From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Diophantine Problem Date: 17 Sep 1998 20:52:26 GMT Roberto Maria Avanzi wrote: >I am stuck on this problem. I have to find the rational values of the >parameter a such that the following polynomial in Z > > 3 2 2 2 2 > Z + 18 ( 5 a - 14 a + 2 ) Z + 81 (19 a - 10 a + 1) (3 a - 10 a + 5) Z > > 5 4 3 2 > - 486 (8 a - 1) (16 a + 233 a - 92 a + 10 a - 4 a + 1) > >has at least one rational root. You are asking for (the projection to one axis of) the rational points on this affine plane curve. For any point on this curve (where the denominator does not vanish) let 2 2 3 2 4 t = 6 (6 a Z + Z + 6 Z + 609 a Z - 432 a Z + 45 a Z + 54 a - 8208 a 3 2 5 / 2 5 + 486 a - 432 a + 34560 a ) / (117 Z - 162 a - 486 a + 108864 a / 4 3 2 3 2 - 6480 a - 10530 a + 162 - 225 a Z - 1017 a Z + 1917 a Z + 19 a Z 2 + 7 Z ) (The only points on the curve where the denominator vanishes are (a,Z)= (-1,-54) and (1/5, -54/25).) Then the point on the curve may be recovered with the formulas 2 5 4 3 t (-144 + 33 t - 12 t + t + 4 t + 6 t ) Z = -54 ------------------------------------------ 3 2 2 (t - 6 t + 33 t - 48) 3 -3 t + 6 + t a = - --------------------- 3 2 t - 6 t + 33 t - 48 Conversely, for any t (making the denominator nonzero, e.g. any rational t) these formulas define a point (a,Z) on the curve. In other words, the curve is birationally equivalent to the rational line (it is of genus zero). In particular, an answer to your question is, "the set of all rational numbers a of the form a=[...] for some rational number t. I don't know if there's an easier description of this set of numbers a. I can't imagine how such an equation came up in the first place. dave Maple input: eq:=Z^3+18*(5*a^2-14*a+2)*Z^2+81*(19*a^2-10*a+1)*(3*a^2-10*a+5)*Z-486*(8*a-1)* (16*a^5+233*a^4-92*a^3+10*a^2-4*a+1): t:= 6*(6*a*Z^2+Z^2+6*Z+609*a^3*Z-432*a^2*Z+45*a*Z+54*a-8208*a^4+486*a^3-432*a^2 +34560*a^5)/(117*Z-162*a-486*a^2+108864*a^5-6480*a^4-10530*a^3+162-225*a*Z -1017*a^2*Z+1917*a^3*Z+19*a*Z^2+7*Z^2): Z:= -54*t*(-144+33*t-12*t^2+t^5+4*t^4+6*t^3)/(t^3-6*t^2+33*t-48)^2: a:= -(-3*t+6+t^3)/(t^3-6*t^2+33*t-48): ============================================================================== From: Dave Rusin Date: Fri, 18 Sep 1998 00:27:42 -0500 (CDT) To: rusin@math.niu.edu Subject: Notes to myself... Note: we expect things to work because CASA reports the genus is zero! eq:=z^3+18*(5*a^2-14*a+2)*z^2+81*(19*a^2-10*a+1)*(3*a^2-10*a+5)*z-486*(8*a-1) *(16*a^5+233*a^4-92*a^3+10*a^2-4*a+1); z0:=coeff(eq,z,2); eq2:=col(subs(z=w-z0/3,eq),w): eq3:=col(subs(a=-1+b,eq2),w); eq4:=col(subs(b=6*c,eq3),w); eq5:=col(subs(w=6*x,eq4/6^3),x); eq6:=col(subs(x=6*y,eq5/6^3),y); eq8:=col(subs(y=-1+z,eq6),z); eq9:=col(subs(z=3*c*v,"/9/c^2),v); eq10:=col(subs(v=19*c+ww,"/3),ww); eq11:=col(factor(subs(ww=c*u,eq10))/c^2,u); eq12:=col(subs(u=-12+t,eq11),t); eq13:=col(subs(t=s*(5*c-1),eq12/(5*c-1)^2),s); eq14:=col(subs(s=1/r,"*r^3),r); eq15:=col(subs(c=d+1/5,eq14),d); eq16:=col(subs(r=d*q,expand(eq15/d)),d); discrim(eq16,d/4); Get: -3q^2+5q=square: q=5/(3+m^2) (i.e. it's a rational curve). Work backwards. ============================================================================== Date: Fri, 18 Sep 1998 12:46:22 +0200 (MESZ) From: Roberto Maria Avanzi To: Dave Rusin Subject: Re: Diophantine Problem Dear Prof. Rusin During the last few days you've been HUGELY helpful to me. As I told you, I am doing "my" part in a work with Prof. Zannier. We have been told by Schinzel of an interesting problem, which I called "the genus 1 problem for separated variables". Essentially we look at the possible pairs (G,H) of polynomials in one indeterminate over an algebraically closed field of characteristic 0 and of degrees m, resp n which are coprime such that the curve C: G(Y)=H(X) is of genus 1. The answer to this problem is a genus 1 analogue of Ritt's second theorem, which is equivalent to the following statement " RITT's #2: If G(t) and H(t) are polynomials of coprime degrees m and n over a field k s.t. char(k) does not divide mn and the curve G(Y)=H(X) (which is irreducible by results of Tverberg) has genus 0, then for suitable nonconstant linear functions L,M,N we have that [here * is composition of polynomials] (i a) L * G * M = t^r P(t)^n and L * H * N = t^n OR (i b) same as (i a) but with G and H, m and n interchanged OR (ii) L * G * M = T_m(t) and L * H * N = T_n(t). where P is a suitable