User paul monsky - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:33:12Z http://mathoverflow.net/feeds/user/6214 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131653/does-this-space-of-mod-2-modular-forms-admit-a-z-8-degree-decomposition Does this space of mod 2 modular forms admit a (Z/8)* degree decomposition? paul Monsky 2013-05-23T21:28:29Z 2013-05-23T21:28:29Z <p>Fix an odd N>0. Let M consist of all odd elements of Z/2[[x]] that are the mod 2 reductions of elements of Z[[x]] arising as the Fourier expansions of modular forms for (Gamma_0)(N); it's easy to see that M is closed under addition. For i in {1,3,5,7} let M_i be the subspace of M consisting of those g in which all the exponents that appear are congruent to i mod 8.</p> <p>QUESTION: Do M_1, M_3, M_5 and M_7 span M?</p> <p>Remark 1: When N=1 the answer is yes. For in this case M is spanned by the odd powers of x+x^9+x^25+x^49+..., and each such power lies in some M_i.</p> <p>Remark 2: When N is prime, I think this will follow from an affirmative answer to another question I've posted on MO--Level p characteristic 2 modular forms and thetas (#121506). But I haven't tried to write things out, and this approach seems awfully complicated. </p> http://mathoverflow.net/questions/127777/a-subring-of-the-serre-swinnerton-dyer-ring-of-level-n-modular-power-series A subring of the Serre Swinnerton -Dyer ring of level N modular power series paul Monsky 2013-04-17T02:05:57Z 2013-05-10T01:10:51Z <p>Suppose ell is prime and (N,ell)=1. Consider those power series over Z that are expansions at infinity of modular forms for gamma_0 (N) of weight a multiple of ell-1. I'll say that an element of (Z/ell)[[x]] is in M#(N) (or more briefly in M#) if it is the mod ell reduction of such a power series.</p> <p>When ell is 2 or 3 there is no weight restriction, and so M# is just the Serre Swinnerton-Dyer ring, M, of question 93059. When ell>3 we can use the fact that the expansion of E_(ell-1) is 1 mod ell to see that M# is closed under addition, and is a subring of M.</p> <p>I'd like to know the structure of M#(N), or more geometrically what the affine curve C(N) over Z/ell attached to M#(N) looks like. My guess is that there's a simple answer in terms of the characteristic ell non-singular projective modular curve X_0 (N) constructed by Igusa.</p> <p>Explicitly let J be the mod ell reduction of the "Laurent expansion" (1/x)+744+... of j(z). Let K(1) be the extension field of Z/ell generated by J(x), and let K(N) be the field of Laurent series generated over Z/ell by the J(x^d) with d dividing N. If my understanding is correct, X_0 (1) is the projective j-line over Z/ell, there is a branched covering phi: X_0(N)-->X_0(1) defined over Z/ell, and the function fields of X_0(1) and X_0(N) identify with K(1) and K(N).</p> <p>Now let P be the monic separable element of Z/ell[x] whose roots are the supersingular j-values. Then it can be shown that M(1) is the subring of (Z/ell)(J) generated by the (J^k)/P(J) with k&lt; deg P. In other words, C(1) is the "ordinary part" of X_0(1); it is the projective j-line with the supersingular j omitted.</p> <p>QUESTION: Is it true that C(N) identifies with the inverse image of C(1) under phi: X_0(N)-->X_0(1)? Alternatively (assuming my description of the function field of X_0 (N) is correct), is it true that M#(N) is the integral closure of M#(1) in the extension field generated by the J(x^d) with d dividing N?</p> <p>EDIT: I've corrected some typos, where I wrote X(N) when I meant X_0(N). Kevin Buzzard, in response to an inquiry, has kindly told me that the answer to my question is yes, and indicated a proof.</p> http://mathoverflow.net/questions/128036/solutions-to-binomn5-2-binomm5/128059#128059 Answer by paul Monsky for Solutions to $\binom{n}{5} = 2 \binom{m}{5}$ paul Monsky 2013-04-19T09:19:31Z 2013-04-21T14:41:09Z <p>This isn't a complete answer, but the problem "reduces" to finding the finitely many rational points on a certain genus 2 hyperelliptic curve. This is often possible by a technique involving a reduction to finding the rational points on a finite set of rank 0 elliptic curves--see for example "Towers of 2-covers of hyperelliptic curves" by Bruin and Flynn in Trans. Amer. Math. Soc. 357 (2005) #11 4329-4347.</p> <p>In your case, the curve is u^2= 9*t^6+16*t^5-200*t^3+256*t+144. There are the following 16 rational points: (t,u) or (t,-u)= (0,12),(1,15),(2,12),(4,204),(-1,9),(-2,36),(-4,180) and (7/4,411/64). If these are the only rational points then the only non-trivial solution to your equation is n=10,m=9. To see this suppose that n(n-1)(n-2)(n-3)(n-4)=2m(m-1)(m-2)(m-3)(m-4). Let y=(n-2)^2 and x=(m-2)^2. Squaring both sides we find that y*(y-1)(y-1)(y-4)(y-4)=4x*(x-1)(x-1)(x-4)(x-4). Suppose y isn't 0. Then 4x/y is t^2 for some rational t with (y-1)(y-4)=t(x-1)(x-4). We replace y by 4x/t^2 in this equation and find that (t^5-16)x^2-(5t^5-20t^2)x+(4t^5-4t^4)=0. So this last quadratic polynomial in x has a rational root and its discriminant is a square. This gives the hyperelliptic curve above. Note that the case n=10, m=9 of your problem corresponds to the point (7/4,411/64) on this curve.</p> <p>EDIT: More generally one can look for rational m and n with [n]_5= 2*([m]_5). If (t,u) is a rational point on the hyperelliptic curve with t non-zero, set x=(5t^5-20*t^2+t^2*u)/(2*t^5- 32) and y=4x/t^2. Then if x is a square in Q, one gets such an m and n with m=2+(a square root of x) and n=2+(a square root of y). Joro's points lead in this way to the solutions (n,m)=(10,9),(10/3,5/3) and (78/23,36/23), the last one being rather unexpected. (And as Francois notes, each (n,m) gives a second solution (4-n,4-m)). Perhaps these solutions and the trivial solutions with m and n in {0,1,2,3,4} are the only rational solutions.</p> http://mathoverflow.net/questions/125877/computing-certain-class-numbers-modulo-4/125904#125904 Answer by paul Monsky for Computing certain class numbers modulo 4 paul Monsky 2013-03-29T11:27:33Z 2013-03-29T12:17:46Z <p>The way Gauss did things was in terms of SL(2,Z) equivalence classes of (primitive) binary quadratic forms over Z. So lets consider such definite forms, axx+bxy+cyy with b^2-4ac=-N. For simplicity suppose N is squarefree and odd. Gauss defines a composition on the classes, making the set of classes into a finite group. For each prime dividing N he defines a genus character from this group to Z/2. The product of these characters is the trivial map, and the joint kernel of them all consists of the squares, a subgroup that he calls the principal genus. In the case N=pq, the character attached to p maps a form to (M/p) where M is any integer prime to p represented by the form. So the squares form, in your case, a subgroup of index 2, and there is a unique non-trivial class of order 2 in the group. Gauss calls the classes of order 2 the "ambiguous classes". They're easily written down in general and are represented by "reduced" forms axx+bxy+cyy with b=0 or with a=b (or c). So in your case the non-trivial ambiguous class is represented by pxx+qyy. The genus character attached to p maps this class to (q/p). So the class is a square if and only if (q/p)=1, giving the result you want. The theory also works for even N not necessarily square-free, though there are 2 genus characters attached to the prime 2 when 32 divides N, and it works for indefinite forms as well. </p> <p>Oops--I should have said that the non-trivial ambiguous class is represented by qxx+qxy+(1/4)(p+q)yy.</p> http://mathoverflow.net/questions/125007/does-there-exist-a-half-integer-weight-theta-function-which-is-is-equivalent-to-1/125128#125128 Answer by paul Monsky for Does there exist a half-integer weight theta function which is is equivalent to 1 modulo 4? paul Monsky 2013-03-21T05:33:01Z 2013-03-25T00:48:33Z <p>This is nothing like a complete answer(EDIT-now it seems to be--see below), but it may suggest a fruitful attack, based on the comment (see below for a version incorporated into this answer) that I made earlier. Your question is related to my question 124243 on this site about a certain ring M consisting of the mod 2 reductions of elements of Z[[x]] that are the Fourier expansions of modular forms for gamma_0 (N). Suppose first that N is an odd prime p.</p> <p>Theorem(?) The element g=x+x^4+x^9+x^16+... of Z/2[[x]] does not lie in M.</p> <p>Proof(?) In the notation of my question, and the other question of mine that it links to, let h=g+g^2+r. Then h(1+h)=(g+g^4)+(r+r^2)=f1+(f1+fp)=fp.(Again, see below). Now I conjecture in my question that in the field of fractions of M, fp has a zero of order p, a zero of order 1 and p+1 poles counted with multiplicity. If this is true then both h and h+1 have (p+1)/2 poles, counted with multiplicity. Furthermore one of them must have a zero of order p, giving a contradiction.</p> <p>Now I'm reasonably sure that this conjecture is correct, is known to the experts, and follows from results proved by Igusa in the 1950's. Unfortunately no expert has yet responded to my question. I also think that similar techniques will show that g is not the mod 2 reduction of any modular form for gamma_0 (N) when N is odd. But the theory, if I understand correctly, is much harder when the level is divisible by the characteristic, and if a proof is to be given based on the idea of my comment this case has to be considered.</p> <p>Edited comment--Suppose that F is a positive definite quadratic form over Z in s variables where s is odd. Here's an idea for showing that the power series theta_F in Z[[x]] attached to F (I'll defy convention and write x instead of q) cannot be congruent to 1 mod 4. Choose S so that s*S=8m-1 and let G be a direct sum of S copies of F and one of the one variable form z^2. Then if theta_F is 1 mod 4, theta G is 1+2x+2x^4+2x^9+2x^16+... mod 4. Now (see Schoeneberg's paper from the 1930's for example) there is an N such that theta_G is the "expansion at infinity" of a weight 4m modular form, phi, for gamma_0 (N). Then (1/2)(phi-(E_4)^m) is a weight 4m modular form for gamma_0 (N) whose expansion at infinity lies in Z[[x]], and has mod 2 reduction equal to g=x+x^4+x^9+x^16+... There are perhaps reasons to believe this can't happen--at any rate it appears to be an approachable problem of a mainstream variety.</p> <p>One more remark. In my proof? given above I use the fact that the mod 2 reduction f1 of the expansion at infinity of the cusp form delta is x+x^9+x^25+... This lovely if well-known fact comes from<br> the infinite product formula for delta and Jacobi's triple product identity.</p> <p>EDIT---MARCH 24 I now believe I have an argument showing that g is not the mod 2 reduction of a modular form for any gamma_0 (N), completing the answer to your question. I'll dispense with any ideas of Igusa, and instead use Hecke operators and classical results of Gauss on ternary quadratic forms. The idea is this. For each odd prime q there is a "Hecke operator" T_q on Z/2[[x]]. If h is the reduction of a modular form for gamma_0 (N) then the T_q(h) span a finite dimensional subspace of Z/2[[x]]. Now let h=g^11. I'll use this last result to show that h cannot be the reduction of a modular form.</p> <p>To this end, fix K and primes p_1,...p_K each of which is 5 mod 8. Then choose primes q_1,...q_K each of which is 7 mod 8 so that the Legendre symbol (q_i/p_j) is -1 if i=j and is 1 otherwise. Now the coefficient of x^p_j in T_(q_i)(h) is the coefficient of x^(p_j)*(q_i) in h. By the last paragraph of and the comments following my answer to MO question 284642-why are there usually an even number of representations as a sum of 11 squares--(an answer based on Disquisitiones Arithmeticae)--, this coefficient is 1 if i=j and 0 otherwise. So the T_(q_i) (h) are linearly independent. Since K may be chosen arbitrarily large, the criterion of the last paragraph gives the result.