Timeline for How large can the gcd of a pair of two-variable polynomials be?
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Dec 15, 2015 at 3:57 | comment | added | Joe Silverman | In case anyone ever looks at this question again and wants the exact reference, the article that Felipe Voloch refers to is: J.H Silverman, Generalized greatest common divisors, divisibility sequences, and Vojta's conjecture on blowups, Monatsch. Math. 145 (2005), 333-350. | |
| Aug 4, 2014 at 22:12 | comment | added | Vesselin Dimitrov | @FelipeVoloch: I see: it follows from Thm. 2 of that paper under Vojta's conjecture for blow-ups of $\mathbb{P}^2$, with $S = \emptyset$ and $(x_0,x_1,x_2) = (1,x,y)$. This is a very neat simultaneous generalization of zeb's statement and the one I knew to be a consequence of Schmidt's theorem, and also of the homogeneous case with $m, n$ restricted to be co-prime (which is a consequence of ABC as noted by Joro in the comments). I deleted the answer that I just gave. | |
| Aug 4, 2014 at 21:39 | comment | added | Felipe Voloch | Such results follow from Vojta's conjectures, see e.g. this paper of Silverman. arxiv.org/pdf/math/0407415.pdf | |
| Aug 4, 2014 at 18:31 | comment | added | Vesselin Dimitrov | @zeb: Right; then the right question would have the exponent $\frac{1}{3}+\epsilon$. | |
| Aug 4, 2014 at 18:21 | comment | added | zeb | @VesselinDimitrov For your question (b), you can take $m = k(n^3-n)+n$ to achieve $\epsilon = \frac{1}{3}.$ | |
| Aug 4, 2014 at 17:00 | comment | added | Vesselin Dimitrov | The following related question also seems attractive to me: Is there an $\epsilon > 0$ such that (a) There are infinitely many pairs of distinct positive integers $n \neq m$ with $\min \Big( \mathrm{GCD}(n-1,m-1), \mathrm{GCD}(n,m) \Big) > \max(n,m)^{\frac{1}{2}+\epsilon}$? (b) The solutions to $\min \Big( \mathrm{GCD}(n-1,m-1), \mathrm{GCD}(n,m), \mathrm{GCD}(n+1,m+1) \Big) > \max(n,m)^{\epsilon}$ are Zariski-dense? | |
| Aug 4, 2014 at 15:58 | comment | added | Vesselin Dimitrov | @joro: As you say above, polynomial automorphisms are irrelevant for this problem, since their inverse has the same degree, thereby only producing examples where the exponent is $1$. | |
| Aug 4, 2014 at 15:42 | comment | added | joro | @VesselinDimitrov if I choose surjective $(f,g)$ and ask about them this appears formulation to me. Btw, the invertible polynomial maps are automorphisms, so they might not be entirely useless. | |
| Aug 4, 2014 at 15:18 | comment | added | Vesselin Dimitrov | @joro: OK, I do not know if such a conjectural generalization could be formulated. Let me just point out an interesting consequence of the restricted assertion (the one involving $x^3-x$ and $y^3-y$): For any $\epsilon > 0$, there are only finitely many pairs of distinct positive integers $n \neq m$ such that $(n^3-n)/(m^3-m)$ is a positive integer smaller than $n^{1-\epsilon}$. This now is a question involving large integral points on elliptic curves. | |
| Aug 4, 2014 at 14:27 | comment | added | joro | @VesselinDimitrov I suspect if $(f,g)$ is surjective on Z^2 things get complicated IMHO. | |
| Aug 4, 2014 at 14:05 | comment | added | Vesselin Dimitrov | @Indeed... and this adds some further evidence to zeb's statement. I will delete my last comment, and will instead post an answer to formulate a possible inhomogeneous Langevin statement which would include zeb's conjecture alongside Roth's theorem and the ABC conjecture. | |
| Aug 4, 2014 at 13:54 | comment | added | joro | @VesselinDimitrov Unfortunately I don't get experimental support for the given map. The inverse map is of degree $5$ which makes $x,y$ large :( | |