polynomial, r is a natural number and T_d is the normalized Chebyshev polynomials of degree d. " We can work w.l.o.g. on algebraically closed fields, The most important case is when the characteristic is zero. The reason lies in the results of Siegel and Faltings that say (better, imply) that if the curve G(Y)=H(X) has rational coefficients and infinitely many rational points, then it has genus at most 1, and if the points have bounded denominators, then the curve has genus 0. In the latter case, Ritt's #2 provides the classification of the polynomials, and it is called "variable separated polynomials: the genus 0 problem" (by Fried and others), but until a few months ago there was no result about the genus 1 case. Also because interest was small. Recently interest has arisen (here my english might be broken) due to some results of Fried, Mueller and Bilu on the old question of Davenport, Lewis and Schinzel, asking for which G, H without restriction on the degrees the polynomials G(Y)-H(X) is irreducble/reducible. These results culminated in a very recent work of Cassou-Nogues and Couveignes giving the classification and explicit factorisations: their machinery includes Dessins d'Enfants and also the (infamous) Classification of Simple Finite Groups. Incidentally CFSG is the briad area where my research work in the IEM fits (loosely). Zannier and I embarked about one year ago in the genus 1 problem. In the meantime I learned a lot, I am only a hobbyist in number theory, but I pursue this research in order to have a broader view of mathematics when (and if) I will be a mature mathematician. Now, we have the complete classification of such polynomials (variable separated, coprime degrees, genus 1) over an algebraically closed field of char 0. This work is already known to some "insiders" like A. Schinzel, M. Fried, Yuri Bilu, Debes, Peter Mueller, and it is also cited in some works or surveys currently being written. Then Zannier told me "good job on the algebraic case [in his language, and in the language of Schinzel's book on polynomials it means over C] but now, I want you to do the arithmetic case [over Q] and maybe the number case [number fields] and we make one big work". We ruled out the third case for the moment, and I was only very recently able to find the right approach to the arithmetic case. The answer (over C, say) to our problem is of course more complicated than Ritt's #2 and gives us a few families of polynomials (infinite familes) plus some sporadic pairs (which was a surprise). In the arithmetic case, ALL the polynomials can be obtained by composition with linear functions and Cebyshev polynomials of a FINITELY described set of polynomials or parametrised polynomials of bounded degree. I do not want to enter into much detail, but we are led to polynomials which are in one-to-one corrispondence with the torsion points of nonsingular elliptic curves. For every curve E with a torsion point P of order N there is a polynomial, say A_{E,P}(x) of degree N/gcd(N,2). If we restrict ourselves to the rationals then N can be 2,3,4,5,6,7,8,9,10,12 by Mazur, and our other conditions can allow us exclude N=2,3. Then N/gcd(N,2) can be 2,3,4,5,6,7,9. I have no problems for degrees up to 6, I have small ones for degree 7 (but I think I can do it easily) and had a bog problem for degree 9. The rational points on the affine curve below parametrize the polynomials of degree 9. > >I am stuck on this problem. I have to find the rational values of the > >parameter a such that the following polynomial in Z > > > > 3 2 2 2 2 > > Z + 18 ( 5 a - 14 a + 2 ) Z + 81 (19 a - 10 a + 1) (3 a - 10 a + 5) Z > > > > 5 4 3 2 > > - 486 (8 a - 1) (16 a + 233 a - 92 a + 10 a - 4 a + 1) > > > >has at least one rational root. This should provide the answer to your question > I can't imagine how such an equation came up in the first place. and also to another question you posed in another email (I could not reply because I have been in Eindhoven one day). The other approach (the "points of order 9 on the curve" one) worked but at the end I obtained the same equation. From which I could infer thet it has a one-parameter set of solutions. But I could not find them. Now, If it does not bother you too much, how did you get the parametrization ? and how did you compute the genus of that curve (from which one infers that a rational parametrisation like the one you provided must