EDIT... I'll present precise results for N=3,5, and 7, saving those for N=11 for later. In each case I'll give a Z/2 basis and a "naive" Z/2[G^2] basis of M(odd) and the traces of the naive basis elements. Then I'll write each C_j as a combination of the naive basis elements.
___But I won't show that the exponents appearing in C_j have the desired properties--this is hard in some cases. For N=3 and 5 my C_j will differ slightly from the ones I gave above but will satisfy the same conditions that I imposed. Let F,G,S, and P be as above. Instead of using modular forms, I'll use the explicit modular equations connecting S and P.

N=3___ Let G=R/S. Since S^4=P, S^2=(G/S)(F/S)=R^2+R. I'll write this as S^2=(1,2), and adopt a similar shorthand for any sum of powers of R. The(k)S are seen by induction to lie in Z/2[F,G]; they form a Z/2 basis of M(odd). Since G=(1)S, (G^2)S=(3,4)S. It follows easily that (0)S, (1)S, (2)S, and (3)S are a Z/2[G^2] basis. Their traces are 0,0,G, and G. The C_j are:

C5=(S^2)G=(2,3)S

N=5___Let R=S^2+(G/S). Then R^2+R=S^2+S^4+(FG/S^2)=S^2. So S^2=(1,2), and once again the (k)S lie in Z/2[F,G] and are a Z/2 basis of M(odd). Since G=RS+S^3=(1)S+(1,2)S=(2)S, it follows that (G^2)S=(5,6)S. Then (0)S, (1)S, (2)S, (3)S, (4)S, and (5)S are a Z/2[G^2] basis. The traces of these elements are 0,0,0,G,G, and G. The C_j are:

C1=F=(0,1)S

C9=(S^4)G+G^3=(4,5)S

---From (*) we deduce that (0)J, (1)J, (2)J, (3)J, (0)S, (1)S, (2)S, and (3)S are a Z/2[G^2] basis of M(odd). Their traces are 0,0,G,0,0,0,0, and G. The C_j are given by:

C9=(S^2)G=(1,2)J+(2,3)S

C11=(S^4)*G=(2)J+(3)S+(G^2)C_1

C13=(5)J=(1,2,3)J+(2,3)S+(G^2)C_3

EDIT... I'll present precise results for N=3,5, and 7, saving those for N=11 for later. In each case I'll give a Z/2 basis and a "naive" Z/2[G^2] basis of M(odd) and the traces of the naive basis elements. Then I'll write each Cj as a combination of the naive basis elements. For N=5 my C7 and C9 will differ slightly from the ones I gave above but will satisfy the same conditions that I imposed. Let F,G,S, and P be as above. Instead of using modular forms, I'll use the explicit modular equations connecting S and P.

N=3___ Let G=R/S. Since S^4=P, S^2=(G/S)(F/S)=R^2+R. I'll write this as S^2=(1,2), and adopt a similar shorthand for any sum of powers of R. The(k)S are seen by induction to lie in Z/2[F,G]; they form a Z/2 basis of M(odd). Since G=(1)S, (G^2)S=(3,4)S. It follows easily that (0)S, (1)S, (2)S, and (3)S are a Z/2[G^2] basis. Their traces are 0,0,G, and G. The Cj are:

C5=(F^2)G=(2,3)S+(G^2)C3

N=5___Let R=S^2+(G/S). Then R^2+R=S^2+S^4+(FG/S^2)=S^2. So S^2=(1,2), and once again the (k)S lie in Z/2[F,G] and are a Z/2 basis of M(odd). Since G=RS+S^3=(1)S+(1,2)S=(2)S, it follows that (G^2)S=(5,6)S. Then (0)S, (1)S, (2)S, (3)S, (4)S, and (5)S are a Z/2[G^2] basis. The traces of these elements are 0,0,0,G,G, and G. The Cj are:

C1=F=(0,2)S

C9=(S^4)G=(4,5)S+G^2*(C1+C5)

---From (*) we deduce that (0)J, (1)J, (2)J, (3)J, (0)S, (1)S, (2)S, and (3)S are a Z/2[G^2] basis of M(odd). Their traces are 0,0,G,0,0,0,0, and G. The Cj are given by:

