Suppose $\ell=2m+1$, $m>0$. Define $[i]$ in $\mathbb{Z}/2\mathbb{Z}[[x]]$ to be $$\sum_{n\equiv i\mod l} x^{n^2}.$$ Note that $[0]=1$, and that $[i]=[j]$ whenever $\ell$ divides $i+j$ or $i-j$.

Now let $u_1,...,u_m$ be indeterminates over $\mathbb{Z}/2\mathbb{Z}$, and $f$ be the homomorphism $\mathbb{Z}/2\mathbb{Z}[u_1,...,u_m]\to \mathbb{Z}/2\mathbb{Z}[[x]]$ 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 $\ell=5$ the curve one desingularizes is $x^5+y^5+xy+(xy)^2=0$, and the genus is 0.

When $\ell=7$, the curve has the following affine plane model of degree 14: $\sum x^iy^j=0$ 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 $\ell=9$ 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 $\ell=5$, $J=P$, and I believe the same holds when $\ell=7$. But when $\ell=9$ 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.

EDIT: Here are answers to question 1 when l=9 and l=11. (As I explained in a comment the genus is 3 when l=7. It now appears that it's 10 when l=9 and 26 when l=11). Remarkably when l=3,5,7,9, or 11 the genus is the same as the genus of the compactification of the quotient of the upper half-plane by the principal congruence group, Gamma(l). I doubt that this is a coincidence, and am interested in what experts in the theory of characteristic p modular forms have to say.

Suppose first l=9. Extend the constant field from Z/2 to its algebraic closure,K. Let C in affine 4-space be the zero-locus of P, and L/K be the function field of C. P is generated by the "quintic relations" together with ab^2+bc^2+ca^2+d+d^2+d^3, where a,b,c,d are the coordinate functions u1,u2,u4 and u3. It follows that P is stabilized by the linear automorphisms (a,b,d,c)-->(b,c,d,a) and (a,b,d,c)-->(ua,ub,d,uc) with u^3=1. These automorphisms generate an order 9 group, G, which acts on L; let L_0 be the fixed field. It can be shown that L_0 is generated over K by abc and d and that (abc)^3=d^7+d^8+d^9. So L_0/K has genus 1. We now use Riemann-Hurwitz to calculate the genus, g, of L/K. (Since G has odd order, L/L_0 is tamely ramified).

The quintic relations all vanish on the line a=b=c=0. It follows that C has 3 points on this line; they are (0,0,d,0) with d+d^2+d^3=0. Each of these points is an ordinary triple point, and G permutes the branches at each of these points in a size 3 orbit. All the other orbits of G acting on the places of the function field L/K (including the places at infinity) are of size 9. Riemann-Hurwitz now tells us that 2g-2=9(2-2)+(9-3)+(9-3)+(9-3), so that g=10.

When l=11, one can argue in like manner. Now P is generated by the quintic relations, and the similar group G, acting on L/K, has order 55. I think one can again show that the genus of L_0/K is 1; this is the one thing I haven't checked completely. Now C sits in affine 5-space, the origin is an ordinary singular point of multiplicity 5, and G permutes the branches at the origin in a size 5 orbit. All other orbits of G acting on the places of L/K are of size 55 and Riemann-Hurwitz tells us that 2g-2=55(2-2)+(55-5), so that g=26.

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-marks around your math LaTeX terms. That'll really simplify reading this. – sleepless in beantown Oct 30 '10 at 23:17