exist) ? In the meantime I will ponder your answer carefully. I am sure I can learn a lot. Thenk you again. Best regards Roberto _/_/ Roberto Maria Avanzi /_/ Institut f=FCr Experimentelle Mathematik / Universit=E4t GHS Essen _/ Ellernstra=DFe 29 / 45326 Essen / Germany / Phone: ++49-201-183-7637 Fax: ++49-201-183-7668 ============================================================================== From: Dave Rusin Date: Fri, 18 Sep 1998 07:44:53 -0500 (CDT) To: mocenigo@exp-math.uni-essen.de Subject: Re: Diophantine Problem Thank you for your interesting email. I would like to spend some time thinking about this carefully! I should tell you that my thesis adviser many years ago was George Glauberman, so the fact that finite simple groups are involved is very interesting to me! After I am done teaching today I will respond in more detail to your question about that genus-0 curve, but let me just comment now that >Now, If it does not bother you too much, how did you get the >parametrization ? and how did you compute the genus of that curve (from >which one infers that a rational parametrisation like the one you provided >must exist) ? this can be done with a Maple package called CASA. You can find it at ftp.risc.uni-linz.ac.at . I would not have pursued the question in your most recent post if CASA had not told me the genus was zero! dave ============================================================================== Date: Fri, 18 Sep 1998 15:16:26 +0200 (MESZ) From: Roberto Maria Avanzi To: Dave Rusin Subject: Re: Diophantine Problem On Fri, 18 Sep 1998, Dave Rusin wrote: > Thank you for your interesting email. I would like to spend some time > thinking about this carefully! I should tell you that my thesis adviser > many years ago was George Glauberman, so the fact that finite simple groups > are involved is very interesting to me! I see. But finite simple groups are in Cassou-Nogues - Couveignes work, not mine-and-Zannier's :( However, I think that also our work is interesting. > After I am done teaching today I will respond in more detail to your > question about that genus-0 curve, Thats' very nice. In the weekend I will go to the computer facilities of the uni to do some work, and I am eager to read your comments. > but let me just comment now that > > > Now, If it does not bother you too much, how did you get the > > parametrization ? and how did you compute the genus of that curve > > (from which one infers that a rational parametrisation like the one you > > provided must exist) ? > > this can be done with a Maple package called CASA. You can find it at > ftp.risc.uni-linz.ac.at . I would not have pursued the question in your > most recent post if CASA had not told me the genus was zero! I can understand. Genus 1 parametrisations are a mess (you basically have to give the variable change leading to the Weierstrass form, which I infer is _practically_ desperate, and then you can already express the points as rational functions of the points of the elliptic curve, which is no rational parametrization at all indeed, I've done it in a few simple cases (by hand !) during my graduate studies and it easily gets obscene). Higher genera, I do not want even start to think... I downloaded CASA and I asked one of the system administrators to install it into the share library of maple. I hope in the next few days I will be able to use it (which is not a sure thing, knowing out sysadms and IBM's broken filesystems - at least they do NOT use NT). Best ! Roberto ============================================================================== From: Dave Rusin Date: Fri, 18 Sep 1998 10:42:24 -0500 (CDT) To: mocenigo@exp-math.uni-essen.de Subject: Re: Diophantine Problem Here are some pointers showing how I parameterized this curve. First, let me note that the CASA software informs me that the genus is zero; if that were not true I would probably not have pursued the problem at all. Note too that CASA's algorithm for determining the genus presumably goes through a resolution of the singularities; this process can be used constructively (as I understand it) to actually produce a birational equivalence with P^1. I don't really remember the details of "blowing up" and so on, so I proceed with somewhat ad-hoc arguments. But these tricks work fairly well to simplify many curves. I'll work with Maple; the script below proceeds directly in MapleVR4. I will think more