MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

2 Further results stated, a conjecture made.

EDIT: Let X be as in the paragraph above. In my question I showed that X is contained in the zero-locus of a set of "quintic relations". In my comment below I showed that this zero-locus imbeds in projective m-space in such a way that it has only finitely many "points at infinity". So all its irreducible components are of dimension 0 or 1. Let Y_i be the 1-dimensional irreducible components and Y their union, so that X is one of the Y_i. All this goes through even when l is composite. But when l is prime I now think I can prove more:

A)... Each Y_i passes through the origin and is the image of X under a certain permutationof coordinates. (The permutation corresponds to the permutation of the theta-series given by [i]-->[ri] for some r prime to l).

B)... deg(Y)= l(l-1)(l+1)/24. So the degree of X divides this number.

I believe that X is the only component of Y. To prove this amounts to showing that if S is the subring of Z/2[[x]] generated by the theta-series, then for each r prime to l there is an automorphism of S taking [i] to [ri] for each i. (This conjecture has a modular forms feel. Can anyone provide a proof of it?)

1

In my question I remarked that when l=2m+1 a modular form argument might be used to show that the field generated by [1], ... ,[m] over Z/2 has transcendence degree 1. Below I give an elementary proof that the transcendence degree is 1 when l is prime, based on the quintic relations. The proof is too long to be a comment or edit, so I'm posting it as an answer.

Lemma----Let F be a field of characteristic 2 and n-->a_n be a function Z/l-->F satisfying:

(1) a_0 =1

(2) a_i =a_-i

(3) The sum of (a_2i)(a_j)^4, (a_2j)(a_i)^4, (a_2i)(a_2j) and (a_(i+j)a_(i-j))^2 is 0.

Then if a_1 =0, each of a_2, ... ,a_(l-1) is 0.

The proof proceeds in 3 steps: First I claim that if a_2 is 0 then a_1 is 0. For suppose the contrary. Since a_2l is 1, there is an odd positive r with a_2r non-zero. Take such an r as small as possible; since a_2 =0,r>1 and so (r+1)/2 is less than r. Taking i=r and j=1 in (3) we find that (a_2r)(a_1)^4 is the square of (a_(r+1))(a_(r-1)). So a_(r+1) and a_(r-1) are non-zero. But one of (r+1)/2, (r-1)/2 is odd. This contradicts the minimality of r.

Observe that if a_2s is 0 then a_s is 0. To see this note that s is not 0 in Z/l, and apply the result of the paragraph above to the function (s)(i)-->a_i.

Suppose finally that a_r and a_s are non-zero while a_(r+s) is 0. Then a_((r+s)/2) is 0. Applying (3) with i=(r+s)/2 and j=(r-s)/2 we find that the square of (a_r)(a_s) is 0, a contradiction. So the n in Z/l with a_n non-zero form a subgroup of the additive group, completing the proof.

Theorem---Let K be an algebraic closure of Z/2 and T be the subring of K[[x]] generated over K by all the [i]. Then the only prime ideal of the affine domain T that contains [1] is the maximal ideal ([1], ... ,[m]).

Note first that T is generated by [1], ... ,[m]. So the ideal of T generated by these elements is indeed maximal. Let I be a prime ideal that contains [1], and F be the field of fractions of T/I. Consider the function Z/l-->F taking the congruence class i+lZ to the image of [i] in T/I. This function clearly satisfies (1) and (2) of the Lemma. The quintic relations show that it satisfies (3) as well. As the function takes 1+lZ to 0, we find by the Lemma that it takes 2+lZ, ... , m+lZ to 0 as well. So I contains each of [1], ... ,[m].

The result I mentioned is an immediate consequence. For let X be the irreducible algebraic set in affine m-space over K corresponding to T. The Theorem tells us that the intersection of X with one of the coordinate hyperplanes consists of the origin alone. So X has dimension 1 and T has transcendence degree 1 over K.