Suppose l=2m+1, m>0. Define [i] in Z/2[[x]] to be sum (x^(n^2)), the sum running over all n congruent to i mod l. Note that [0]=1, and that [i]=[j] whenever l divides i+j or i-j.

Now let u_1,...,u_m be indeterminates over Z/2, and f be the homomorphism Z/2[u_1,...,u_m]-->Z/2[[x]] taking u_i to [i]. Using the theory of modular forms I think I can show that the kernel, P, of f is a dimension 1 prime ideal.

Question 1----What is the genus of(a non-singular projective model) of the curve corresponding to P?

Examples: When l=5 the curve one desingularizes is x^5+y^5+xy+(xy)^2=0, and the genus is 0. When l=7, the curve has the following affine plane model of degree 14: sum((x^i)(y^j))=0 where (i,j) runs over the 10 pairs (14,0) (12,1) (10,2) (7,7) (6,4) (5,8) (5,1) (4,5) (1,10) and (0,14). (Perhaps someone with access to Singular or time on their hands can work out the genus?). When l=9 the curve has an affine plane model of degree 27; this time one gets the 20 pairs (27,0) (24,3) (21,6) (20,1) (15,3) (13,2) (12,15) (12,6) (11,10) (11,1) (9,18) (9,9) (7,17) (6,21) (5,16) (5,7) (4,20) (4,11) (1,23) and (0,27).

One has the following curious but easily proved relations between the various [i]. Let a,b,c,d,e,f be [i],[j],[2i],[2j],[i+j],[i-j]. Then d(a^4)+c(b^4)+cd+(ef)^2=0. Each such identity gives rise to a "quintic relation" lying in P. (I used these relations to get the curves in the above examples). Let J be the ideal contained in P that is generated by these quintic relations.

Rather vague Question 2----What can be said about J? For example--Are all the minimal primes of J of dimension 1? If so, what are the associated primes other than P? Is J a radical ideal?

Examples: When l=5, J=P, and I believe the same holds when l=7. But when l=9 one needs to add the element a(b^2)+b(c^2)+c(a^2)+d+(d^2)+(d^3), where a,b,c,d are u_1,u_2,u_4,u_3 to J in order to get P. Let K be the ideal (a+ad,b+bd,c+cd,ab+c^2,ac+b^2,bc+a^2). Then K is the intersection of three dimension 1 primes, and I believe that J is the intersection of P and K.

@sleepless--I hope you like this orthography better.