carefully about the problem at the heart of your project and write to you again later. dave ############################################################################## #I like to use this procedure instead of "collect" #because it factorizes the coefficients. coll:=proc(f, x) convert(factor(taylor(f, x, degree(f, x) + 1)), polynom) end: #OK, here's a Maple session. #Enter your equation eq:=z^3+18*(5*a^2-14*a+2)*z^2+81*(19*a^2-10*a+1)*(3*a^2-10*a+5)*z-486*(8*a-1) *(16*a^5+233*a^4-92*a^3+10*a^2-4*a+1); #It's a cubic in z, so my first instinct is to eliminate the z^2 term. z0:=coeff(eq,z,2); eq2:=coll(subs(z=y-z0/3,eq),y); #Let's look for a reasonable translation of the other variable, too factor(discrim(eq2,y)); #Something interesting happens at a=-1, so we set eq3:=coll(subs(a=-1+b,eq2),y); #So much for translations, how about scalings? factor(discrim(eq3,y)); #suggests that perhaps b should be divisible by 6? eq4:=coll(subs(b=6*c,eq3),y); #Now there are lots of 6's in [seq(ifactor(content(coeff(eq4,y,i))),i=0..1)]; #So we try scaling y by 6's too. eq5:=coll(subs(y=36*x,eq4/36^3),x); #Looking better already! #Now we return to the issue of translation in x. Recall I had tried to #avoid any x^2 term, but maybe that's not idea. factor(discrim(eq5,c)); #Clearly shows x=-1 is interesting. So let's translate there. eq6:=coll(subs(x=-1+w,eq5),w); #Aha! Now we see it looks like c (and 3, too) should divide u. This #suggests a _quadratic_ transformation eq7:=coll(subs(w=3*c*v,eq6/9/c^2),v); #This is getting simpler! #OK, now there are still singularities at e.g. (0,0). To determine the #nature of those singularities, look at the terms of lowest degree (quadratic) eq8:=coll(subs({c=c*lambda, v=v*lambda},eq7),lambda); #This shows that we should rotate our coordinates to focus on v-19c. eq10:=coll(subs(v=19*c+u,eq7/3),u); #Another "Aha!": clearly c should divide u, too! eq11:=coll(factor(subs(u=c*t,eq10))/c^2,t); #Well, now, we've done a non-diagonal linear change of variables, and some #nonlinear ones, too. Maybe it's time to try translations and scalings again. factor(discrim(eq11,c)); #Certainly something interesting happens at t=-12, so we do eq12:=coll(subs(t=-12+s,eq11),s); #Again we see some interesting factors of s, so we try eq13:=coll(subs(s=r*(5*c-1),eq12/(5*c-1)^2),r); #At this point I have worked so hard to remove singularities at the origin #that I have higher-degree polynomials only for high powers of r. Let's #turn that back around: eq14:=coll(subs(r=1/q,eq13*q^3),q); #This is very good! We now have a polynomial of degree 3 in two variables. #This certainly represents a curve of degree 0 or 1. Moreover, it is #standard elliptic-curves material to try to move a cubic into Weierstrass #normal form. See e.g. Cassel's book. #However, here I just use some old tricks again: factor(discrim(eq14,q)); eq15:=coll(subs(c=d+1/5,eq14),q); eq16:=coll(subs(q=d*p,expand(eq15/d)),d); #But wait -- now we see that this new variable d appears only to the #second degree. Thus (q,d) is a rational point on the square iff q makes #the discriminant of this quadratic a square. coll(eq16,d); #Not very interesting discrim(eq16,d)/4; #Very interesting! expand("); # 2 # -3 p + 5 p #So we see that the curve is indeed birationally equivalent to the conic # -3*p^2+5*p=e^2 ! Clearly one rational point here is (0,0); we then #parameterize the whole curve in the usual way: p=0+s, e=0+t*s imply #p = 5/(3+t^2). #So we should be able to parameterize the original curve in terms of t alone. pp:= 5/(3+t^2); dd:=factor(solve(subs(p=pp,eq16),d)[1]); #Now work backwards to get qq,rr,ss,tt,uu,vv,ww,xx,yy,zz and cc,bb,aa. cc:=factor(dd+1/5); qq:=factor(dd*pp); rr:=factor(1/qq); ss:=factor(rr*(5*cc-1)); tt:=factor(-12+ss); uu:=factor(cc*tt); vv:=factor(19*cc+uu); ww:=factor(3*cc*vv); xx:=factor(ww-1); yy:=factor(36*xx); bb:=factor(6*cc); aa:=factor(-1+bb); zz0:=factor(subs(a=aa,z0)); zz:=factor(yy-zz0/3); #Now it's time to check our work: factor(subs({a=aa,z=zz},eq)); #We get 0 -- hurray! our parameterizaton a=aa, z=zz is correct. #Now, you might be concerned that we simply have a map t |-> (a,z) which #maps into the variety; but is it one-to-one and onto? In other words, can #we recover the parameter from (a, z)? readlib(eliminate): eliminate({z=zz,a=aa},t); #Another double check, first: expand("[2][1]-eq); # "[2][1] should be equal (or scalar multiple of) eq inv:=""[1][1]; #That gives the