Timeline for Separable and algebraic closures?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 11, 2013 at 8:19 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
added 337 characters in body
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| Jun 9, 2010 at 2:11 | comment | added | David Roberts♦ | whoops, $k^{alg}$! | |
| Jun 9, 2010 at 2:10 | comment | added | David Roberts♦ | @Martin: Hmm, of course the algebraic closure is not unique up to unique isomorphism, because the set of k-isomorphisms between $k^alg_1$ and $k^alg_2$ is (morally) a torsor for the absolute Galois group! Thanks for that. | |
| Jun 7, 2010 at 15:28 | comment | added | Bugs Bunny | BCnrd, it is only normal to suggest that complex numbers come with complex structure but sometimes they don't:-)) | |
| Jun 7, 2010 at 10:54 | comment | added | BCnrd | Bugs, according to Bourbaki, the boldface font is reserved for objects which are unique up to unique isomorphism (with the exception of $\mathbf{F}_ q$), and in particular Bourbaki regards $\mathbf{C}$ as equipped with a choice of $i$. That is, the notation $\mathbf{C}$ is meant to encode more structure than just an algebraic closure of $\mathbf{R}$. I think this viewpoint on $\mathbf{C}$ is misguided, but just keep it in mind if you find yourself discussing $i$ with Serre. (He says that there is a canonical $\sqrt{-1}$ in $\mathbf{C}$, due to this convention.) | |
| Jun 7, 2010 at 8:49 | comment | added | Torsten Ekedahl | @Martin: I did as you suggested (with some elaboration to further justify the move). I put this here so that your answer makes sense as I delete my comments here. | |
| Jun 7, 2010 at 8:29 | comment | added | Bugs Bunny | Doc, you can see this already in the case of $\mathbb R$. You choose your complex numbers and I choose mine. We won't be able to decide algebraically whether my $i$ is your $i$ or your $-i$. | |
| Jun 7, 2010 at 7:43 | comment | added | Martin Brandenburg | @David: Algebraic closure is not unique in the sense that $k \to k^{alg}$ cannot be extended to a functor such that $k \subseteq k^{alg}$ is a natural transformation. This shows that everytime you have to choose an algebraic closure. Two algebraic closures are isomorphic, but the isomorphism is not unique and not constructive. Thus the lnab-the does not fit here. Concerning your other question: I think that the canonical homomorphism $Gal(k^{alg} / k) \to Aut(k^{sep} / k)$ is an isomorphism. @Torsten: You may post this as an answer, I'll upvote it. | |
| Jun 7, 2010 at 6:47 | comment | added | David Roberts♦ | whoops, ncatlab.org/nlab/show/generalized+the | |
| Jun 7, 2010 at 6:47 | comment | added | David Roberts♦ | Can I cheekily say I was using the <a href="ncatlab.org/nlab/show/generalized+the">generalised 'the'</a>? The algebraic closure of k is unique up to k-isomorphism. More seriously, we might say 'Galois theory can't work' for the algebraic closure, but there is still a group Gal(k^alg/k) of k-automorphisms. How does it relate to Gal(k^sep/k)? | |
| Jun 7, 2010 at 6:34 | history | answered | Martin Brandenburg | CC BY-SA 2.5 |