Normal subgroups of the Galois group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T07:06:26Z http://mathoverflow.net/feeds/question/33077 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33077/normal-subgroups-of-the-galois-group Normal subgroups of the Galois group Shamim 2010-07-23T10:42:15Z 2010-07-23T11:12:05Z <p>I am trying to teach myself Galois theory. Is it true that every for a field extension K->L, that every normal subgroup of Gal(L:K) is of the form Gal(L:M) for some intermediate field M, ie K->M->L?</p> http://mathoverflow.net/questions/33077/normal-subgroups-of-the-galois-group/33078#33078 Answer by Pete L. Clark for Normal subgroups of the Galois group Pete L. Clark 2010-07-23T10:51:14Z 2010-07-23T11:12:05Z <p>Let $L/K$ be an arbitrary extension of fields, and put $G = \operatorname{Aut}(L/K)$.</p> <p>For a subgroup $H$ of $G$, put <code>$L^H = \{x \in L \ | \ \forall \sigma \in H, \ \sigma(x) = x\}$</code>.</p> <p>We say that a subgroup $H$ of $G$ is <strong>closed</strong> if $H = \operatorname{Aut}(L/L^H)$. </p> <p>Theorem (Artin-Kaplansky): Every finite subgroup $H$ of $G$ is closed. </p> <p>Moreover, a closed subgroup $H$ of $G$ is normal iff for all $\sigma \in G$, $\sigma L^H = L^H$. If $L/K$ is algebraic, this says that $L^H/K$ is normal. </p> <p>Also, if $[L:K]$ is finite, then $|G| \leq [L:K]$. So for finite extensions, every subgroup is closed. </p> <p>If $[L:K]$ is infinite, it need not be the case that every subgroup $H$ of $G$ is closed. In fact, if $L/K$ is algebraic and Galois of infinite degree, then there are always nonclosed subgroups, since any closed subgroup $H$ of $G$ is profinite, hence finite or uncountably infinite. But the group $G$ itself is uncountably infinite, so has countably infinite subgroups. </p> <p>To see a particular example of a nonclosed subgroup which is moreover <em>normal</em>, take $K = \mathbb{F}_p$ and $L = \overline{\mathbb{F}_p}$, an algebraic closure. Then $\operatorname{Aut}(L/K) \cong \widehat{\mathbb{Z}}$, and the closed subgroups are those of the form $n \widehat{\mathbb{Z}}$: they are all open, and in particular uncountably infinite. Then the subgroup of $G$ generated (not topologically generated!) by the Frobenius automorphism $\operatorname{Fr}: x \mapsto x^p$ is isomorphic to $\mathbb{Z}$, and is a nonclosed normal subgroup whose closure is $\widehat{\mathbb{Z}}$. </p>