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

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