</p> http://mathoverflow.net/questions/124243/are-these-empirical-discoveries-about-the-serre-swinnerton-dyer-ring-of-prime-lev Are these empirical discoveries about the Serre Swinnerton-Dyer ring of prime level modular power series actual theorems? paul Monsky 2013-03-11T16:39:13Z 2013-03-19T12:35:04Z <p>In <a href="http://mathoverflow.net/questions/93059" rel="nofollow">this question</a> Joel Bellaiche constructed an algebra, M, of modular forms for gamma_0 (N) in finite characteristic (which he called p, but I'll call ell) and asked to know its structure. Matt Emerton gave a "comment-answer" that showed, among other things, that M is integrally closed.</p> <p>I've made some explicit calculations (particularly when ell=2 and 3; see <a href="http://mathoverflow.net/questions/121506" rel="nofollow">my question</a> for some of these when ell=2). The empirical "discoveries" unveiled by these seem likely to be theorems. I'll present a couple of these here, saving the remaining ones for edits. I'll assume that the level N is a prime, p. M contains f1 and fp; the mod ell reductions of the Fourier expansions of the cusp forms delta(z) and delta(pz).</p> <p>Discovery 1---M is integral over Z/ell[f1,fp]</p> <p>Discovery 2---When ell=2 or 3, then M is the integral closure of Z/ell[f1,fp] in its field of fractions.</p> <p>Are these really true? I'll also hazard the following guess when l>3. The mod ell reductions of the Eisenstein series E_4 and E_6 generate an extension of Z/ell(f1,fp) of degree (l-1)/2, and M is the integral closure of Z/ell[f1,fp] in this extension.</p> <p>EDIT: To describe further observations I'll introduce some notation. If k is even and non-negative, M[k] will be the Z/ell subspace of M consisting of the mod ell reductions of modular power series in Z[[x]] corresponding to weight k forms for gamma_0 (p). C will be a non-singular projective curve over Z/ell with function field the field of fractions of M, and D will be the divisor of poles of the element fp of M[12].</p> <p>For example, when ell=2 and p=11, then in the notation of my question referenced above, M[k] has dimension k, while M<a href="http://mathoverflow.net/questions/121506" rel="nofollow">2</a> is spanned by 1 and t, and M[4] is spanned by 1,t,t^2 and r. We have the relation r^2+r=t^3+t, and C is the curve r^2+r=t^3+t with the point O at infinity adjoined. f11=r^3+r^4+t^3 has zeros of orders 11 and 1 at (t,r)=(0,0) and (0,1) and D=12(O). Furthermore M[12] is spanned by 1,t,t^2,t^3,t^4,t^5,t^6,r,t*r,t^2*r,t^3*r and t^4*r and is the complete linear series attached to the divisor D.</p> <p>Here's what I think is true in general. Suppose ell=2. Then:</p> <p>1---fp has one zero of order p, and one of order 1 on C.</p> <p>2a--When p is 11 mod 12, fp has (p+1)/12 poles of order 12.</p> <p>2b--When p is 5 mod 12, fp has 1 pole of order 6 and (p-5)/12 of order 12.</p> <p>2c--When p is 7 mod 12, fp has 2 poles of order 4 and (p-7)/12 of order 12. </p> <p>2d--When p is 13 mod 12, fp has 1 pole of order 6, 2 of order 4 and (p-13)/12 of order 12.</p> <p>3---For each k there is a divisor D, easily describable in terms of D and k such that M[k] is the complete linear series attached to this divisor; in particular D&lt;12m>=m(D).</p> <p>I think that entirely similar results hold when ell=3. My belief is that all of this is known, but I'd appreciate proofs and/or references.</p> http://mathoverflow.net/questions/121506/level-p-characteristic-2-modular-forms-and-thetas Level p characteristic 2 modular forms and thetas paul Monsky 2013-02-11T17:17:04Z 2013-03-04T21:29:24Z <p>BACKGROUND</p> <p>Let p be an odd prime. An element of Z/2[[x]] is "modular of level p" if it is the mod 2 reduction of a g in Z[[x]] with g the Fourier expansion of a modular form for gamma_0(p). In various cases when f is modular of level p then f(x^p) can be expressed as a polynomial of a special shape in "characteristic 2 theta series".</p> <p>I've developed a connection between modular function theory and theta series. Namely let [i] in Z/2[[x]] be the sum of the x^(n^2), where n runs over the integers that are congruent to i mod p. Then the field of fractions of the ring generated over the algebraic closure of Z/2 by these theta series identifies with Igusa's field of modular functions for gamma(p). But computer calculation suggests further connections, involving modular forms for gamma_0(p). Here are some examples:</p> <p>1.<em>_</em> If f is x+x^9+x^25+x^49+..., the mod 2 reduction of the expansion of delta, then f(x^p) lies in the ring B generated over Z/2 by the [i]. Indeed if we let B' be the subring of B consisting of those elements that are power series in x^p, and are fixed by the automorphisms of B taking [i] to [ni] when n is in (Z/p)*, then f is in B'. (Proofs are in my MO questions and answers).</p> <p>2.<em>_</em> If the exponents appearing in the element r of Z/2[[x]] are just the products of the non-zero squares by 1,2,p and 2p, then r is the mod 2 reduction of the expansion of a weight 4 Eisenstein series, and once again r(x^p) is in B'. (Again, proofs are in my MO questions and answers).</p> <p>3.<em>_</em> If p=7 and s is the modular power series of shape x^2+... coming from a weight 4 modular form, then s(x^7)=[1][2][3], and so is in B'.</p> <p>4.<em>_</em> If p=11 and t=x+... comes from the weight 2 cusp form, then t(x^11) is the sum of [1][1][3]+[2][2][5]+[4][4][1]+[3][3][2]+[5][5][1] and [1][1][2][4]+[2][2][4][3]+[4][4][3][5]+[3][3][5][1]+[5][5][1][2], and so lies in B'.</p> <p>QUESTION</p> <p>I can show that when p is 3,5,7 or 11, then f is modular of level p if and only if f(x^p) lies in B'. To what extent does this generalize to larger p?</p> <p>EDIT: It seems likely that the answer to my question is always yes--f is in the ring A of level p characteristic 2 modular power series if and only if f(x^p) is in the subring, B', of the ring B generated by the "theta series". In this edit I'll present further conjectures, some in part known, that would imply this. In a later edit I'll give explicit descriptions of A and B' for p&lt;20. Fix an odd prime p.</p> <p>THE RING A</p> <p>A consists of the mod 2 reductions of all elements of Z[[x]] that are Fourier expansions of modular forms, of arbitrary weight, for gamma_0 (p). A is closed under multiplication. Using the fact that the expansions of the normalized Eisenstein series of weights 4 and 6 lie in 1+2*Z[[x]] we see that A is closed under addition as well, and is a ring. f1=x+x^9+x^25+..., and fp=f1(x^p) are, as I remarked, in A, and so Z/2[f1,fp] is a subring of A.</p> <p>Conjecture 1: A is the integral closure of Z/2[f1,fp] in its field of fractions. (There are 3 separate questions here. Is A integrally closed? Is A integral over the subring? And is the field of fractions of A equal to Z/2(f1,fp)? I suspect that the first, at least, of these is known. My investigations for p&lt;20 support the conjecture).</p> <p>THE RING B'</p> <p>Let L be the quotient of (Z/p)* by {1,-1}, and P be a polynomial ring over Z/2 in the variables x_i, with i running over L. There is a gradation of P by Z/p, with the "degree" of x_i being i^2. Also, L acts on P by permutation of variables with m taking x_i to x_mi, and the effect of m is to multiply degrees by m^2. Let P' be the subring of P consisting of L-stable elements of "degree" 0. For example when p=13, (x_1)(x_5)+(x_2)(x_3)+(x_4)(x_6) is in P', while when p=7 the same is true of (x_2)(x_1)(x_1)(x_1)+(x_3)(x_2)(x_2)(x_2)+(x_1)(x_3)(x_3)(x_3). Now if i is prime to p, let [i] in Z/2[[x]] be the sum of the x^(n^2) where n runs over the integers congruent to i mod p. [i] only depends on the image of i in L.</p> <p>We define B' to be the image of P' under the ring homomorphism P-->B taking x_i to [i]. There's a simple compact notation for elements of B' that I'll use when presenting my results. For example the image [1][5]+[2][3]+[4][6] of the p=13 element of the last paragraph will be called C(1,5) (or C(2,3) or C(4,6)), while that of the p=7 element is called C(2,1,1,1) (or C(3,2,2,2) or C(1,3,3,3)). Now the answers I've given to other questions on this site show that f1(x^p) and fp(x^p) lie in B'. Evidently my conjecture would follow from Conjecture 1 combined with:</p> <p>Conjecture 2: B' is the integral closure of the ring Z/2[f1(x^p),fp(x^p)] in its field of fractions.</p> <p>I find myself on firmer ground here--I think my MO answers come close to establishing Conjecture 2, though there may be some separability questions when p is 1 mod 4. The key to showing that B' is integrally closed should lie in the facts I've established about the curve attached to the ring B and the action of PSL_2(Z/p) on this curve.</p> <p>EDIT---EXPLICIT FORMULAS</p> <p>For each odd prime p&lt;20 I'll write down:</p> <p>(a). Explicit generators of A, the polynomial relations they satisfy, and a description of the affine curve attached to A. f1 and fp will be as in the last edit. r in A will be the reduction of the expansion x+... of that weight 4 Eisenstein series for gamma_0 (p) having a zero at infinity. Classical results show that r+r^2=f1+fp. I'll take r as one of the generators of A, and express fp (and consequently f1=r^2+r+fp) as polynomials in my generators.</p> <p>(b). Let R=r(x^p). More generally I'll use this lower case --> upper case convention in passing from an element of Z/2[[x]] to its image when x is replaced by x^p. For each generator g of A as given above, I'll give a formula for G in terms of thetas, using the notation of the last edit. When I have a particularly nice formula for fp(x^p) I'll give it as well.</p> <p>I start with the genus 0 cases.</p> <p>p=3</p> <p>(a) A=Z/2[r]. The curve is the affine line. f3=r^3+r^4</p> <p>(b) R=C(1,1,1)</p> <p>p=5</p> <p>(a) A=Z/2[r]. The curve is the affine line. f5=r^5+r^6</p> <p>(b) R=C(1,2)</p> <p>p=7</p> <p>(a) Let u be the reduction of the expansion x^2+... of that weight 4 form with an order 2 zero at infinity. (This expansion lies in Z[[x]]). Then A=Z/2[r,u], and r^2+ru+u^2+u=0. Since the affine curve y^2+xy+x^2+x=0 has, at infinity, two points conjugate over Z/2, our curve is obtained by removing such a pair of points from the projective line. f7=r^3+r^4+ru.</p> <p>(b) R=C(1,1,1,2)+C(1,2,3) U=C(1,2,3)</p> <p>p=13</p> <p>(a) Let u be the reduction of the rational weight 4 newform. Then A=Z/2[r,u] and u(1+r+r^2)=r+r^2. So A is generated by r and 1/(1+r+r^2) and the curve is the affine line with the points with x^2+x=1 removed. f13=(r^13)(1+r)(1+u)^4.</p> <p>(b) R=C(2,3) U=R+C(1,2,3,5)</p> <p>I next turn to genus 1:</p> <p>p=11</p> <p>(a) Let t be the reduction of the weight 2 newform. Then A=Z/2[r,t], and r^2+r=t^3+t. So the curve is an elliptic curve with the origin removed. f11=r^4+r^3+t^3.</p> <p>(b) R=C(1,1,3) T=R+C(1,1,2,4) and f11(x^11) is T*C(1,2,3,4,5)^2.</p> <p>p=17 </p> <p>(a) The Fourier expansions x-3*x^2+.. and x+9*x^2+... of the rational weight 4 newform and the weight 4 Eisenstein series vanishing at infinity are congruent mod 4. Let u be the reduction of (1/4)*(their difference). Then A=Z/2[r,u] and (1+r)(u^2+u)=r^2. Let x=1+r and y=ux. Then y^2+xy=x^3+x. So the curve is the affine curve y^2+xy=x^3+x with the point (0,0) removed. f17=(r^2+r)*u^8.</p> <p>(b) R=C(1,4) I have only a horrible formula for U: U=(1+R)(C(1,2,3,4,5,6,7,8)+(R+R^3)(C(1,2,4,8)+C(1,3,4,5))+C(1,3,4,5)+R^2+R^5. But f17(x^17)=C(1,2,4,8)C(1,2,3,4,5,6,7,8).</p> <p>p=19</p> <p>(a) Let t and u be the reductions of the rational newforms of weights 2 and 4. Then u=t/(1+t). Set v=r+u. Then A=Z/2[r,t,u]=Z/2[v,t,1/(1+t)]. And v^2+v=t^3. So the curve is the cubic curve y^2+y=x^3+x with the points with x=1 removed. f19=t*(v^6)*(1+u^4).</p> <p>(b) R=C(1,3,3) V=C(1,2,4,6) T=R+C(2,3,3,4) U=R+V</p> <p>Caveat: My MO results show that my alleged generators of B' given above really generate B'. But for p=11,13,17 and 19 I haven't checked that the alleged generators of A really generate A. But this can no doubt be done with sage.</p> http://mathoverflow.net/questions/117904/elementary-examples-of-the-weil-conjectures/118011#118011 Answer by paul Monsky for Elementary examples of the Weil conjectures paul Monsky 2013-01-04T01:29:56Z 2013-01-04T01:29:56Z <p>It's possible to give a semester course in number theory, free from overt algebraic geometry, that handles the zeta functions of curves over finite fields, proving the functional equation and Weil's RH for them. (And also does Mordell-Weil for elliptic curves over number fields). I taught such a course to second year grad students some 20 or 30 years ago.