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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 $\ell+1$).

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 $f$?

Examples--(See pages 127-132 of Borwein and Borwein which allow one to calculate $f$ for $\ell<29$):

  • $\ell=3$: $f=XY+(X+Y)^4$
  • $\ell=5$: $f=XY+(X+Y)^6$
  • $\ell=7$: $f=XY+(XY)^2+(X+Y)^8$.

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.

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.

EDIT: (1/12/14)

(1)--- There's a much simpler proof than the one I sketch in my comments that there is an f of total degree ell+1 in Z/2[X,Y] with f(B,A)=0. Rather than using the Jacobi thetas one notes that A and B are the mod 2 reductions of the modular forms delta(z) and delta(ell*z). Now there is a monic polynomial of degree ell+1 whose roots are delta(ell*z) and the delta((z+j)/ell). One finds that the coefficients of this polynomial lie in Z[[x]] and are modular forms of level 1. So they lie in the ring generated by delta(z), E_4 and E_6, and we use the fact that the mod 2 reductions of E_4 and E_6 are 1, (and that delta(ell*z) is a root of our polynomial) to get the desired result. This argument gives related results in characteristics 3,5,7 and 13--see the edits to my question 153787.

(2)--- Here's one more fact about f. If g is the degree ell+1 part of f then g=(X+Y)^(ell+1) as in the examples given in my question. I had thought this was shown in Borwein and Borwein, but this seems not to be so. Surely the result is known--I'd be interested in a reference. At any rate here's a complicated argument based on my theory of theta series in characteristic 2; see my various questions and answers. Let C=B(x^ell), so that f(C,B)=0. Write f as g+h where g is homogeneous of degree ell+1, and h has degree at most ell. Let W be 1/B. Dividing the relation f(C,B)=0 by B^(ell+1) we find that C/B satisfies a monic equation of degree ell+1 with coefficients in Z/2[W], and that the mod W reduction of this equation is g(X,1). So (B+C)/B satisfies such an equation, but now the reduction is g(X+1,1).

---Now in my MO questions and answers I study the field L generated over an algebraic closure, K, of Z/2 by A,B, and C. I show that L is Galois over K(B) and has as a model an affine curve with ell*(ell+1)*(ell-1)/24 points at infinity. I show further that these points are just the points lying over the zero of W, and that the rational function (B+C)/C has a zero at each of these points. It follows that the irreducible monic equation satisfied by (B+C)/B over K(B) has all its coefficients, apart from the coefficient of X^(ell+1), in W*Z/2[W] and so has W-reduction equal to X^(ell+1). So g(X+1,1)= X^(ell+1), g(X,1)= (X+1)^(ell+1), and g(X,Y)=(X+Y)^(ell+1).

---This is discussed further in the edits to my question 153787. Computer experiment suggests results of a similar character in characteristics 3,5,7 and 13, with A and B replaced by the reductions of delta(z) and delta(ell*z) in these characteristics. I don't know if such results are known--again I'd appreciate a reference if they are.

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I haven't fully explained why the equation relating the modular functions u and v gives a non-trivial relation between A and B. To see this note that u (as described in "Pi and the AGM" or classical papers) is a square root of (theta_2)/(theta_3), and that v=u(lz). Let r be exp((pi)(iz)/8). Then (theta_2)(z/2)=2*(r+r^9+r^25+...), while (theta_3)(z)=1+2*r^8+2*r^32+2*r^72+... So the quotient, U, of (theta_2)(z/2) by 2(theta_3)(z) has an expansion in powers of r, with integer coefficients, whose mod 2 reduction is A(r). And V=U(lz) has (to be continued)... –  paul Monsky Jan 28 '11 at 21:19
an expansion with mod 2 reduction B(r). Since the square of (theta_2) (z/2) is 2(theta_2)(theta_3), u=(root 2)U and v=(root 2)V. The u-v modular equation, Omega_l, gives, on replacing u and v by (root 2)U and (root 2)V, a relation between U and V. The monomials of smallest "2-ord" in the relation are those for which c_i,j +(1/2)(i+j) is least.Scaling and reducing mod 2 we get a non-trivial f with f(A,B)=0. –  paul Monsky Jan 28 '11 at 21:50
I'll give a proof that X^(l+1), Y^(l+1) and XY appear in f and that f is irreducible. When i=l+1 and j=0, c_i,j=0. So the least value of c_i,j +(1/2)(i+j) is at most (1/2)(l+1) and the total degree of f is at most l+1. Note that some monomial (X^i)(Y^j) in f has i at least l. For (A^i)(B^j)=x^(i+lj)+ higher degree terms, and these (A^i)(B^j) sum to 0. So f has total degree l or l+1. Furthermore some monomial X^s appears in f. For otherwise Y divides f, and by symmetry f=XYg for some g. Since g(A,B)=0, g has total degree l or l+1, a contradiction. Choose s (to be continued).. –  paul Monsky Jan 28 '11 at 22:07
as small as possible. Since f(A,B)=0, arguing as above we see s is l or l+1. If s=l so that X^l appears in f, then the monomial Y must also appear in f. But then X appears, a contradiction. We've shown that X^(l+1) (and Y^(l+1)) appear in f. Since f(A,B)=0, XY must appear too. Let g be the irreducible relation between A and B. Then g divides f, and since g(A,B)=0, g has degree l or l+1. So either f=g or f= (X+Y)g. Since XY appears in f, the first possibility holds, and f is irreducible. –  paul Monsky Jan 28 '11 at 22:15
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