Is the absolute Galois group the same as the automorphism group? [closed]

Is the absolute Galois group $\mathrm{Gal}(\overline{\mathbf{Q}}|\mathbf{Q})$ the same as the group $\mathrm{Aut}_{\mathbf{Q}}(\overline{\mathbf{Q}})$ the automorphism group in the category of $\mathbf{Q}$-algebras?

The first group is profinite and the second one seems to be an ordinary group.

Edit: The question generates a very interesting comments and discussion. At some point I don't understand why it was closed and why so many down votes. Anyway, Thank you for your clarifications and useful comments!

closed as off-topic by YCor, Alex Degtyarev, Qiaochu Yuan, Joël, Noah SnyderMar 17 '16 at 11:34

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• What is the definition of the Galois group, if it is not the automorphism group? – user1688 Mar 15 '16 at 14:54
• By definition, $\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ is the set of field automorphisms of $\overline{\mathbf{Q}}$. The latter is the same as the set of $\mathbf{Q}$-algebra automorphisms of $\overline{\mathbf{Q}}$. So, unless I am missing something, the answer is "yes" by definition. – GH from MO Mar 15 '16 at 15:15
• What do you mean by "ordinary"?... – Vladimir Dotsenko Mar 15 '16 at 15:24
• @VladimirDotsenko "Profinite" often carries connotations of being endowed with a topology (by taking an inverse limit of discrete spaces), as opposed to being "ordinary". In other words, I think Myshkin has correctly guessed that OP was thinking of the absolute Galois group as defined by taking an inverse limit of groups of finite Galois extensions. – Todd Trimble Mar 15 '16 at 15:51
• The group $\mathrm{Aut}(A)$, when $A$ is a ring, is naturally a topological group (as a closed subgroup of the group of all permutations of $A$, itself endowed with the pointwise convergence topology). In the case $A$ is an algebraic extension of $\mathbf{Q}$, this is indeed a compact group. – YCor Mar 15 '16 at 21:37

This has already been answered in the comments, but perhaps you can see it more clearly like this.

Take the isomorphism $\mathrm{Gal}(\bar{\mathbb{Q}}|\mathbb{Q}) \cong \varprojlim(K|\mathbb{Q})$ as your definition of the absolute Galois group.

The usual proof of this proposition (Neukirch, Lang...) amounts to show that

$$\mathrm{Gal}(\bar{\mathbb{Q}}|\mathbb{Q})\to \mathrm{Aut}_\mathbb{Q}(\bar{\mathbb{Q}})$$

is both injective and surjective.

So $\mathrm{Aut}_\mathbb{Q}(\bar{\mathbb{Q}})$ is by no means ordinary. Perhaps you were thinking about $\mathrm{Aut}(\mathbb{Q}) \cong \mathbb{Q}^\times$, or something along those lines.

• @oxeimon, the topology on a profinite group comes from taking the inverse limit over all finite quotient groups, working modulo all open normal subgroups. So there is no "choice" of a way to view the group as an inverse limit. – KConrad Mar 16 '16 at 4:52
• @oxeimon, a remarkable theorem of Nikolov--Segal asserts that the topology on a finitely generated profinite group is uniquely determined by the abstract group structure. This is false in the infinitely generated case. – HJRW Mar 16 '16 at 15:20
• @KConrad, I don't think I agree with your answer to oxeimon's question in comments. For infinitely generated profinite groups $G$, one can make different choices of inverse systems of finite quotients, which can induce different profinite topologies on $G$. – HJRW Mar 16 '16 at 20:38
• @HJRW, reread my comment again: I am only talking about a given profinite group (e.g., I talk about open normal subgroups), not an abstract group without a topology in advance. – KConrad Mar 16 '16 at 21:07
• @KConrad: but there is little sense in your interpretation of oxeimon's question (which starts with "does its topology depend..."), because you suppose the topology is given (so you answer: if the topology is given, I can retrieve the topology; more precisely, you assume that the open normal subgroups are specified). HJRW's interpretation is that the underlying abstract group is known to be profinite. In this case the answer is no (one example being the group given as a vector space of continuum dimension over a field of prime order); it has plenty of distinct topologies making it profinite. – YCor Mar 17 '16 at 9:50

The automorphism group of $\overline{\mathbf Q}$ is equipped with a topology, induced by viewing it as a subset of the permutation group of an infinite set. There is a basis of open neighborhoods in which the stabilizer of a finite subset of $\overline{\mathbf Q}$ is considered to be open. This is the same as the profinite topology.

• It's not just an analogy, if you view it (as I said in my comment) as a closed subgroup of the group of permutations of $\overline{\mathbf{Q}}$. – YCor Mar 17 '16 at 12:35