Timeline for Are $\mathbb{C}$ and $\overline{\mathbb{Q}}_p$ isomorphic?
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| S Sep 6, 2022 at 20:18 | history | suggested | Hagi | CC BY-SA 4.0 |
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| May 4, 2010 at 10:49 | comment | added | Joel David Hamkins | H. Hasson, that is not quite said correctly. In the countable case the transcendence degree DOES have to be countable, but the issue is that there are infinitely many different countable cardinalities: all the finite cardinalities, plus the countably infinite cardinalitiy. Each of these occurs as a transcendence degree, leading to non-categoricity. | |
| May 3, 2010 at 18:47 | history | edited | Ilya Nikokoshev |
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| May 2, 2010 at 22:46 | comment | added | H. Hasson | Yes, I apologize. As Petersen noted, the theory is only uncountably categorical. The reason is that in the countable case the transcendence degree doesn't have to be countable. | |
| May 2, 2010 at 22:13 | answer | added | Tilman | timeline score: 13 | |
| May 2, 2010 at 20:44 | comment | added | François G. Dorais | @Georges: Yes, the correct invariant is the transcendence degree. The theory ACF_p is only uncountably categorical. | |
| May 2, 2010 at 20:32 | comment | added | Georges Elencwajg | Dear H.Hasson: you write "In fact any two fields of a given characteristic that are algebraically closed and share the same cardinality are isomorphic". This is not true in the denumerable case. For example the algebraic closures of $\mathbb Q$ and $\mathbb Q(X)$ ($X$ an indeterminate) are not isomorphic. | |
| May 2, 2010 at 19:07 | answer | added | Emerton | timeline score: 21 | |
| May 2, 2010 at 18:44 | answer | added | François G. Dorais | timeline score: 29 | |
| May 2, 2010 at 18:40 | history | edited | Minhyong Kim | CC BY-SA 2.5 |
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| May 2, 2010 at 17:57 | comment | added | BCnrd | @H.Hasson: I agree, though I was trying to convey that specializing between different chars is less incredible if one focuses on the algebraic mechanism that makes it work: same mechanism that is used (implicitly) all the time to relate properties of a generic fiber to properties of special fibers. We can spread out over Z[1/N] as we do over C[t][1/f], etc; specializing to C lets us use alg. closedness & topology, specializing to F_p gives other tools. Perhaps the incredible part is that geometric/topological problems over C really can descend to finitely generated Z-subalgebras! | |
| May 2, 2010 at 17:42 | comment | added | H. Hasson | @BCnrd: You're right, it doesn't involve AC heavily. The question is what is the issue in discussion here? The consistency of ZF implies the consistency of ZFC, and in the ZFC models the theorem is correct. So I took the discussion to be about whether we should allow things that are completely (unbelievably) different to be related. My answer was: we could and we should. | |
| May 2, 2010 at 17:39 | comment | added | BCnrd | @H.Hasson: the arguments involving descent through direct limits and specialization seem different from the issue which Minhyong is asking about: a "finite" version of the same sort of "countable AC" as in my previous remark. That is, you are invoking that (by work in EGA) finitely presented structures over a direct limit occur (with their "interesting properties") down in the limit process, and there we specialize, etc. It is a fantastic technique, but equally useful in equi-char; the mixed-char aspect is why it seems more exotic than it is. No serious AC is being used. | |
| May 2, 2010 at 17:34 | comment | added | Dan Petersen | @H.Hasson: the property of ACF_p that you mention is called being "uncountably categorical". (this corrects my previous comment which just said categorical, which is not true.) For the connection to completeness, se the last paragraph of en.wikipedia.org/wiki/Morley's_categoricity_theorem | |
| May 2, 2010 at 17:28 | answer | added | S. Carnahan♦ | timeline score: 6 | |
| May 2, 2010 at 17:17 | comment | added | H. Hasson | In fact, that it is "unbelievable" is part of what makes it a nice theorem. For example, in algebraic geometry it is often the case that we do "topology" on varieties over F_p, by moving to Z_p, and then to Q_p, and then to an algebraic subfield over Q in Q_p, and then to C. There we can do actual topology, and then pull the results back to F_p. It is amazing, but that's what makes it good. | |
| May 2, 2010 at 17:15 | comment | added | BCnrd | Considering injective resolutions, I never really understood AC objections if one uses derived functors in a substantial way. I agree with Minhyong about the way one uses trasncendence bases, so I have no problem with that field isomorphism. However, as a matter of style it shouldn't be invoked if not actually needed. In the case of Weil II, the issue can be avoided at the end of the proof because only countably many complex numbers actually arise (from stalks at closed points for countably many Weil sheaves on finitely many schemes of f.t. over Fbar_p): so only countable AC arises. | |
| May 2, 2010 at 17:09 | comment | added | H. Hasson | The theorem is true given the axiom of choice. In fact any two fields of a given characteristic that are algebraically closed and share the same cardinality are isomorphic (this is a property of the "theory" - I forget what this property is called, but it has something to do with completeness, which this theory is also. I'll let the model theorists be more specific). Is the problem that you didn't find a proof that was... rigorous enough? It's surely "believable". You can find much less believable iso., like: C is iso. to the ultraproduct of all the F_p bars. | |
| May 2, 2010 at 16:57 | history | asked | Minhyong Kim | CC BY-SA 2.5 |