Timeline for product of all F_p, p prime
Current License: CC BY-SA 2.5
24 events
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| Feb 13, 2010 at 4:40 | comment | added | Joel David Hamkins | Thanks, Francois. At en.wikipedia.org/wiki/Pseudo_algebraically_closed_field, it is stated at least that every nonprincipal ultraproduct of distinct finite fields is pseudo algebraically closed, as a consequence of the Riemann Hypothesis for function fields. | |
| Feb 13, 2010 at 4:31 | comment | added | François G. Dorais | @Joel: No. Every F_p has an irreducible quadratic, hence so does the ultrapower. | |
| Feb 13, 2010 at 4:05 | comment | added | Joel David Hamkins | But now I wonder: is it conceivable that we can actually make the ultraproduct algebraically closed? For this, one would need to know whether for each n, there are infinitely many primes p such that polynomials over F_p of degree n have zeros in F_p. Is this true? If so, we could choose the filter to make R/U algebraically closed. | |
| Feb 13, 2010 at 2:15 | comment | added | Joel David Hamkins | And Thanks, Kevin, for supplying Cebotarev's theroem, which is critical for this application! | |
| Feb 13, 2010 at 1:35 | comment | added | Joel David Hamkins | Ultraproducts are great, and I highly recommend them. You can form the ultraproducts of any type of structure---groups, rings, graphs, partial orders, whatever, and Los says that in every case, truth in the ultraproduct is the same as truth on a large set of coordinates, for any first order statement. | |
| Feb 13, 2010 at 1:27 | vote | accept | Wanderer | ||
| Feb 13, 2010 at 1:25 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
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| Feb 13, 2010 at 1:14 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
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| Feb 13, 2010 at 0:57 | comment | added | Wanderer | Ok, I think I sort of get what happens - it is my first encounter with ultraproducts :) so I shouldn't feel bad about not being able to solve the problem, I guess. I like the problem very much, and I guess using ultraproducts is really the right way (and the only way) to do it. | |
| Feb 13, 2010 at 0:37 | comment | added | Kevin Buzzard | You have countably many things to say: you alternate between saying statements of the form "this prime isn't 0" and "this monic polynomial with integer coefficients factors into linear factors". For any finite initial sequence of this countably infinite list of statements, the set of primes satisfying all of them is infinite (but this is a non-trivial fact). So if I've understood you correctly, we're home! | |
| Feb 13, 2010 at 0:26 | comment | added | Joel David Hamkins | If you have countably many things to say, and the set of p for which they are true give rise to a descending sequence of infinite sets, then these sets all lie in a nonprincipal filter, which can be extended to an ultrafilter. So you'll get it in M/U. | |
| Feb 13, 2010 at 0:22 | comment | added | Wanderer | @Mr Buzzard: it does help, but still, I need some time for all this... :) It doesn't look like there will be a method for this problem which doesn't use this kind of constructions from logic? | |
| Feb 13, 2010 at 0:21 | comment | added | Joel David Hamkins | All you need is that the statements that you want to be true in R/U are true in most F_p. Then they will be true in the ultrapower. | |
| Feb 13, 2010 at 0:19 | comment | added | Kevin Buzzard | @Joel: here are some statements you might need to finish the job. (1) a field of char 0 contains Q-bar iff every monic polynomial with integer coefficients factors into linear factors in the field. (2) if f is a monic poly with Z coeffts and S(f) is the set of primes p for which f factors completely mod p then S(f) is infinite (and indeed has positive density---this is Cebotarev, or at least a consequence, and is non-formal). (3) Any finite intersection of the S(f)'s is also infinite (because S(f) cap S(g)=S(fg)). (4) For any p there's an f with p not in S(f). Is that enough for you? | |
| Feb 13, 2010 at 0:15 | comment | added | Kevin Buzzard | I do the ultra-stuff in quite a concrete way in my answer, if that helps any... | |
| Feb 13, 2010 at 0:14 | comment | added | Wanderer | Ok, thanks. I am currently trying to understand all this ultra-stuff :p | |
| Feb 13, 2010 at 0:11 | comment | added | Kevin Buzzard | @AS: then Joel's answer hasn't done the job yet---all he has done is constructed a quotient that has characteristic zero. | |
| Feb 13, 2010 at 0:09 | comment | added | Wanderer | Yes, that's exactly what I mean, it should contain all algebraic numbers! | |
| Feb 13, 2010 at 0:07 | comment | added | Kevin Buzzard | PS Joel: if I'm right and he did want Q-bar in, can you glance over my answer and instantly tell me exactly what statements about number fields one needs to finish the job? | |
| Feb 13, 2010 at 0:05 | comment | added | Kevin Buzzard | Joel: I read the question as "...and contains an algebraic closure of Q". Am I right in thinking that you read it some other way? | |
| Feb 13, 2010 at 0:03 | comment | added | Joel David Hamkins | This is a method that comes from logic. Ultraproducts and ultrapowers appear all over model theory and set theory. Many large cardinal axioms are stated in terms of various ultrapower of the universe. | |
| Feb 13, 2010 at 0:01 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
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| Feb 12, 2010 at 23:56 | comment | added | Wanderer | Interesting. I must confess that I do not know the techniques you are using. I "met" the problem in the context of Chebotarev's density theorem... | |
| Feb 12, 2010 at 23:52 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |