Timeline for Can a non-surjective polynomial map from an infinite field to itself miss only finitely many points?
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 1, 2009 at 4:12 | comment | added | Pete L. Clark | Wan's result is very impressive: I didn't know about it (although the record reflects that I was inclined in that direction, for whatever little that's worth). So Wan's argument uses the key "model-theoretic fact" that a polynomial over a finite field is injective iff it is surjective. (This is a tongue-in-cheek allusion to the standard model-theoretic proof that injective polynomial maps from C^n to C^n are surjective.) Altogether this is turning out to be quite a fascinating question... | |
| Nov 30, 2009 at 19:24 | comment | added | Kevin Buzzard | I second that: definitely don't delete. You didn't say anything wrong. You only said "if I can do X, I can do Y" and someone else has said "It's not immediately obvious, but it's impossible to do X". If these facts were lost then they might end up being rediscovered. | |
| Nov 30, 2009 at 16:40 | comment | added | D. Savitt | Jason, certainly don't delete. It can be valuable to know what ideas don't work out, and why. | |
| Nov 30, 2009 at 15:40 | comment | added | Jason DeVito - on hiatus | @ Dror: Given this, it seems as though my entire construction is useless. Should I just delete it? | |
| Nov 30, 2009 at 15:20 | comment | added | Dror Speiser | @Pete: There does not exist such a sequence. A non-surjective polynomial of degree n over a finite field of q elements has image smaller than q-(q-1)/n. If you bound n and m, then for all but a finite number of finite fields, the bound is smaller than q-m. This was proved by D. Wan in "A p-adic lifting lemma and its applications to permutation polynomials". I found this immediately in the second result of the google search "size of polynomial finite fields". | |
| Nov 30, 2009 at 14:35 | comment | added | Greg Kuperberg | Something does not feel right. First, I think I learned somewhere that (ii) and (iii) don't happen. Second, how do you get around the fact that if you adjoin one root of any irreducible polynomial of degree $n$ to a finite field $F$, then suddenly all irreducible polynomials of degree $n$ over $F$ factor completely? | |
| Nov 30, 2009 at 13:46 | history | edited | Jason DeVito - on hiatus | CC BY-SA 2.5 |
Editted for clarity based on comments
|
| Nov 30, 2009 at 13:41 | comment | added | Jason DeVito - on hiatus | To Dror - I meant for F_q and F_p to be any two finite fields (of any characteristic, and not neccesarily of prime order) - perhaps "p" was a bad choice here ;-) To buzzard - Reading your summary, I agree completely, but I don't consider myself good enough in logic to have come up with it on my own ;-) | |
| Nov 30, 2009 at 7:26 | comment | added | Kevin Buzzard | I think it could be more succinctly summarised by saying that "the standard ultraproduct trick will produce an example, if one can get uniform bounds in the prodands" (yes yes, I know no-one calls them prodands). | |
| Nov 30, 2009 at 5:46 | comment | added | Philipp Lampe | It's an interesting approach. But I don't see why such a sequence exsists either. | |
| Nov 30, 2009 at 5:27 | comment | added | Pete L. Clark | No, (i) just means that there is an infinite sequence of primes. The argument is correct, I think -- in fact, I had thought of something very similar myself a few days ago. If (ii) and (iii) hold, then you get an infinite "pseudofinite" field K and a polynomial P such that K \setminus P(K) is finite and nonempty. The trouble is that it is not at all clear that a sequence (p,f_p) exists satisfying (ii) and (iii) above. It seems doubtful to me, in fact. | |
| Nov 30, 2009 at 4:56 | comment | added | Dror Speiser | Is Fp a finite field as well? Then no, there does not exist an infinite sequence of finite fields Fq of same characteristic such that Fq is smaller in cardinality than Fp. So T itself is finite, and the field constructed will not be infinite. | |
| Nov 30, 2009 at 3:01 | history | answered | Jason DeVito - on hiatus | CC BY-SA 2.5 |