</p> <p>One gets around the geometry of curves in 19th century Dedekind fashion, working with their function fields--finite extensions of k(t)-- and valuations, just as one does with number fields. Riemann-Roch is done as in Chevalley's book using "repartitions". Rationality and the functional equation follow directly from Riemann-Roch. The RH is done by Bombieri's elegant technique--first one uses Riemann-Roch to get a good upper bound for the number of rational points, then one combines this upper bound with the functional equation to get a good lower bound. (Ayanta thinks the proof is miraculous but uninformative; this may be true of Stepanov's original version, but I find Bombieri's argument to be natural).</p> <p>Of course there are defects to this approach. Algebraic geometry is far more enlightening. But it can be learned later, in whatever form the student finds appropriate. And there are also advantages. To do Weil-style algebraic geometry one would have to worry about "fields of definition". And Grothendieck's version (which continues to intimidate me), would only appeal to the rare student at this level. That such beautiful mathematics can be presented in such an accessible fashion seems to me a boon.</p> http://mathoverflow.net/questions/117558/are-there-heronian-triangles-that-can-be-decomposed-into-three-smaller-ones/117790#117790 Answer by paul Monsky for Are there Heronian triangles that can be decomposed into three smaller ones? paul Monsky 2013-01-01T17:30:09Z 2013-01-01T19:54:57Z <p>There should be many examples where the interior vertex lies on the perpendicular bisector of one of the sides. It's best to redefine a Heronian triangle to be one with rational sides and rational area. I'll call such a T "standard" if its vertices are at (-2,0), (2,0) and (r,s) where r and s are rational. Every Heronian triangle is similar to a standard one.</p> <p>Now let T be standard and P be (0,(x)-(1/x)) where x is rational. Then P has rational distance from 2 vertices of T. The condition that it have rational distance from the third is that there exists a rational y such that (xx-sx-1)^2 +(rx)^2 =y^2. If the elliptic curve one gets in this way has positive rank then there will be a dense set of points (0,(x)-(1/x)), all lying on the perpendicular bisector of the base of T, each giving the desired decomposition of T.</p> <p>I worked out the case r=-2, s=3. Unfortunately the curve one gets is one of conductor 15 with 8 rational torsion points and rank 0. But there must be lots of choices of r and s where the rank is positive.</p> <p>EDIT: Here's another construction which should give many examples where the interior point lies on an altitude. Consider a Heronian triangle with the base extending from (0,0) to (a+b.0), and the foot of the altitude to the base at (a,0). Let P be (a,x) where x is rational. Then P is at a rational distance from one vertex, and is at a rational distance from the other two when there are rational u and v with xx+aa=uu and xx+bb=vv. These equations again define an elliptic curve and one will get a dense set of points (a,x) on the altitude, each giving a desired decomposition, when the curve has positive rank.</p> <p>The interesting question then seems to be the existence of a point, that lies neither on an altitude nor on the perpendicular bisector of a side, and that yields the desired decomoposition. </p> http://mathoverflow.net/questions/108171/the-mod-3-reduction-of-some-powers-of-delta The mod 3 reduction of some powers of delta paul Monsky 2012-09-26T15:22:38Z 2012-12-13T04:36:20Z <p>Let f in Z/3[[x]] be the mod 3 reduction of the Fourier expansion of the normalized weight 12 cusp form delta for the full modular group. The exponents appearing in f are all 1 mod 3. Fix k>0 and prime to 3. If j is 1 or 2, let S(j) consist of all primes p for which the coefficient of x^p in f^k is j. It's well known that if k=1, S(1) is empty and S(2) consists of all p that are 1 mod 3.</p> <p>I find experimentally that similar results hold when k is 2,4,5 or 10. Indeed it seems:</p> <ol> <li><p>When k=2, S(1) is all p that are 2 mod 9, S(2) all p that are 5 mod 9</p></li> <li><p>When k=4, S(1) is all p that are 4 or 7 mod 9, S(2) is empty</p></li> <li><p>When k=5, S(1) is all p that are 5,7 or 20 mod 27, S(2) all p that are 8,11, or 23 mod 27</p></li> <li><p>When k=10, S(1) is all p that are 13 or 25 mod 27, S(2) all p that are 16 or 22 mod 27.</p></li> </ol> <p>I have little doubt that these results hold. But are they known, and is there a reference? </p> <p>EDIT: For the case of reduction mod 2 rather than mod 3, see my recent question, "Does this theorem of Hasse....?" . But it seems less likely that techniques from the theory of binary and ternary quadratic forms can yield a proof of the above "results".</p> <p>FURTHER EDIT: Here's a sketch of a proof of the first result. The space of Fourier series of weight 2 cusp forms for gamma_0 (9) has a basis of Eisenstein elements F,G, and H lying in Z[[x^3]], xZ[[x^3]], and x^2 Z[[x^3]] respectively. In Z[[x]], F is congruent to 1 mod 12 x^3. Furthermore the coefficient of x^n in G is sigma_1(n) when n is 1 mod 3, while the coefficient of x^n in H is (1/3)(sigma_1 (n)) when n is 2 mod 3.</p> <p>Let C=x-8x^4+20x^7+.. be the Fourier expansion of the weight 4 form (eta(3z))^8 for gamma_0(9). A comparison of the coefficients of x^n for small n gives the identities C=FG-27H^2, and G^2=FH. So mod 3, C^2=G^2=H, and the coefficient of x^p in C^2, when p is a prime congruent to 2 mod 3 is, modulo 3, equal to (1/3)(sigma_1(p))=(p+1)/3. Now the cube of C^2 is the square of f(x^3), where f is the Fourier expansion of delta. It follows that mod 3, the coefficient of x^p in f^2 is (p+1)/3 when p is a prime congruent to 2 mod 3. This is precisely 1. above.</p> http://mathoverflow.net/questions/106267/does-this-variant-of-a-theorem-of-hasse-really-due-to-gauss-have-an-elementary Does this variant of a theorem of Hasse (really due to Gauss) have an "elementary" proof? paul Monsky 2012-09-03T19:27:05Z 2012-10-11T23:04:35Z <p>BACKGROUND</p> <p>Here are 3 theorems of varying difficulty. Let $M$ be the $Z/2$ subspace of $Z/2[[x]]$ spanned by $f^k$, with the $k>0$ and odd, and $f=x+x^9+x^{25}+x^{49}+\cdots$. For $g$ in $M$, let $S(g)$ consist of the primes, $p$, for which the coefficient of $x^p$ in $g$ is 1. Note that each $p$ in $S(f^k)$ is congruent to $k$ mod 8.</p> <p>T1.----- If $k=3 {\rm\ or\ } 5$, $S(f^k)$ consists of the $p$ that are $k$ mod 8</p> <p>T2.----- $S(f^7)$ consists of the $p$ that are 7 mod 16</p> <p>T3.----- If $k=19 {\rm\ or\ } 21$, then $S(f^k)$ consists of the $p$ that are $k$ or $k+8$ mod 32.</p> <p>To prove T1 when $k=3$, we write $f^k$ as $f*f^2$ and use the fact that if $p$ is 3 mod 8, then $p$ is uniquely the sum of a square and twice a square. When $k=5$ we argue similarly using Fermat's two square theorem.</p> <p>As I indicated in a comment on a recent MO question of Joel Bellaiche, "<a href="http://mathoverflow.net/questions/100701/primes-and-x22y24z2" rel="nofollow">Primes and x^2+2y^2+4z^2</a>" ,T2 follows from a result of Hasse on the class number of $Q(\sqrt{-2p})$, using Gauss' theorem that the number of representations of $2p$ as a sum of 3 squares is 12*(this class number). Hasse's proof is an application of the Gauss theory of genera and ambiguous forms.</p> <p>T3 is thornier. Because $f$ is the mod 2 reduction of (the Fourier expansion of) the normalized weight 12 cusp form for the full modular group, each $g$ is the mod 2 reduction of a modular form of integral weight. A profound result of Deligne, relating Hecke eigenforms to Galois representations, then shows that $S(g)$ is a "Frobenian set". Nicolas, Serre and Bellaiche, continuing in this vein, developed a theory of level 1 modular forms in characteristic 2 that led to more precise results. Their investigations motivated me to try to determine $S(f^k)$ empirically for small $k$, and I was led to conjecture T3. Joel then applied his methods to give a proof. But this is very hard, and so I ask:</p> <p>QUESTION</p> <p>Does there exist an "elementary proof" of T3, using the theory of binary quadratic forms, along the lines of the Hasse-Gauss argument?</p> <p>EDIT: Motivated by my recent simple proof of T2 (see my answer to the question of Joel cited above), I've found arguments that ought to reduce the proof of T3 to Sage calculations. The point is that forms of weight 2 are easier to deal with than forms of weight 3/2, so one should work with quadratic forms in 4 variables rather than in 3, even when the genera that arise have more than 1 class in them. Here's the idea of my argument for f21.</p> <p>Let p be a prime that is 5 mod 8. Writing $f^{21}$ as $(f)(f^2)(f^2)(f^{16})$ we find that if $R$ is (1/16)*(the number of representations of $p$ by $G_1=x^2+2y^2+2z^2+16t^2$ with $x,y,z$ and $t$ all odd), then $p$ is in $S(f^21)$ if and only if $R$ is odd. Now since $p$ is 5 mod 8, in any representation of $p$ by $G_1$, $x,y$ and $z$ must be odd. So if we set $G_2=x^2+2y^2+2z^2+64t^2$ then $R=(N1-N2)/16$, where $N_1$ and $N_2$ are the numbers of representations of $p$ by $G_1$ and $G_2$ respectively. Now write $p$ as $a^2+4b^2$ with $a$ and $b$ congruent to 1 mod 4. Computer calculations indicate:</p> <p>Conjecture 1. $N_1=p+1+2a$</p> <p>Conjecture 2. $N_2=((p+1)/2)+a+4b$</p> <p>If these conjectures hold then $R=(p+1+2a-8b)/32$. The numerator here is $4(b-1)^2 +(a+3)(a-1)$, which mod 64 is $(a+3)(a-1)$. So $R$ has the same parity as $(a+3)(a-1)/32$ and is odd just when $a$ is $5$ or $9$ mod $16$. Now mod $32$, $p=a^2+4$. So $R$ is odd just when $p$ is $29$ or $85$ mod $32$, and so the conjectures imply Joel's result for $S(f^21)$.</p> <p>How does one attack the conjectures? The theta series attached to $G_1$ and $G_2$ are modular forms for $\Gamma_0 (64)$ and $\Gamma_0 (256)$ respectively. If the conjectures are to hold it seems that each of these theta series should be a linear combination of Eisenstein series and cusp forms attached to Grossencharaktere for $\mathbb Q(i)$. It should be possible, using Sage, to get an explicit formulation of this, and prove the conjectures.</p> <p>My proposed treatment of $S(f^{19})$ is entirely similar. Suppose $p$ is $3$ mod $8$. Writing $f^{19}$ as $(f)(f)(f)(f^{16})$ and arguing as above we find that if we take $H_1$ and $H_2$ to be $x^2+y^2+z^2+16t^2$ and $x^2+y^2+z^2+64t^2$ respectively, and let $N_1$ and $N_2$ be the number of representations of $p$ by $H_1$ and $H_2$, then $p$ is in $S(f^{19})$ just when $R=(N_1-N_2)/16$ is odd. Now Jacobi's 4 square theorem, (see the argument in my answer to Joel's question), shows that $N_1$ is $2(p+1)$. Write $p$ as $a^2+2b^2$ with $a =1$ or 3 mod 8. The computer suggests:</p> <p>Conjecture 3. $N_2=p+1+4a$</p> <p>So if the conjecture holds, $R=(p+1-4a)/16$, and one sees easily that this is odd just when $p$ is 19 or 27 mod 32. Once again the theta series attached to H2 is a modular form for $\Gamma_0 (256)$. The conjecture indicates that it should be a linear combination of Eisenstein series and cusp forms attached to Grossencharaktere for $\mathbb Q(\sqrt{-2})$; all this should admit a proof using Sage.</p> http://mathoverflow.net/questions/100701/primes-and-x22y24z2/109275#109275 Answer by paul Monsky for Primes and $x^2+2y^2+4z^2$ paul Monsky 2012-10-10T06:00:46Z 2012-10-10T06:00:46Z <p>Here's a simpler argument. We may assume p is 7 mod 8. Let N be the number of triples of squares (r,s,u) with r+2s+4u=p. We will show that N is odd if p is 7 mod 16 and even if p is 15 mod 16. Let M be (1/64)(the number of representations of p by xx+yy+zz+tt). Jacobi's 4 square theorem (which has elementary proofs using quaternions for example) shows that M=(p+1)/8. So it suffices to show that M and N have the same parity. Now if p=xx+yy+zz+tt, then just one of x,y,z,t is even. So M=(1/16)(the number of representations of p by xx+yy+zz+4tt). In other words, M is the number of ordered quadruples of squares (r,s,t,u) with r+s+t+4u=p. Now the involution (r,s,t,u)-->(r,t,s,u) on this set of quadruples has the N fixed points (r,s,s,u), giving the result.