inverse map (a,z) |-> t. #Oh yes, the denominator -- where is it zero on the curve? eliminate({denom(rhs(inv)),eq},z); #shows only a=-1, a=1/5 are a problem, and likewise eliminate({denom(rhs(inv)),eq},a); #shows there are THREE bad z-coordinates (I missed one before). #The bad points are (a,z)= (-1,-54), (-1,-162), (1/5, -54/25). ============================================================================== From: Dave Rusin Date: Fri, 18 Sep 1998 11:42:08 -0500 (CDT) To: mocenigo@exp-math.uni-essen.de Subject: Re: Diophantine Problem I have had a chance to consider some of your explanation of your project. This sounds very interesting! >RITT's #2: If G(t) and H(t) are polynomials of coprime degrees m and n >over a field k s.t. char(k) does not divide mn and the curve G(Y)=H(X) >(which is irreducible by results of Tverberg) has genus 0, then for >suitable nonconstant linear functions L,M,N we have that Presumably if we drop the condition gcd(m,n)=1 we get a similar conclusion with L, M, N now allowed to be non-linear? I suppose this result is related to a theorem (which I cannot completely remember) that if P, Q are polynomials with P*Q=Q*P then P, Q are powers of t, or else Tchebyshev polynomials (or something like that)? >...for which G, H without restriction on the degrees >the polynomials G(Y)-H(X) is irreducble/reducible. These results >culminated in a very recent work of Cassou-Nogues and Couveignes >giving the classification and explicit factorisations These must be too new for Math Reviews. Do you know if preprints exist on a preprint server or something? I would really like to investigate the connection with finite group theory. I am only dimly aware of what les dessins des enfants are all about. Perhaps this is a good time to learn! >Now, we have the complete classification of such polynomials (variable >separated, coprime degrees, genus 1) over an algebraically closed field of >char 0. Again I would be interested in learning of some details. Here is something for you to consider: you have a procedure which translates torsion points on elliptic curves to information about polynomials. Suppose this process could be reversed: from information about polynomials you could obtain some restrictions about the possible torsion on elliptic curves. This would be very useful -- it is conjectured that there is an extension of Mazur's theorem to any number field, but this is not known to be true. So the connection between torsion on curves and classifications of polynomials is possibly of great interest. (I assume the sporadic examples from finite group theory are not related to this, but instead give a separate connection to polynomials; but if the groups _are_ related to the elliptic curves -- what fun!) >I pursue this research in order to have a broader view of mathematics >when (and if) I will be a mature mathematician. Of course this is good. I recently took the project of considering the broadest view possible; I got a "picture" of mathematics. See welcome.html dave ============================================================================== Date: Wed, 23 Sep 1998 16:29:43 +0200 (MESZ) From: Roberto Maria Avanzi To: Dave Rusin Subject: Re: Diophantine Problem Dear Prof. Rusin First of all I want to thank you again for your help. I remembered at once the foggy discussion made some years ago in a undergraduate course in Padova about Blowing-Up of singularities, then I embarked on pondering what you have sent me. Reworking my problem, I found that my polynomials are better parametrised by looking at the diophantine equation 5 4 3 2 4 3 2 -(7 a - 1) (7 a - 5 a + 40 a - 10 a - 7 a + 2) + (36 a + 66 a + 108 a - 90 a + 15) b 2 2 3 + (-15 a - 33 a + 12) b + 3 b and following your reasoning I started translating, scaling, rotating and blowing-up singularities until I got this parametrisation 3 t + 3 t - 2 a = ------------------- 3 2 t - 3 t + 6 t - 5 5 3 2 (2 t - 1) (t + 2 t + t + 3 t - 4) b = 3 ------------------------------------ 3 2 2 (t - 3 t + 6 t - 5) which works for t different from -1,1/2 and 2. I think I got the smallest coefficients possible or anyway very close. A minute achievement, but it is nontheless the last step in a more important thing. I keep this parametrisations because it leads itself to be connected later with the Tate form for elliptic curves with a torsion point of order 9 (clear later). This is a little messy. I say now WHICH