Suppose l is odd and >1. For each i in the quotient of (Z/l)* by {1,-1} let U_i in Z/2[[x]] be sum(x^s) where s in Z runs over the squares congruent to i^2 mod l.
Does the extension field of Z/2 generated by the U_i have transcendence degree 1? And if it does, what is the genus of the corresponding non-singular projective curve?
Can one describe the ideal of polynomial relations between the U_i explicitly? For example, suppose l is prime. Then if U_i and U_j are different one has:
(U_2i)(U_j)^4 +(U_2j)(U_i)^4 +(U_2i)(U_2j)+ ((U_k)^2)(U_l)^2 =0, where k and l are i+j and i-j. Do these relations generate the full ideal of polynomial relations?
Example: When l=5 let A and B be U_2 and U_1. Then the ideal of relations is generated by A^5+B^5+AB+(AB)^2, and we get a curve of genus 0.
Remark: Each (U_i)*(U_j) is the mod 2 reduction of the Fourier expansion of a weight 1 modular form for a congruence group. So perhaps one is getting something like the mod 2 reduction of some modular curve that has been studied?
EDIT: l should be an odd prime or a power of an odd prime--otherwise the remark isn't right.