</p> http://mathoverflow.net/questions/106859/beautiful-theorems-with-short-proof/106885#106885 Answer by paul Monsky for Beautiful theorems with short proof paul Monsky 2012-09-11T05:23:13Z 2012-09-11T05:23:13Z <p>The proof(via the pigeon-hole principle--continued fractions would need too much preparation) that when D>0 is not a square then the "Pellian equation" xx-Dyy=1 has a non-trivial solution.</p> http://mathoverflow.net/questions/106859/beautiful-theorems-with-short-proof/106866#106866 Answer by paul Monsky for Beautiful theorems with short proof paul Monsky 2012-09-11T01:02:58Z 2012-09-11T01:02:58Z <p>Fermat's proof, by infinite descent, that there is no Pythagorean right triangle whose area is a square might qualify.</p> http://mathoverflow.net/questions/61348/simple-groups-with-the-same-cardinality-as-psl-2z-p Simple groups with the same cardinality as PSL_2(Z/p) paul Monsky 2011-04-11T23:44:23Z 2012-08-23T07:39:15Z <p>In an undergrad honors algebra course it's sometimes shown that when $p$ is prime and $>3$ then $PSL_2(Z/p)$ is simple of of order $p(p-1)(p+1)/2$. But that this is the "only" simple group having that order is seldom or never (except when $p=5$) proved.</p> <p>I've worked out a fairly simple proof, using the Burnside transfer theorem, of this last result, but it's perhaps a little too intricate to present in class.</p> <p>QUESTION: Are there proofs of this result, on-line or in texts, that are appropriate for an undergrad honors algebra class? (If not, I might post an argument on arXiv).</p> <p>EDIT: Since no simple available proof has yet been found, I'll sketch the argument that I culled from classification arguments for Zassenhaus groups.</p> <p>Suppose G is simple of order p(p-1)(p+1)/2. First one shows that G has p+1 p-Sylows. Let S be the union of Z/p and {infinity}. An easy study of the conjugation action of G on the p-Sylows allows one to identify G with a doubly transitive group of (even) permutations of S, containing the p-cycle z-->z+1. Then the subgroup of G fixing both 0 and infinity is cyclic generated by z-->cz for some c. Once this is done, the key is in showing:</p> <p>A.--- The subgroup of elements that either fix 0 and infinity or interchange them is dihedral.</p> <p>Once A is shown it's not hard to show that z-->-1/z is in G, thereby identifying G with a fractional linear group. The proof of A is a counting argument when p=1 mod 4. But when p=3 mod 4 the situation is more delicate, and one uses Burnside tranfer.</p> http://mathoverflow.net/questions/100265/not-especially-famous-long-open-problems-which-anyone-can-understand/101112#101112 Answer by paul Monsky for Not especially famous, long-open problems which anyone can understand paul Monsky 2012-07-02T02:12:16Z 2012-07-02T02:12:16Z <p>A few decades ago Sherman Stein asked whether a trapezoid whose parallel sides are in the ratio 1:root 2 can be dissected into triangles, all of the same area. This remains open--it's a mystery which trapezoids admit such dissections./</p> http://mathoverflow.net/questions/100265/not-especially-famous-long-open-problems-which-anyone-can-understand/100448#100448 Answer by paul Monsky for Not especially famous, long-open problems which anyone can understand paul Monsky 2012-06-23T11:51:16Z 2012-06-23T11:51:16Z <p>Here's another Birch Swinnerton-Dyer related problem. Sylvester conjectured that every prime that is 4,7 or 8 mod 9 is a sum of two rational cubes. Elkies (unpublished?) settled the first two cases. As far as I know, the third is still open.</p> http://mathoverflow.net/questions/68247/existence-of-certain-identities-involving-characteristic-2-thetas Existence of certain identities involving characteristic 2 "thetas" paul Monsky 2011-06-19T23:11:15Z 2012-01-25T20:10:37Z <p>Let l=2m+1 be prime. In my previous MO question, "What are the polynomial relations between these characteristic 2 thetas?", I defined a subring of Z/2[[x]] as follows:</p> <p>The subring, S, is generated by [1],...,[m] where [i] is the sum of the x^(n^2), n running over all integers congruent to i mod l.</p> <p>QUESTION...... Let F=x+x^9+x^25+x^49+...,G=F(x^l), and H=G(x^l). Are G and H in S?</p> <p>The answer is yes when l=3,5 or 7. When l=7, if we set a=[1],b=[2] and c=[3], we have the curious identities H=(abc)^3*(abc+ba^3+cb^3+ac^3), and G=(abc)^2+a^7+b^7+c^7+H.</p> <p>Remark 1... Kevin Buzzard explained to me that one can decide whether an explicitly given identity such as the ones we've displayed holds by using the theory of characteristic 2 modular forms and computer calculation. But how does one produce these putative identities?</p> <p>Remark 2... For all l one can show in an elementary way that H is in the field of fractions of S. In fact if a=[i], b=[2i] and c=[4i], then H is the quotient of a^8(a^8+b^2) by b^4+c. Furthermore for l at most 13, H is in S. (One shows that the quotient lies in S, by combining the "quintic relations" of my MO question cited earlier with Groebner basis computer calculations.)</p> <p>I'll sketch an argument giving the l=7 identities. Let C be the curve in affine 3-space defined by the ideal of quintic relations. C has 3 linear branches at the origin and 3 linear branches at each of the seven points (r,r^4,r^9) with r^7=1. Passing to projective 3-space we find that (the Zariski closure of) C has 14 simple points at infinity. The formula for H as a quotient shows that H has zeros of order 49 at the branches at the origin, simple zeros at the branches at the other singular points, and poles of order 12 at infinity. This leads to the identity for H. To get the identity for G one notes that (GH)+(GH)^2+(G+H)^8=0--see my MO question, "What's known about the reduction...?" It follows from this that if G is in the field of fractions of S then G+H has zeros of order 7 at the branches at the origin, of order 3 at the branches at the other singular points, and poles of order 6 at infinity. This suggests that G+H=(abc)^2+a^7+b^7+c^7. To verify this we set J=(abc)^2+a^7+b^7+c^7+H, and use Groebner basis computer calculations to show that JH+(JH)^2+(J+H)^8=0; it then follows that J=G.</p> <p>EDIT: I think I can now show that when l=11, G is NOT in the field of fractions of S, even though H is in S. I'll make this an answer once I'm surer of it.</p> <p>EDIT #2: My supposed counterexample when l=11 is incorrect; G like H is in S. I had the wrong modular equation of degree 11 relating G and H. Once I found the correct equation, in Cayley's article, I was able to argue as in the case l=7.</p> <p>FINAL(?) EDIT: As I've shown in my answer, G and H are indeed always in S. And I've produced a simple conjectural explicit formula for G+H that holds for l&lt;1500. Whether there is anything comparably simple for H isn't clear. At any rate here are formulas for H when l&lt;24. I write C(a,b,c) for the sum of the [ra]<em>[rb]</em>[rc] where r runs from 1 through (l-1)/2; more generally (a,b,c) can be replaced by any multi-set. P is the product of the [r] where r runs from 1 through (l-1)/2. The identity when l=17 is striking.</p> <p>l=3.......... [1]^9 +[1]^12</p> <p>l=5.......... P^5 +P^6</p> <p>l=7.......... (P^3)(P+C(1,1,1,2))</p> <p>l=11.........(P^2)(C(1,1,3)+C(1,1,2,4))</p> <p>l=13.........P*(P+[1][2][3][5]+[1][4][5][6]+[2][3][4][6]+C(1,1,2,2,2,5))</p> <p>l=17.........P*([1][2][4][8]+[3][5][6][7])</p> <p>l=19.........P*([2][3][5]+[4][6][9]+[1][7][8]+C(3,3,2,4))</p> <p>l=23.........P*(C(1,2,3,3)+C(1,2,4,5)+C(1,4,4,6)+C(1,2,2,5,9))</p> http://mathoverflow.net/questions/52781/whats-known-about-the-mod-2-reduction-of-the-level-l-jacobi-modular-equation What's known about the mod 2 reduction of the level l Jacobi modular equation? paul Monsky 2011-01-21T18:08:30Z 2011-12-27T03:55:08Z <p>Motivation:</p> <p>Let $\ell$ be an odd prime. Let $A$ in ${\mathbb Z}/2[[x]]$ be $x+x^9+x^{25}+x^{49}+...$, and $B=A(x^\ell)$. One can use the level $\ell$ Jacobi modular equation to get a polynomial relation between $A$ and $B$ over ${\mathbb Z}/2$. I'm curious as to what is known about this relation. To be precise, let $\Omega_\ell$ in ${\mathbb Z}[u,v]$ be the modular equation in $u-v$ form; see page 126 of Borwein and Borwein, "Pi and the AGM". Write this polynomial as a sum of monomials $2^{c_{i,j}} d_{i,j} (u^i) (v^j)$ with the $d_{i,j}$ odd. Let $f \in {\mathbb Z}/2[X,Y]$ be the sum of the $(X^i)(Y^j)$, the sum extending over the pairs $(i,j)$ for which $(c_{i,j})+(1/2)(i+j)$ takes its minimal value. (It appears that this minimal value is $\ell+1$).</p> <p>It's not hard to see that $f(A,B)=0$. And the theory of the modular equation shows that $f$ is symmetric in $X$ and $Y$. Question---What more is known about $f$?</p> <p>Examples--(See pages 127-132 of Borwein and Borwein which allow one to calculate $f$ for $\ell&lt;29$):</p> <ul> <li>$\ell=3$: $f=XY+(X+Y)^4$</li> <li>$\ell=5$: $f=XY+(X+Y)^6$</li> <li>$\ell=7$: $f=XY+(XY)^2+(X+Y)^8$.</li> </ul> <p>EDIT: A few simple remarks. The l+1 at the end of the first paragraph above should have been (1/2)(l+1); see my comment below. Also problem 6a on page 135 of Borwein and Borwein says that in our notation, c_1,1 =(l-1)/2. So XY, X^(l+1) and Y^(l+1) all appear in f. Finally the "octicity "result of page 134 problem 3 puts a restriction on the monomials appearing in f.</p> <p>EDIT 2: The revised comments below form an edit. (When I tried to put them up as such, a bug intervened). In them I define the modular functions u and v in terms of Jacobi's thetas, and indicate why one can derive relations between A and B over Z/2 from relations between u and v over Z. I also show that the relation f(X,Y) derived from Omega_l is irreducible.</p> http://mathoverflow.net/questions/30501/variations-on-a-theme-of-obryant-cooper-and-eichhorn-concerning-power-series-ov Variations on a theme of O'Bryant, Cooper and Eichhorn concerning power series over $\mathbb Z/2\mathbb Z$ paul Monsky 2010-07-04T11:06:32Z 2011-12-27T03:53:16Z <p>Define 2 power series over the field $\mathbb Z/2\mathbb Z$ by $f=1+x+x^3+x^6+\dots$, the exponents being the triangular numbers, and $g=1+x+x^4+x^9+\dots$, the exponents being the squares. Write $f/g$ as $c_0+c_1x+c_2x^2+\dots$ with each $c_n$ in $\mathbb Z/2\mathbb Z$.</p> <p><strong>Question.</strong> Is it true that when $n$ is even then $c_n$ is 1 precisely when $n$ is in the set of even triangular numbers $\lbrace 0,6,10,28,36,\dots\rbrace$? <a href="http://mathoverflow.net/users/935/kevin-obryant" rel="nofollow">Kevin O'Bryant</a> has verified that this holds when $n$ is 512 or less.</p> <p><strong>Remark.</strong> If one writes $1/g$ as $b_0+b_1x+b_2x^2+\dots$, then $n\mapsto b_n$ is the characteristic function $\bmod 2$ of the set $B$ studied by O'Bryant, Cooper and Eichhorn (see <a href="http://mathoverflow.net/questions/26839/" rel="nofollow">this</a> and <a href="http://mathoverflow.net/questions/28462/" rel="nofollow">this</a> questions of O'Bryant on MO); they show that when $n$ is even then $b_n$ is 1 precisely when $n$ is twice a square. A positive answer to my question would give a nice characterization of those elements of $B$ that are congruent to $7 \bmod 16$.</p> <p>(I've used the modular forms tag because of the formal similarity of $f$ and $g$ to Jacobi theta functions, and the motivation of O'Bryant, Cooper and Eichhorn in looking at $B$).</p> http://mathoverflow.net/questions/33711/if-p-is-a-prime-congruent-to-9-mod-16-can-4-divide-the-class-number-of-qp1-4 If p is a prime congruent to 9 mod 16, can 4 divide the class number of Q(p^(1/4))? paul Monsky 2010-07-28T21:01:07Z 2011-12-27T03:52:51Z <p>When $p$ is a prime $\equiv9\bmod16$, the class number, $h$, of $\mathbb Q(p^{1/4})$ is known to be even. In </p> <p><a href="http://dx.doi.org/10.1515/crll.1980.314.40" rel="nofollow">[Charles J. Parry, A genus theory for quartic fields. <em>Crelle's Journal</em> <strong>314</strong> (1980), 40--71]</a></p> <p>it is shown that $h/2$ is odd when 2 is not a fourth power in $\mathbb Z/p\mathbb Z$. Does this still hold when 2 is a fourth power?