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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.

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.

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 $K$.

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.

1. 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?

2. 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?

3. 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.

4. 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.

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.