polynomials I was hunting. I was looking at polynomials F(x) such that F(x) = x (x^4 + 9*(a*x^3+b*x^2+c*x+d) )^2 F(x)+L = q(x) p_0(x)^2 where q and p_0 are degree 3 polynomials in particular is H(x) = (4/L * F(X)) + 2 then we get H(X)^2 - 4 = x q(x) * square of a polynomial R of degree 7 mmmm, if we have something like C(x)^2 - 4 = poly of degree 2 with distinct roots * square of poly then C(x) is a normalised chebyshev poly. Well, our H(x) is related (also for degrees other than 9) to elliptic integrals and weierstrass functions in a similar way to the relation between chebyshev polynomials and trigonometric functions (and their inverses). in particular q(x) must be square free. (if it has a double root, we get Chebyshev polynomials again, more or less). now, I use maple to get a relation between a, b, c, d above (multiplied some by 9 to get better coefficients later). a and b are related by the equation above (which must be solved over the rationals because we are looking at rational solutions), then there are expressions for c and d as functions of a and b. so we get everything in only one parameter (we must take into account also t=infinity (slightly imprecise) because the substitution t=1/s also leads to a solution when s=0 - the parametrisation is on the projective line, not the affine). by the way, we can consider the H above as well as R as functions on the elliptic curve E: Y^2 = X^3 q(1/X) =: r(X) and look at their divisors we infer that the points (0,+-1) on it have order 9 (the degree of H). On the other hand, by Abel-Jacobi the converse is also true - if there are such points we have polynomials, if the points are rational, then also our coefficients are. with the above coefficients, the curve E is immediately brought to Tate form with the coefficients already in the usual (I should have known it, grrrr) parametrization. At the moment I do not need it, but it is nice to show it. if the order of (0,1) is _exactly_ 9 the polys are (in a weak sense) indecomposable and we call them primitive solutions. if the order of the point is 3, then we show that mf+n = T_3(g(x)) where m,n are constants, T_3 is the chebyshev polynomial of degree 2 and g(x) is a primitive solution for the analogous problem in degree 3. cute, isn't it ? good. done it time ago and the work is already known (only a draft of a pre-pre-preprints available, but I hope a real preprint will be available soon). now we have that for all t's except the three values above, and also t= infty (say t\in P^1_Q) f = x ( x^4 + 9 a(t) x^3 + 9 b(t) x^2 + 9 c(t) x + 9 d(t) ) ^2 are the solutions (for all of them. these are all of them, indeed the case equivalent to f= x ^ 3 * (square of a poly of order 3) seems to be considered yet, but it can ruled out since we can show that these polynomials (for the right coefficients) come from torsion points of order 3 on the curve, and thus are not primitive solutions. I hope I answered your questions. We use Mazur's result here and there in a crucial way. So I do not expect that we can use the polynomials to prove extensions of it. Nontheless, the classification of polynomials as above is now of interest (I do not dare say "great" interest, but there is ferment). We got it (I do not owe only the _problem_ itself to Zannier, but also the way of attacking the algebraic part, he also proved a few lemmas better than me - after all is a 4-hand work and I am quite proud to work with him). Last, the connection between Ritt's #2 and the result about P*Q = Q*P ==> that the polys are cyclic or Chebyshev is a known fact. The original formulation of Ritt's #2 is something like "the polys G,H,A,B such that |G|=|B|=m, |H|=|A|=n with m and n coprime and such that G(A)=H(B), are given by ...." this leads immediately to the genus 0 formulation of the Theorem. I find this area of number theory quite fascinating, in particular because it deals with "concrete" mathematical objects (one might object that a sheaf in not more abstract than a straight line, because both are mathematical objects, and the line IS a sheaf, albeit very simple, but somehow I feel NOT to interested in things like Drinfeld modules and the like which I look at suspiciously...) Thanks for your attention. Roberto _/_/ Roberto Maria Avanzi /_/ Institut f=FCr Experimentelle Mathematik / Universit=E4t GHS Essen _/ Ellernstra=DFe 29 / 45326 Essen / Germany / Phone: ++49-201-183-7637 Fax: ++49-201-183-7668