</p> <p>Some years ago I gave an (unpublished) proof that this is true provided the elliptic curve $y^2=x^3-px$ has positive rank, and in particular that it is true on the B. Sw.-D. hypothesis. It's known that the above curve has positive rank for primes that $\equiv5$ or $7\bmod16$, but to my knowledge $p\equiv9\bmod16$ remains untouched. But perhaps there's an elliptic-curve free approach to my question?</p> http://mathoverflow.net/questions/80944/vanishing-constant-term-in-powers-of-a-laurent-polynomial/84278#84278 Answer by paul Monsky for Vanishing constant term in powers of a Laurent polynomial paul Monsky 2011-12-25T19:02:19Z 2011-12-25T19:02:19Z <p>I'll give an algebraic argument, which is I think essentially the same as Duistermatt's, but substitutes partial fractions and some valuation theory for complex analysis. K will be algebraically closed of characteristic 0; f will be in K[x,1/x]. We suppose f is not in K[x] or K[1/x]. Let r and -s be the largest and smallest exponent of x appearing in f; r and s are >0.</p> <p>Lemma:---Let M be a finite extension of the field of fractions of K[[t]]. Extend the obvious valuation on the field of fractions to M. Then if a is in M with f(a)=t, ord f'(a) must be &lt; 1. (To see this note that ord a is 0. Then a has a Newton-Puiseux expansion a0 +(a1)(t^p)+..., with a0 and a1 non-zero, and p a positive rational. The derivation D=d/dt extends to M, and 1=D(f(a)) is the product of f'(a) by (a1)(p)*(t^(p-1))+... So ord f'(a)=1-p which is &lt; 1.)</p> <p>Theorem 1:---Let S be a subset of the algebraic closure of K(z). Suppose that for each a in S, f(a)=1/z. Then the sum over S of the 1/(af'(a)) cannot be z. (To see this let t=1/z. Then K(z)= K(t) which imbeds in the field of fractions of K[[t]], and we may view S as a subset of a finite extension, M, of this field. By the lemma, each 1/(af'(a)) has ord > -1. But z=1/t has ord equal to -1.)</p> <p>Now let c_n be the constant term in f^n, and W=1+... be the element sigma (c_n)(z^n) of K[[z]]. Combinatorialists know that a partial fraction argument shows that W is algebraic over L=K(z). Carrying out the partial fraction argument explicitly one finds:</p> <p>Theorem 2:---There are a lying in a finite extension of L with each f(a) equal to 1/z, such that zW is the sum of the 1/(af'a)). (So by Theorem 1, W is not 1, and c_1, c_2,... cannot all be 0.)</p> <p>I'll sketch a proof of Theorem 2. Let U be the element (x^s)(1-zf) of L[x]. If a is a root of u, then f(a)=1/z. So by the lemma f'(a) is not 0. Then U'(a) is not 0, and U is separable. It follows that 1/(1-fz)=(x^s)/U is a sum of c/(x-a) where a runs over the roots of f(x)=1/z. It's easy to see that the c corresponding to a given a is -1/(zf'(a)). I now refer to Lemmas 3.6 and 3.7 of my article "Generating functions attached to some infinite matrices"---see arXiv 0906.1836 or Elec. J. of Comb, v.18 (1) 2011:</p> <p>If we take U1=x^s and U2=U=(x^s)(1-zf) we are in the situation of Lemma 3.7. W is "the coefficient of x^0 in the element (U1)/(U2)". In the language of the lemma, W is the sum of the l_0(c/(z-a)). The proof of Lemma 3.6 shows that l_0(1/(z-a)) is either 0 or -1/a. So z l_0(c/(z-a)) is either 0 or 1/(af'(a)). This completes the proof.</p> http://mathoverflow.net/questions/39635/density-stability-questions-for-those-who-like-computer-calculation Density stability; questions for those who like computer calculation paul Monsky 2010-09-22T16:17:41Z 2011-12-22T03:34:45Z <p>BACKGROUND: The question, which has its roots in <a href="http://mathoverflow.net/questions/26839/how-thick-is-the-reciprocal-of-the-squares" rel="nofollow">a question asked on MO</a> by O'Bryant, concerns the relative density of certain subsets, $B$, of ${\mathbb N}$ in congruence classes modulo a power of 2. Let $I$ be such a congruence class. I'll say that $B$ is "stable in $I$" if there is a $c$ such that $B$ has relative density $c$ in $J$ whenever $J$ is a congruence class contained in $I$ whose modulus is also a power of 2.</p> <p>Suppose $B$ consists of all $n$ such that the coefficient of $x^n$ in the reciprocal of the element $g=1+x+x^4+x^9+x^{16}+\dots$ of ${\mathbb Z}/2[[x]]$ is 1. Cooper et al. showed that $B$ has density 0 in 12 of the mod 16 congruence classes. I extended the result to 3 of the 4 remaining classes. But calculations by O'Bryant suggest that in the class 15 mod 16, $B$ is stable with relative density $1/2$. For a detailed account see my note <a href="http://arxiv.org/abs/1009.3985" rel="nofollow">Disquisitiones Arithmeticae and online sequence A108345</a>.</p> <p>These QUESTIONS pertain to sets introduced by Cooper et. al:</p> <ol> <li><p>Replace the exponents $0, 1, 4, 9, \dots$ in $g$ by the numbers $3n^2-2n$, $n \in {\mathbb Z}$, to get a new $B$. This $B$ has density 0 in 7 of the classes mod 8. Does the computer suggest that it is stable with relative density 1/2 in the class 0 mod 8?</p></li> <li><p>Suppose the exponents are $5n^2-4n$, $n \in {\mathbb Z}$. Does the computer suggest that there's a $q$ such that the new $B$ you get is stable in each mod $q$ congruence class? And if so, what do the relative densities appear to be? (The density is provably 0 in some mod 8 classes).</p></li> <li><p>Answer the same question as 2. when the exponents are the $5n^2-2n$, $n \in {\mathbb Z}$.</p></li> </ol> <p>EDIT: I'll give a modified and generalized version of the question (and an expansion of my answer) using notation and ideas from my MO question on characteristic 2 thetas. Let $L$ be the field of formal Laurent series in $x$ over ${\bf Z}/2$. If $f$ (not zero) is in $L$, $B(f)$ consists of all $n$ for which the coefficient of $x^n$ in $1/f$ is 1. Now fix $l=2m+1$, $m>0$, and for $i$ in $\lbrace1,...,m\rbrace$ let $[i]$ be the element of ${\bf Z}/2[[x]]$ defined in the "thetas" question.</p> <p>Question: For $q$ a power of 2, what does the computer suggest about the relative density of $B([i])$ in the various mod $q$ congruence classes? (Since all elements of $B[i]$ are congruent to $-(i^2)$ mod $l$, these relative densities are at most $1/l$).</p> <p>Example: When $l=3$, it can be shown that $B([i])$ has density 0 in all congruence classes mod 8, with the possible exception of 7. And the computer (perhaps) indicates that in the 7 mod 8 class (or any class contained therein) the relative density is 1/6.</p> <p>My "answer" generalizes the first sentence of the example. I made no computer calculations--indeed the computer evidence is at first sight contrary to my results because of the slow approach to zero. Let $L(q)$ contained in $L$ be the field of formal Laurent series in $x^q$. Then $L$ is the direct sum of the $(x^k)L(q)$, $k$ in $\lbrace0,...,q-1\rbrace$. Let $p_{(q,k)}$ be the obvious projection map $L\to(x^k)L(q)$. Let $S$ contained in</p> <p>${\bf Z}/2[[x]]$ </p> <p>be the smallest ring that contains all the $[i]$ and is stable under the $p_{(q,k)}$ for all $q$ and $k$. It can be shown that every element of $S$ is the mod 2 reduction of the Fourier series of an integral weight modular form for a congruence group. A theorem of Serre then shows that if $\sum((c_n)(x^n))$ is in $S$ then the set of $n$ for which $c_n$ is 1 has density 0.</p> <p>As a corollary one finds: Let $p$ be a $p_{(q,k)}$. If $p(1/[i])$ is in $S$ then $B([i])$ has density 0 in the class $k$ mod $q$.</p> <p>By making use of the quintic relations from my theta question I can show that the hypothesis of the theorem holds in various cases. In particular suppose $i$ is prime to l. When $l=5$, $B=B([i])$ has density 0 in each mod 32 class except perhaps the 5 classes $n=7$ mod 8 and $n=28$ mod 32. When $l=7$, $B$ has density 0 in each mod 32 class except perhaps the 7 classes 7 mod 8, 14 mod 16 and 28 mod 32. When $l=9$, $B$ has density 0 in each mod 64 class except perhaps the 19 classes 1 and 7 mod 8, 28 mod 32, and 48 mod 64.</p> <p>In the various classes qualified by "except perhaps" in the above paragraph (and the subclasses contained therein) it seems plausible that the relative densities are $1/(2l)$. But this may be wishful thinking. I hope that someone will make further calculations.</p> <p>FURTHER EDIT: Heres a more explicit and more speculative version of my question. Let n_j be the negative exponents appearing in the Laurent series 1/[i], 1/[2i], 1/[4i], i/[8i],..., and q_j be the largest power of 2 dividing n_j.</p> <p>QUESTION: Does computer evidence support the following speculations?</p> <p>(1) The relative density of B([i]) in each congruence class n_j mod 8q_j, and in all congruence classes modulo a power of 2 contained therein, is 1/(2l).</p> <p>(2) Outside of these congruence classes B([i]) has density 0.</p> <p>For example when l=9 and i=1 the n_j are -16,-7,-4 and -1, and the classes in (1) are 1 mod 8, -1 mod 8, -4 mod 32 and -16 mod 128. The technique I indicated in my earlier edit shows that (2) holds in this case, so one gets 128-37 classes mod 128 where B has density 0. The technique also shows that (2) holds when l=3,5 or 7. This isn't much evidence, and there's far less for (1). But as these are the simplest answers one might hope for, I'd be interested in any calculations concerning them.</p> http://mathoverflow.net/questions/39635/density-stability-questions-for-those-who-like-computer-calculation/84069#84069 Answer by paul Monsky for Density stability; questions for those who like computer calculation paul Monsky 2011-12-22T03:34:45Z 2011-12-22T03:34:45Z <p>For prime l I've now proved (a corrected version of) the speculation (2) made in the further edit to my question. (See Theorem I below). The proof avoids the extensive computer verifications made in arXiv NT 1107.4137. So let K be an algebraic closure of Z/2. Call an element g=a_0+a_1(x)+a_2(x^2)+... of K[[x]] "sparse" if the n with a_n non-zero form a set of density 0.</p> <p>Lemma 1----Suppose g is in the subring R of K[[x]] generated by K and the [i]. Then g is sparse. (This follows from the fact that the elements of R are the mod 2 reductions of Fourier expansions of modular forms of integral weight, and the theorem of Serre mentioned in my previous answer).</p> <p>We have shown in another question that the above ring R is the co-ordinate ring of an affine curve C. Let m_0 be the maximal ideal of R generated by [1],...,[l-1], and p_0 be the point of C corresponding to m_0. We showed in addition (using the fact that l is prime) that m_0 is the only maximal ideal of R containing any of [1],...,[l-1], and that there are (l-1)_/2 linear branches at p_0 with distinct branch tangents.</p> <p>Lemma 2---Suppose g=a_0+a_1(x)+... is in K[[x]], that g is the quotient of an element of R by a product of powers of [i], and that g has positive ord at every branch of C centered at p_0. Then g is sparse. (For g is in the localization of R at every maximal ideal other than m_0. Furthermore if n is large g^n is in the localization of R at m_0. So for n large, g^n is in R, and is sparse by Lemma 1. Take n to be a large power of 2 to get the result.</p> <p>---------Now let U=U_2 be the operator K[[x]]-->K[[x]] taking sum(a_n)(x^n) to sum(a_2n)(x^n). Note that U([i][i]g)=[i]U(g).</p> <p>Lemma 3---The subring of Z/2[[x]] generated by the [i] is stable under U. (It suffices to show that U takes a product of terms, [ ], to an element of this ring. We argue by induction on the number of terms in the product. We may assume that the first 2 terms are [2i] and [2j]. Then [2i][2j] is the sum of [2i][j]^4, [2j][i]^4 and ([i+j][i-j])^2. Multiplying by the remaining terms in the product, applying U, and using induction we get the result).</p> <p>Now let L be the field of Laurent series in x over Z/2, and L(q) be the field of Laurent series in x^q, where q is a power of 2. We write p_(q,k) for the obvious projection map L-->(x^k)L(q). Note that for g in Z/2[[x]], U(g) is the square root of p_(2,0)(g). So if g is in the ring of Lemma 3, then p_(q,0)(g) is a qth power in that ring.</p> <p>Theorem I---Let q be a power of 2, and suppose that k is in {0,1,2,3,4,5,6,7}. Suppose that g_i=p_(8q,kq)(1/[i]) is in Z/2[[x]] for all i--that is to say that no negative exponents appear in any g_i. Then each g_i is sparse. (In other words each B([i]) has density 0 in each congruence class kq mod 8q).