| Aug 4, 2014 at 13:35 | comment | added | joro | @VesselinDimitrov I am not sure I understand correctly, but aren't invertible polynomial maps of sufficiently large degree counterexample? This map has polynomial inverse and clearly covers Z^2: $(x,y) \mapsto (x-(x+y)^5,y+(x+y)^5) $ ? | |
| Aug 4, 2014 at 13:05 | comment | added | Vesselin Dimitrov | @joro: Because GL requires (of course) that $(m,n) = 1$, and zeb does not assume this restriction. | |
| Aug 4, 2014 at 13:01 | comment | added | joro | @VesselinDimitrov I am pretty sure GL can't be relaxed to drop the homogeneous condition. Why in a comment zeb appears concerned about "if f, g were both products of many homogeneous linear polynomials." | |
| Aug 4, 2014 at 12:47 | comment | added | Vesselin Dimitrov | Indeed, @joro 's remark proves that for the case that $f,g$ are homogeneous polynomials with $fg$ splitting into distinct linear factors over $\mathbb{C}$, ABC implies the affirmative answer to zeb's first question, at least if we restrict to co-prime pairs $(m,n)$. E.g. this solves the problem (under ABC) for $f = x^3-xy^2$ and $g = y^3-x^2y$. But I don't see how this could apply to the inhomogeneous case of $f = x^3-x$, $g = y^3-y$. | |
| Aug 4, 2014 at 12:29 | comment | added | joro | @VesselinDimitrov might be wrong, but for homogeneous $f,g$ and coprime $x,y$, if the gcd is large enough (which the question asks) then $F(x,y)=f(x,y)g(x,y)$ will have small radical since the gcd will contribute square. | |
| Aug 4, 2014 at 12:23 | comment | added | Vesselin Dimitrov | @joro: The conjecture you refer to (which incidentally is a vast generalization of Roth's theorem), is actually equivalent to the ABC conjecture; this is due to Langevin after work of Elkies, cf. Theorem 12.2.12 in Bombieri and Gubler's book Heights in diophantine approximation. (This also answers your linked question.) But how is this related to zeb's questions? | |
| Aug 4, 2014 at 12:16 | comment | added | joro | This is related to Granville-Langevin conjecture. Check: mathoverflow.net/questions/149631/… | |
| Aug 4, 2014 at 10:15 | comment | added | Vesselin Dimitrov | In my first remark I was assuming $f,g$ have non-zero constant terms (resp. $(0,0) \notin S$); if $(0,0)$ is a simple intersection point of $f = 0$ and $g=0$ (resp. $S$ contains $(0,0)$), and when $x,y$ are restricted to be $T$-units, the same remark applies with the exponent $1+\epsilon$. (And by the way, you do not need in your two questions to include the constant $c_{\epsilon}$: it is absorbed by the exponent $2+\epsilon$ while you may always delete from $U_{\epsilon}$ a finite number of exceptional points.) | |
| Aug 4, 2014 at 9:31 | comment | added | zeb | yeah, I just noticed that (and edited to fix it). Another bad case with the old version would come up if f, g were both products of many homogeneous linear polynomials. | |
| Aug 4, 2014 at 9:30 | history | edited | zeb | CC BY-SA 3.0 |
fixed first question
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| Aug 4, 2014 at 9:28 | comment | added | Vesselin Dimitrov | Also your first question has to be revised. (What happens to the statement when you replace $f,g$ by $f^m,g^m$?) Perhaps you want to assume that $f,g$ are irreducible? | |
| Aug 4, 2014 at 9:16 | comment | added | Vesselin Dimitrov | (And this remark concerns both variants of the question.) How is question 1 implies by question 2 (assuming $\{f = g = 0\} \subset \mathbb{Z}^2$)? Aren't you assuming $f$ and $g$ univariate (or homogeneous) for this implication? | |
| Aug 4, 2014 at 9:04 | comment | added | Vesselin Dimitrov | Just a minor remark: If you restrict $x,y$ to be powers of $2$ (or to have prime factors only from a given finite set), Schmidt's Subspace theorem yields this even with exponent $\epsilon$. | |
| Aug 4, 2014 at 8:17 | history | asked | zeb | CC BY-SA 3.0 |