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BACKGROUND

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".

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:

1._ 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).

2._ 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).

3._ 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'.

4._ 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'.

QUESTION

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?

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<20. Fix an odd prime p.

THE RING A

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.

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<20 support the conjecture).

THE RING B'

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.

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:

Conjecture 2: B' is the integral closure of the ring Z/2[f1(x^p),fp(x^p)] in its field of fractions.

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.

EDIT---EXPLICIT FORMULAS

For each odd prime p<20 I'll write down:

(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.

(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.

I start with the genus 0 cases.

p=3

(a) A=Z/2[r]. The curve is the affine line. f3=r^3+r^4

(b) R=C(1,1,1)

p=5

(a) A=Z/2[r]. The curve is the affine line. f5=r^5+r^6

(b) R=C(1,2)

p=7

(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.

(b) R=C(1,1,1,2)+C(1,2,3) U=C(1,2,3)

p=13

(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.

(b) R=C(2,3) U=R+C(1,2,3,5)

I next turn to genus 1:

p=11

(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.

(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=17

(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.

(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=19

(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).

(b) R=C(1,3,3) V=C(1,2,4,6) T=R+C(2,3,3,4) U=R+V

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.

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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. –  paul Monsky Mar 5 '13 at 1:34
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