</p> <p>To prove this note that p_(q,0)(1/[i]) is the quotient of p_(q,0)([i]^(8q-1)) by [i]^8q. The paragraph before Theorem I shows that this is the quotient of v^q by [i]^8q for some v in R. Applying p_(8q,kq) we find that g_i=(1/[i]^8q)(w^q), where w=p_(8,k)(v). Since p_(8,k) stabilizes R, g_i is the quotient of an element of R by a power of [i]. The exponent restriction tells us that if g is a g_j, then g has positive ord at each branch of C centered at m_0, and we invoke Lemma 2.</p> <p>Example: Suppose l=13, q=16 and k=3. The negative exponents appearing in the i/[i] are -36,-23,-10, -25, -16,-9,-4,and -1. Since none of these is congruent to 48 mod 128, all the g_i are in Z/2[[x]]. It follows from Theorem I that each B([i]) has density 0 in the congruence class 48 mod 128, a result that had eluded me.</p> http://mathoverflow.net/questions/68247/existence-of-certain-identities-involving-characteristic-2-thetas/68573#68573 Answer by paul Monsky for Existence of certain identities involving characteristic 2 "thetas" paul Monsky 2011-06-23T01:14:29Z 2011-11-23T16:18:17Z <p>In the first version of this answer I gave a (necessarily incorrect) proof of the false statement that when the prime,l, is 11, then G is not in the field generated over Z/2 by the [j]. In the second version I found my error, and gave a computer-aided proof that for this l, G is in the ring generated over Z/2 by the [j].</p> <p>In this completely rewritten answer I state the following conjecture and explain why it holds when l is congruent to 1 mod 4 or to 3 mod 8.</p> <p>CONJECTURE: Let l be an odd prime. Then there is a C in the ring generated by the [j] such that C^2+C=G+H. In particular, G like H is in the field generated by the [j], and if H is in the ring generated by the [j], the same is true of G.</p> <p>Proofsketch when l=1 mod 4 or l=3 mod 8.------When l=1 mod 4, take r with r^2=-1 mod l. Then [j][rj] only depends on the coset of {1,r,-1,-r} in (Z/l)* that contains j. Take C to be the sum of the [j][rj] where j runs over a set of representatives of the cosets. For example when l=13, C=[1][5]+[2][3]+[4][6]. It's an exercise in the arithmetic of Z[i] to show that C^2+C=G+H. When l=3 mod 8, take r with r^2=-2 mod l, and let C be the sum of the [rj][j][j] where j runs over representatives of the cosets of {1,-1} in the multiplicative group of Z/l. Now the result is proven using the arithmetic of Z[Root(-2)]. </p> <p>Remark: When l=7 mod 8, I may present evidence for the truth of the conjecture in a separate question. But now it seems that ternary rather than binary quadratic forms enter the picture.</p> <p>EDIT(11/23/11)</p> <p>I believe I can now prove the above conjecture. But since my proof uses the fact that G is a polynomial (over the algebraic closure, K, of Z/2) in my theta series, it doesn't supersede my other (self-accepted) answer.</p> <p>Here's the idea. Let q=x^l, and E be the elliptic curve Y^2+XY=X^3+(q+q^9+q^25+...) defined over the field of fractions of Z/2[[q]]. The j-invariant of E is 1/(q+q^9+q^25+...) "=" (E_4)^3/(Delta). Using this fact one shows that E is the characteristic 2 Tate curve. The study I've performed of the field, L, generated over K by the theta-series shows that L is the field generated over K by the x co-ordinates of the l-division points of E. (In the proof of this I use the fact that G and H are in L). But one can write these x co-ordinates explicitly as power series, using a characteristic 2 analogue of the Weierstrass P-function (see Roquette's book). It turns out that there are (l-1)/2 of these division points for which, when their x co-ordinates are summed, one gets a power series C with C^2+C=G+H. So C is in L. Once this is known it's straightforward to see that C is a polynomial in the theta-series. But why the remarkable empirical formulas for C in terms of the theta-series hold when l is 7 mod 8 remains a mystery.</p> http://mathoverflow.net/questions/43925/what-are-the-polynomial-relations-between-these-characteristic-2-thetas What are the polynomial relations between these characteristic 2 "thetas" ? paul Monsky 2010-10-28T01:47:32Z 2011-10-16T08:44:49Z <p>Suppose $\ell=2m+1$, $m>0$. Define $[i]$ in $\mathbb{Z}/2\mathbb{Z}[[x]]$ to be $$\sum_{n\equiv i\mod l} x^{n^2}.$$ Note that $[0]=1$, and that $[i]=[j]$ whenever $\ell$ divides $i+j$ or $i-j$.</p> <p>Now let $u_1,...,u_m$ be indeterminates over $\mathbb{Z}/2\mathbb{Z}$, and $f$ be the homomorphism $\mathbb{Z}/2\mathbb{Z}[u_1,...,u_m]\to \mathbb{Z}/2\mathbb{Z}[[x]]$ taking $u_i$ to $[i]$. Using the theory of modular forms I think I can show that the kernel, $P$, of $f$ is a dimension 1 prime ideal.</p> <p>Question 1: What is the genus of (a non-singular projective model) of the curve corresponding to $P$?</p> <p>Examples: When $\ell=5$ the curve one desingularizes is $x^5+y^5+xy+(xy)^2=0$, and the genus is 0.</p> <p>When $\ell=7$, the curve has the following affine plane model of degree 14: $\sum x^iy^j=0$ where $(i,j)$ runs over the 10 pairs $(14,0)$, $(12,1)$, $(10,2)$, $(7,7)$, $(6,4)$, $(5,8)$, $(5,1)$, $(4,5)$, $(1,10)$ and $(0,14)$. (Perhaps someone with access to Singular or time on their hands can work out the genus?).</p> <p>When $\ell=9$ the curve has an affine plane model of degree 27; this time one gets the 20 pairs $(27,0)$, $(24,3)$, $(21,6)$, $(20,1)$, $(15,3)$, $(13,2)$, $(12,15)$, $(12,6)$, $(11,10)$, $(11,1)$, $(9,18)$, $(9,9)$, $(7,17)$, $(6,21)$, $(5,16)$, $(5,7)$, $(4,20)$, $(4,11)$, $(1,23)$ and $(0,27)$.</p> <p>One has the following curious but easily proved relations between the various $[i]$. Let $a$,$b$,$c$,$d$,$e$,$f$ be $[i]$,$[j]$,$[2i]$,$[2j]$,$[i+j]$,$[i-j]$. Then $d(a^4)+c(b^4)+cd+(ef)^2=0$. Each such identity gives rise to a "quintic relation" lying in $P$. (I used these relations to get the curves in the above examples). Let $J$ be the ideal contained in $P$ that is generated by these quintic relations.</p> <p>Rather vague Question 2: What can be said about $J$? For example: Are all the minimal primes of $J$ of dimension 1? If so, what are the associated primes other than $P$? Is $J$ a radical ideal?</p> <p>Examples: When $\ell=5$, $J=P$, and I believe the same holds when $\ell=7$. But when $\ell=9$ one needs to add the element $a(b^2)+b(c^2)+c(a^2)+d+(d^2)+(d^3)$, where $a$,$b$,$c$,$d$ are $u_1$,$u_2$,$u_4$,$u_3$ to $J$ in order to get $P$. Let $K$ be the ideal $(a+ad,b+bd,c+cd,ab+c^2,ac+b^2,bc+a^2)$. Then $K$ is the intersection of three dimension 1 primes, and I believe that $J$ is the intersection of $P$ and $K$.</p> <p>@sleepless--I hope you like this orthography better.</p> <p>EDIT: Here are answers to question 1 when l=9 and l=11. (As I explained in a comment the genus is 3 when l=7. It now appears that it's 10 when l=9 and 26 when l=11). Remarkably when l=3,5,7,9, or 11 the genus is the same as the genus of the compactification of the quotient of the upper half-plane by the principal congruence group, Gamma(l). I doubt that this is a coincidence, and am interested in what experts in the theory of characteristic p modular forms have to say. </p> <p>Suppose first l=9. Extend the constant field from Z/2 to its algebraic closure,K. Let C in affine 4-space be the zero-locus of P, and L/K be the function field of C. P is generated by the "quintic relations" together with ab^2+bc^2+ca^2+d+d^2+d^3, where a,b,c,d are the coordinate functions u1,u2,u4 and u3. It follows that P is stabilized by the linear automorphisms (a,b,d,c)-->(b,c,d,a) and (a,b,d,c)-->(ua,ub,d,uc) with u^3=1. These automorphisms generate an order 9 group, G, which acts on L; let L_0 be the fixed field. It can be shown that L_0 is generated over K by abc and d and that (abc)^3=d^7+d^8+d^9. So L_0/K has genus 1. We now use Riemann-Hurwitz to calculate the genus, g, of L/K. (Since G has odd order, L/L_0 is tamely ramified).</p> <p>The quintic relations all vanish on the line a=b=c=0. It follows that C has 3 points on this line; they are (0,0,d,0) with d+d^2+d^3=0. Each of these points is an ordinary triple point, and G permutes the branches at each of these points in a size 3 orbit. All the other orbits of G acting on the places of the function field L/K (including the places at infinity) are of size 9. Riemann-Hurwitz now tells us that 2g-2=9(2-2)+(9-3)+(9-3)+(9-3), so that g=10.</p> <p>When l=11, one can argue in like manner. Now P is generated by the quintic relations, and the similar group G, acting on L/K, has order 55. I think one can again show that the genus of L_0/K is 1; this is the one thing I haven't checked completely. Now C sits in affine 5-space, the origin is an ordinary singular point of multiplicity 5, and G permutes the branches at the origin in a size 5 orbit. All other orbits of G acting on the places of L/K are of size 55 and Riemann-Hurwitz tells us that 2g-2=55(2-2)+(55-5), so that g=26.</p> http://mathoverflow.net/questions/43925/what-are-the-polynomial-relations-between-these-characteristic-2-thetas/78258#78258 Answer by paul Monsky for What are the polynomial relations between these characteristic 2 "thetas" ? paul Monsky 2011-10-16T08:44:49Z 2011-10-16T08:44:49Z <p>Felipe Voloch referred me to two 1959 papers of Igusa in v. 81 of Amer. J. of Math. pages 453-475 and 561-577. Results from these papers and techniques I've developed on MO give an answer to my question when l is prime.</p> <p>As I suggested in the edit to the question, the genus is (l-3)(l-5)(l+2)/24. More is true. In the first of the above-cited papers Igusa constructs, for each prime p and each N prime to p, a "field of modular functions of level N", finite and Galois over k(j) where k is the algebraic closure of Z/p. When N is a prime, l, he shows that this field has Galois group PSL_2(Z/l) over k(j) and is the splitting field of the "invariant transformation equation" Phi(X,j). In Lemma 2 of the second paper cited above he shows that this symmetric 2 variable Phi is the mod p reduction of the classical modular equation. I use these results to show that the field of my question, generated over the algebraic closure, k, of Z/2 by the theta series, identifies with Igusa's field of modular functions of level l and characteristic 2.</p> <p>A key observation(the key observation according to Kevin Buzzard--it allows me to pass from modular functions that appear to be of even level to ones evidently of level l) is that Phi(1/G,1/F)=0 where F=x+x^9+x^25+... and G=F(x^l). To see this recall Jacobi's identity (1-q)(1-q^2)(1-q^3)...=1-3q+5q^3-7q^6+9q^10... where the exponents are the triangular numbers. Raising to the power 8, multiplying by q, and reducing mod 2 shows that the mod 2 reduction of the Fourier expansion of Delta(z) is F(q). Since j(z) is the quotient of (E_4(z))^3 by Delta(z), the mod 2 reduction of the Fourier expansion of j(z) is 1/F(q), while that of j(lz) is 1/G(q). This together with the result from Igusa's second paper gives the observation.</p> <p>It now suffices to show that my field is the field generated over the algebraic closure, k, of Z/2 by G together with the l+1 conjugates of F over k(G). In various answers to other MO questions I've sketched a proof that my field admits PSl_2(Z/l) as an automorphism group, that it contains all of the above elements, and that the elements of PSL_2 all fix G. I now look at G sitting inside the fixed field of PSL_2, and show that it has exactly one zero (counted with multiplicity) in that field. So the fixed field is precisely k(G). It follows that my field is generated over k(G) by F and its k(G)-conjugates concluding the proof.