## What’s known about the mod 2 reduction of the level l Jacobi modular equation?

Motivation:

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.

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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 2011 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 2011 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 2011 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 2011 at 22:15