</p> http://mathoverflow.net/questions/78077/function-fields-of-characteristic-p-modular-curves-and-mod-p-reductions-of-the-c Function fields of characteristic p modular curves, and mod p reductions of the classical modular equation paul Monsky 2011-10-13T23:02:28Z 2011-10-13T23:02:28Z <p>Let l and p be distinct primes, l>2. There are "characteristic p modular curves" X_0(l) and X(l), defined over an algebraic closure, K, of Z/p, solving moduli problems for elliptic curves with some additional level-l structure. Each of these curves has the same genus as the corresponding characteristic 0 object; in particular the genus of X(l) is (l-3)(l-5)(l+2)/24. </p> <p>There is also an irreducible symmetric f in Z[x,y] with f(j(lz),j(z))=0, where j is the elliptic modular function. This is the "classical modular equation". Let f* be the mod p reduction of f. I'm looking for a proof that certain well-known relations between f and the function fields of the characteristic zero X_0(l) and X(l) persist when f is replaced by f*, and X_0 and X are replaced by their characteristic p counterparts. I'd like an argument showing:</p> <p>1) f* is irreducible in K[x,y]</p> <p>2) The Galois group of f* over K(y) identifies with PSL_2(Z/l).</p> <p>3) The function field (over K) of the curve defined by f* identifies with the function field of the characteristic p X_0.</p> <p>4) If L is the splitting field (over K(y)) of f*, then L identifies with the function field (over K) of the characteristic p X.</p> <p>Remarks:</p> <p>a) I would guess that 1)---4) somehow follow from the existence of moduli schemes defined over Z[1/l]. But can someone provide a reference and details?</p> <p>b) A weaker form of 4) whose statement doesn't involve the theory of modular forms in characteristic p, is that the genus of L/K equals the genus of the classical X(l). As an old dog who has trouble with new tricks, I'd be happiest with a classical proof of this result.</p> <p>c) I'm mostly interested in the case p=2, where I can prove 1) and 2). This is all related to an MO question of mine about the genus of a curve coming from the theory of characteristic 2 theta functions.</p> http://mathoverflow.net/questions/74480/more-questions-involving-characteristic-2-theta-series-identities More questions involving characteristic 2 theta series identities paul Monsky 2011-09-04T02:53:26Z 2011-09-29T01:54:54Z <p>In my answer to my earlier question, "Existence of certain identities involving characteristic 2 thetas", I established some curious identities when the thetas have prime "level" congruent to 1 mod 4 or to 3 mod 8. This question concerns the case when the level is 7 mod 8.</p> <p>I reprise notation from earlier questions. l is an odd prime and [j] is the sum of the x^(n^2), where n runs over the integers congruent to j mod l; we view the "theta series" [j] as elements of Z/2[[x]]. F is the power series x+x^9+x^25+x^49+x^81..., G=F(x^l) and H=G(x^l). My identities involve G,H and the various [j].</p> <p>There is evidently a unique C in Z/2[[x]], having constant term 0, with C^2+C=G+H. I showed that when l is 1 mod 4 or 3 mod 8 (or when l=7), then C can be written explicitly as a polynomial in the [j]. Here is what the computer suggests when l=7 mod 8 and is &lt; 50. First some notation. If (r,s,t) is a triple of integers, we define C(r,s,t) to be the sum of the power series [rj][sj][tj] where j runs from 1 to (l-1)/2. Define C(r,s,t,u) similarly. (When l is 3 mod 8, I showed that C is C(1,1,t) where t^2 is congruent to -2 mod l).</p> <p>(1) When l=7, I can show that C=C(1,1,1,2)+C(1,2,3)</p> <p>(2) When l=23 I think that C=C(3,3,1,2)+C(1,3,6)</p> <p>(3) When l=31 I think that C=C(3,3,2,3)+C(2,3,7) (In my original post I wrote C(2,5,8), but C(2,3,7)=C(2,5,8))</p> <p>(4) When l=47 I think that C=C(3,3,2,5)+C(2,3,9)</p> <p>(Note that the sum of the squares of 3,3,2 and 5 is 47, etc.)</p> <p>QUESTION 1: Can one establish the truth of (2),(3) and (4)? Kevin Buzzard explained to me that it's enough to show that the power series expansions agree up to a certain exponent, but I'm not sure what that exponent is, and I doubt that I have the computer power.</p> <p>QUESTION 2: Are there identities like those above for l>50? And if so, what are these identities explicitly?</p> <p>EDIT: Let V be the space spanned by the C(r1,r2,r3,r4) with r1=r2 and l dividing the sum of the squares of r1,r2,r3 and r4, together with the C(s1,s2,s3) with l dividing the sum of the squares of s1,s2 and s3. When l=7 mod 16 I can use Jacobi's 4-square theorem to show that C is in V. It's then possible to prove identities like those of (2) above by exploiting the geometry of of Spec R where R is the subring of Z/2[[x]] generated by the theta series [j].</p> <p>-----One can show that an element of V has at most l(l-1)(l+1)/6 poles, counted with multiplicity, on the obvious projective completion of this curve. So if it has a zero of large enough order at the origin, it vanishes. I applied this technique for various l congruent to 7 mod 16; the results boggled my mind. It's only necessary to use 2 terms in the power series expansion of each theta series. When l=23, I got (2) above. </p> <p>When l=71, I found that C=C(3,3,2,7)+C(5,6,9)</p> <p>When l=103, I got 5 different expressions for C! Explicitly:</p> <p>a----C(3,3,6,7)+C(2,9,11)</p> <p>b----C(7,7,1,2)+C(5,9,10)</p> <p>c----C(5,5,2,7)+C(1,3,14)</p> <p>d----C(3,3,2,9)+C(6,7,11)</p> <p>e----C(1,1,1,10)+C(1,6,13)</p> <p>It seems possible to me that in general, for l=7 mod 8, one gets h/4 formulae of this sort where h is the class-number of Q(Root(-2l)). I've discussed the case l=31 in the comment to ARupinski. When l=47, I can show that C(3,3,2,5)+C(2,3,9)=C(1,1,3,6)+C(3,6,7). So if (4) above holds, there's a second formula for C in this case, just as in the case l=31. But I can't prove that C is in V when l=15 mod 16.</p> <p>UPDATE_<em>_</em> Suppose l is 7 mod 8; consider the vectors W in Z^3 with(W,W)=2l. There is a group of order 48 operating on the set of such W by permutation and sign change of co-ordinates; the group operates without fixed points. So if there are 12h such W there are h/4 orbits under the group action.</p> <p>----Ira Gessel's calculations, carried out for l&lt;1500, indicate that there is an involution, O-->O' on the set of orbits, which has the following property. Let O be any of the (h/4) orbits and (r1,r2,r3) be a representative of O with r1 even (so that r2 and r3 are odd). Then if (s1,s2,s3) is a representative of O', we have the explicit identity C((r1)/2,(r1)/2,(r2+r3)/2,(r2-r3)/2)+C(s1,s2,s3)=C.</p> <p>----But to know what these conjectured(but true beyond possibility of doubt) equations are for l>1500, we need to describe the involution. Franz Lemmermeyer suggested that the involution comes from an involution on a set of equivalence classes of quadratic forms of discriminant -8l. This is surely the case; I'll explain what the involution on the forms is, and how to transfer it to the orbits.</p> <p>----Consider positive quadratic forms rx^2+2sxy+ty^2 with s^2-rt=-2l. Gauss showed that these fall in exactly h equivalence classes under the action of SL_2(Z), where 12h is the number of W with (W,W)=2l; we'll be interested in GL_2 equivalence however. Since rt=2l+s^2, we find that mod 16, rt is 2,7,14 or 15. This can be used to show that one of the following possibilities must occur:</p> <p>a.--- Every non-zero n represented by the form is the product of an integer that is 1 or 7 mod 8 by a power of 2.</p> <p>b.---Every non-zero n represented by the form is the product of an integer that is 3 or 5 mod 8 by a power of 2.</p> <p>----In the first case we say that the form is in the principal genus, while in the second that it is in the non-principal genus. There are (h/4) GL_2 classes in the non-principal genus. Furthermore there is an involution on this set of classes taking the class of rx^2+2sxy+2ty^2 to the class of 2rx^2+2sxy+ty^2. I'll call this involution "composition with 2x^2+ly^2".</p> <p>----I now describe a map from the set of (h/4) orbits to the set of (h/4) classes. The map can be shown to be onto, and so is bijective. When we transfer composition with 2x^2+ly^2 to the set of orbits, we get our desired involution; one which is in complete accord with Gessel's calculations. Suppose (W,W)=2l. Let W# consist of all elements of Z^3 orthogonal to W. We attach to W the class of the form (xU+yV,xU+yV), where U and V are a basis of W#. This class is evidently independent of the choice of basis; one can show that it consists of forms of discriminant -8l and lies in the non-principal genus. This gives the desired map from orbits to classes of forms; as I've indicated it is bijective.</p> <p>EXAMPLE____Take l=1567, and W=(3,25,50) so that (W,W)=2l. Let O be the orbit of W. I'll calculate O', and write down the conjectured equations coming from O and O'. A basis for W# consists of U=(0,2,-1) and V=(25,1,-2). Then (U,U)=5, (U,V)=4, (V,V)=630, and a form attached to O is 5x^2+8xy+630y^2. Composition with 2x^2+1567y^2 takes this to 10x^2+8xy+315y^2. So we seek U' and V' with (U',U')=10, (U',V')=4, and (V',V')=315. Take U'=(3,1,0). A little experimenting, writing 315 as a sum of 3 squares, shows that we should take V'=(5,-11,13). Then W' which is orthogonal to U' and V' can be taken to be their vector product (13,-39,-38). So O' is the orbit of (13,38,39). And one of our predicted expressions for C is C(25,25,11,14)+C(13,38,39), while another is C(19,19,13,26)+C(3,25,50). </p> http://mathoverflow.net/questions/68247/existence-of-certain-identities-involving-characteristic-2-thetas/75368#75368 Answer by paul Monsky for Existence of certain identities involving characteristic 2 "thetas" paul Monsky 2011-09-14T02:14:25Z 2011-09-24T22:44:58Z <p>I suppose it's bad form to answer one's own MO question, but I now have an almost complete solution to this one. I can prove:</p> <p>1.----H is always in the ring S generated by the [j].</p> <p>2.----The same holds for G except perhaps when l=15 mod 16. (In "More questions involving characteristic 2 theta series identities" I provide some experimental evidence when l=15 mod 16.)</p> <p>To prove 1. note that I gave a formula in my question expressing H as a quotient of elements of S. Now I have made a study of the variety V consisting of the zeros of the polynomial relations between the various [j]. V is a curve; when l>3 it has exactly l+1 singular points, each of which is an ordinary multiple point of multiplicity (l-1)/2. Using my formula for H, I can show that it has ord at least 0 at every non-singular point of V, and ord> 0 at every branch centered at every singular point. So it lies in all the local rings of S.</p> <p>EDIT:NOT SO--the condition of being in the local ring at a singular point is more stringent. For a correct argument see the FINAL EDIT below.</p> <p>To prove 2. let C be the sum of the x^(ln) where n runs over all (non-zero) integers of the form (square) or 2(square) or l(square) or 2l(square). Note that C^2+C is G+H. So in view of 1. it suffices to show that C is in S. In my previous answer I indicated why this is true when l=1 mod 4 or l=3 mod 8, writing C explicitly as a polynomial in the [j]. I will edit this answer shortly to handle the more difficult case l=7 mod 16.</p> <p>EDIT: Suppose now l=7 mod 16. Here's a proof that C lies in S. Let T, contained in a product of 4 copies of Z/l, consist of all (r1,r2,r3,r4) other than (0,0,0,0) with (r1^2)+(r2^2)+(r3^2)+(r4^2)=0. There is a group of order (24)(16)=384 acting on T by permutation of co-ordinates and sign changes of co-ordinates. Using the fact that l=7 mod 8, we find that every orbit has size 384 or 192 or 64. Call an orbit "small" if it has size 192 or 64. We shall show that C is a sum of terms attached to the small orbits. To each small orbit attach the power series [r1][r2][r3][r4] where (r1,r2,r3,r4) is an orbit representative; this is independent of the representative. I'll show that C is the sum of these contributions. Clearly every exponent appearing in the sum of these power series is divisible by l. It remains to show that x^ln appears in the sum if and only if n is the product of a non-zero square by 1,2,l or 2l.</p> <p>Now the coefficient of x^ln in [r1][r2][r3][r4] is the mod 2 reduction of M where M is the number of integer 4-tuples (a,b,c,d) satisfying:</p> <p>(1)---(a^2)+(b^2)+(c^2)+(d^2)=0</p> <p>(2)---(a,b,c,d) reduces to (r1,r2,r3,r4) mod l</p> <p>Modulo 2, M is (1/64)(the number of (a,b,c,d) satisfying (1) and reducing to an element in the orbit of (r1,r2,r3,r4)). So the sum of the M, modulo 2, is the number of (a,b,c,d) satisfying (1) and reducing to a point in some small orbit. Also the number of (a,b,c,d) satisfying (1) and reducing to a point in an orbit of size 384 is clearly a multiple of 384. So the coefficient of x^ln in our sum is the mod 2 reduction of (1/64)(the number of (a,b,c,d) that satisfy (1) and do not reduce to (0,0,0,0)).</p> <p>Let R(n) be the number of representations of n as a sum of 4 squares. We have just shown that the coefficient of x^ln in our sum is the mod 2 reduction of (1/64)(R(ln)-R(n/l)). It remains to show that (1/64)(R(ln)-R(n/l)) is odd precisely when n is the product of a square by 1,2,l or 2l. Jacobi proved that R(n) is 8(the sum of the divisors of n) when n is odd, and 24(the sum of the odd divisors) when n is even. So (1/8)(R(ln)-R(n/l)) is a product of local factors, one from each prime. The factor attached to 2 is 1 or 3. That attached to l is l^t(1+l) where t=(ord_l)(n). Since l=7 mod 16, this is 8(an odd number). Finally that attached to an odd prime p other than l is the sum of the divisors of p^s where s is (ord_p)(n). This factor is odd just when s is even, and the result follows.</p> <p>A couple of remarks. When l=15 mod 16 the same argument shows that the sum we've constructed is not C, but 0. Also an orbit is small precisely when it has a representative with r1=r2 or a representative with r4=0.</p> <p>FINAL EDIT: I now have an answer I'm prepared to accept, unless some spoilsport finds a flaw; it shows that G,H (and F) all lie in the subring S of Z/2[[x]] generated by the [j] irrespective of l. Unlike the approach taken in the last edit which exhibited G+H explicitly as a polynomial in the [j], (except when l is 15 mod 16), this one doesn't seem to give nice explicit formulas. I'll be using results from other MO questions of mine, and some further results in manuscript. Let K be an algebraic closure of Z/2, and S' be the subring of K[[x]] generated over K by the [j]. It,s enough to show that G,H and F lie in S'.</p> <p>First I show that they're all in the field of fractions, L, of S'. In another MO post I wrote H as a quotient of 2 elements of S. To handle F I use the following:</p> <p>(1)___For l>3, Spec(S') is a curve with l+1 singular points, among them the maximal ideal m generated by [1],...,[l-1]. These are ordinary singular points of multiplicity (l-1)/2.</p> <p>(2)___There is a group of automorphisms of S'/K isomorphic to PSL_2(Z/l). These automorphisms stabilize the space spanned by [0],...,[l-1] and act transitively on the (l+1)(l-1)/2 valuation rings in L/K containing the local rings at the singular points. The group is generated by the maps [j]-->[rj], r prime to l, [j]-->a^(j^2) [j] where a is an l'th root of unity in L, and a sort of characteristic 2 "Fourier transform".</p> <p>Now the maps [j]-->[rj] and [j]-->a^(j^2) [j] generate a subgroup B of PSL_2 of order l(l-1)/2, and my "quotient formula for H" shows that B fixes H. So the orbit of H under PSL_2 has size at most l+1. A rather formal calculation with the "Fourier transform" shows that the orbit consists of H and the F(ax) where a^l=1. I claim that each of these elements lies in the local ring of m on S'. For H this is easy; H has ord l^2 at each valuation ring containing m. Taking E to be the sum of [1],...[(l-1)/2] we find that E+E^4=F+H. So F is in this local ring as well, and the result follows easily for each F(ax). The fact that PSL_2 acts transitively on the singular points now shows that H and the F(ax) lie in the local ring at every singular point. Also the quotient formula for H shows that H has ord 0 at every non-singular point, and the same then holds for the F(ax). Thus H and the F(ax) are in S'; this corrects the argument I gave earlier.</p> <p>I now turn to G. There is a degree l+1 2-variable symmetric polynomial P over Z/2 with P(F,G)=0. Furthermore P(z,G) is monic of degree l+1, and has H and the F(ax) as roots. Also the constant term of P(z,G) is G^(l+1), while the coefficient of z is G+ higher degree terms. Since the product of H and the F(ax), as well as the l'th symmetric function of H and the F(ax), are in S', both G^(l-1) and G+... are in S'. Now over K these 2 elements generate a field between K(G^(l+1)) and K(G); since G+... is in this field it is all of K(G), and G is in L. Also G^(l+1), as the product of H and the F(ax), is fixed by PSL_2. Since every homomorphism from PSL_2 to the l+1 th roots of unity is trivial, G is fixed by PSL_2.</p> <p>At the valuation rings lying over m, G has ord l. So G is in the local ring of m, and consequently in the local ring at every singular point. Furthermore, like H and the F(ax), G has ord 0 at the non-singular points. So it is in S'. (Note also that like H and the F(ax), G has poles of order 12 at every valuation ring in L/K that doesn't contain S').</p> http://mathoverflow.net/questions/125877/computing-certain-class-numbers-modulo-4 Comment by paul Monsky paul Monsky 2013-03-29T13:37:48Z 2013-03-29T13:37:48Z @Sarah--In the answer to your earlier question I could have used the seventh power of g= x+x^4+x^9+x^16+... rather than the eleventh. Then it would turn out that the coefficient of x^pq in the expansion is odd or even according as the class number of Q(root(-2pq)) is 4 mod 8 or 0 mod 8. And the Gauss theory of forms axx+bxy+cyy with b^2-4ac=-8pq would show that the class number is 4 mod 8 when (q/p)=-1 and 0 mod 8 when (q/p)=1. So my argument would show that g^7 isn't the reduction of the expansion at infinity of any modular form for any gamma_0. http://mathoverflow.net/questions/125007/does-there-exist-a-half-integer-weight-theta-function-which-is-is-equivalent-to-1/125128#125128 Comment by paul Monsky paul Monsky 2013-03-27T22:52:14Z 2013-03-27T22:52:14Z Alternatively, if p=5 mod 8 and q=7 mod 8 then the number of ways of writing pq as s_1+2s_2+8s_3 with the s_i squares is the same as the number of ways of writing pq as s_1+2s_2+8s_3 with the s_i non-zero squares. So the coefficients of x^pq in g^11 and in (1+g)^11 are the same. http://mathoverflow.net/questions/125007/does-there-exist-a-half-integer-weight-theta-function-which-is-is-equivalent-to-1/125128#125128 Comment by paul Monsky paul Monsky 2013-03-27T21:38:25Z 2013-03-27T21:38:25Z @Sarah. That's right. Another way to say it--instead of using (1/2)*(phi-(E_4)^m) in my edited comment, use (1/2)*(phi+(E_4)^m). http://mathoverflow.net/questions/125007/does-there-exist-a-half-integer-weight-theta-function-which-is-is-equivalent-to-1/125128#125128 Comment by paul Monsky paul Monsky 2013-03-25T11:48:32Z 2013-03-25T11:48:32Z Here's the argument that if u is the mod 2 reduction of an element of Z[[x]] that is the expansion at infinity of some modular form phi of weight w for gamma_0 (N) then the space spanned by the image of u under the formal Hecke operators &quot;T_q&quot; ,q prime, has finite dimension. For fixed N and w the Z-module of such elements of Z[[x]] has finite rank and is stable under the Hecke T_q where (N,q)=1. So the image of this module under mod 2 reduction contains u and has finite Z/2 dimension. The T_q with (2N,q) reduce to &quot;T_q&quot;. These &quot;T_q&quot; stabilize the image. And only finitely many q divide N. http://mathoverflow.net/questions/125007/does-there-exist-a-half-integer-weight-theta-function-which-is-is-equivalent-to-1/125128#125128 Comment by paul Monsky paul Monsky 2013-03-22T03:49:04Z 2013-03-22T03:49:04Z @Will-My proof? is more or less a version of this idea, as I think my conjectures when the level is odd should follow from results of Igusa on the modular curve. But I have no feeling for what happens when the level is even. (I took Katz and Mazur out of the library last year, but it went unread--it's not for amateurs). http://mathoverflow.net/questions/125007/does-there-exist-a-half-integer-weight-theta-function-which-is-is-equivalent-to-1 Comment by paul Monsky paul Monsky 2013-03-19T23:03:51Z 2013-03-19T23:03:51Z Maybe the following approach might be helpful.(I'm guessing that the answer to your question is no.) If there is such a theta, raising it to an appropriate odd power, multiplying what you get by 1+2(x+x^4+x^9+...), subtracting off an Eisenstein series and dividing by 2 would give a modular form of integral weight whose mod 2 reduction, g, is x+x^4+x^9+.... Then a theorem of Serre would imply that for any k almost all the coefficients of g^k are 0. I wonder what the evidence is for or against this claim about g. http://mathoverflow.net/questions/124243/are-these-empirical-discoveries-about-the-serre-swinnerton-dyer-ring-of-prime-lev Comment by paul Monsky paul Monsky 2013-03-11T19:45:05Z 2013-03-11T19:45:05Z @Alberto: Thanks. I followed your suggestion. http://mathoverflow.net/questions/121506/level-p-characteristic-2-modular-forms-and-thetas Comment by paul Monsky paul Monsky 2013-03-05T01:34:21Z 2013-03-05T01:34:21Z There's a typo in my treatment of p=19. I should have written y^2+y=x^3, not y^2+y=x^3+x. http://mathoverflow.net/questions/118512/n22m4-2p2e4-hasnt-solution-in-integers Comment by paul Monsky paul Monsky 2013-01-13T19:52:28Z 2013-01-13T19:52:28Z Since the OP hasn't accepted Franz Lemmermeyer's simple answer on stackexchange, I'll give an expanded version of it here. It's enough to show that 2K^2=(M^4)-(p^2)(e^4) has no integer solutions with (M,e)=1. Suppose on the contrary there's a solution. If M or e is even then the right-hand side is odd, a contradiction. So M and e are odd, and (M^2)+p(e^2) is 6 mod 8. But since the gcd of the positive integers (M^2)+p(e^2) and (M^2)-p(e^2) divides 2p, while the product is twice a square, (M^2)+p(e^2) is the product of an odd square by 2 or 2p and is 2 mod 8. http://mathoverflow.net/questions/118512/n22m4-2p2e4-hasnt-solution-in-integers Comment by paul Monsky paul Monsky 2013-01-12T20:00:58Z 2013-01-12T20:00:58Z This question has satisfactory answers on stackexchange--it could be closed as no longer relevant http://mathoverflow.net/questions/118117/can-every-curve-be-written-as-fxgy/118148#118148 Comment by paul Monsky paul Monsky 2013-01-06T16:38:48Z 2013-01-06T16:38:48Z Suppose you restrict attention to curves defined over Q? Is the result still true, and can you exhibit an example? What about the Klein quartic for example? I believe its Jacobian is a factor of the Jacobian of the Fermat curve of degree 7, but it's not clear to me whether it is a model of some f(x)=g(y). http://mathoverflow.net/questions/117904/elementary-examples-of-the-weil-conjectures/117909#117909 Comment by paul Monsky paul Monsky 2013-01-03T00:33:39Z 2013-01-03T00:33:39Z But the most elementary proof of RH for curves is that of Bombieri, which uses nothing more than RR for curves, and avoids any higher dimensional algebraic geometry. http://mathoverflow.net/questions/108171/the-mod-3-reduction-of-some-powers-of-delta Comment by paul Monsky paul Monsky 2012-12-14T12:40:11Z 2012-12-14T12:40:11Z Ramanujan determined the mod 27 reduction of the Fourier expansion of delta, and perhaps others have worked on my particular powers of delta. http://mathoverflow.net/questions/108171/the-mod-3-reduction-of-some-powers-of-delta Comment by paul Monsky paul Monsky 2012-12-14T12:37:20Z 2012-12-14T12:37:20Z @Will--I agree. Let g in Z/2[[x]] be the characteristic 2 analogue of f. Joel Bellaiche proved a conjecture of Nicolas and Serre, and using this found just which linear combinations of the g^k with k odd corresponded to abelian Galois representations. (In particular the g^k with k=3,5,7,19 and 21 are &quot;abelian&quot;--I write a little about this on other MO questions). I've experimentally confirmed a characteristic 3 analogue of Joel's result on linear combinations, but have no proofs. However my question about f^k where k=2,4,5 or 10 perhaps admits a more elementary answer--(to be continued) http://mathoverflow.net/questions/106859/beautiful-theorems-with-short-proof Comment by paul Monsky paul Monsky 2012-09-11T15:50:45Z 2012-09-11T15:50:45Z This one is probably on your list--Schur's proof of the Euler pentagonal number theorem by comparing partitions of n into an even number of distinct parts and partitions into an odd number of distinct parts.