C9=(F^2)G=(1,2)J+(2,3)S+(G^2)C7

C11=(S^4)*G=(2)J+(3)S+(G^2)C1

C13=(5)J=(1,2,3)J+(2,3)S+(G^2)C3

N=11___I claim that that there is a t in Z/2[[x]] with t^12=P and t^3+t=S, and that furthermore t lies in M(odd). To see this note that P=FG=x^12+..., and that all the exponents that appear are 0 mod 4. It follows that P=t^12 for some t in Z/2[[x]]. So by the equation connecting S and P, (S^4+P)^3=P=t^12, so that S^4+P=t^4, (S+t)^4=P=t^12, and S=t^3+t. Since t(t^2+1)=S is odd, so is t. It remains to show that t is in Z/2(F,G). I'll use [2,6] to denote t^2+t^6, and adopt a similar shorthand for sums of powers of t. Then t=[3,5,7,9]/[2,4,6,8]. But [3,5,7,9]=[1,3]^3=S^3, while [2,6]=S^2 and [4,8]=(S^4+P)+(S^8+P^2).

___Next I define R to be (t+F)/S so that R also lies in Z/2(F,G). Then R^2+R=(t+F)(t+G)/S^2= (1/S^2)(t^2+S*t+P)= (1/S^2)(t^2+t^2+t^4+t^12)= (S^4)/(S^2)= S^2 once again! So R is integral over Z/2[F,G] and lies in M. One finds that a Z/2 basis of M(odd) is given by the [k]R and the [k] with k odd. Now G=F+S=RS+S+t=[1,3]R+[3]. Since R^2=R+S^2=R+[2,6], we see that:

(*)___(G^2)[1]=[3,7]R+[5,7,13], and (G^2)[1]R=[3,5,13]R+[5,13]

---It follows from (*) that a Z/2[G^2] basis of M(odd) is given by the [k]R and the [k] with k in {1,3,5,7,9,11}. Then [9]R, [9] and [11] have trace G, while the other elements in this basis have trace 0. The Cj are:

C1=[1,3]*R+[1]

C3=[1,5,7]R+[7]

C5=[5,7]R+[5]

C7=[5,9,11]R+[11]

C9=[3,5,7]R+[5]

C11=[1,3]R+[3]

C13=[7,9,11]R+[9]

C15=[5,7]R+[7]

C17=[3,5,7,9,11]R+[5,9]

C19=[9,11]R+[9]

C21=[3,5,11]R+[5]+(G^2)C11

CORRECTIONS AND ADDITIONS(8/11)___Above I've corrected the typos and errors of my earlier edit, and added the results for N=11. One still has to show that each Cj that I present lies in C(plus) or C(minus), and that all exponents appearing in it are j mod 8. When this isn't obvious it can be carried out with the help of Hecke operators. For example, when N=7 the operator T_3 takes (F^2)G, (G^2)F and(F^4)(G^5) into C3,C5 and C13, so these elements have the desired properties. I've also succeeded in showing that M_0=C when N=11.

EDIT... I'll present precise results for N=3,5, and 7, saving those for N=11 for later. In each case I'll give a Z/2 basis and a "naive" Z/2[G^2] basis of M(odd) and the traces of the naive basis elements. Then I'll write each C_j as a combination of the naive basis elements.
___But I won't show that the exponents appearing in C_j have the desired properties--this is hard in some cases. For N=3 and 5 my C_j will differ slightly from the ones I gave above but will satisfy the same conditions that I imposed. Let F,G,S, and P be as above. Instead of using modular forms, I'll use the explicit modular equations connecting S and P.

N=3___ Let G=R/S. Since S^4=P, S^2=(G/S)(F/S)=R^2+R. I'll write this as S^2=(1,2), and adopt a similar shorthand for any sum of powers of R. The(k)S are seen by induction to lie in Z/2[F,G]; they form a Z/2 basis of M(odd). Since G=(1)S, (G^2)S=(3,4)S. It follows easily that (0)S, (1)S, (2)S, and (3)S are a Z/2[G^2] basis. Their traces are 0,0,G, and G. The C_j are:

C1=F =(0,1)S

C3=G =(1)S

C5=(S^2)G=(2,3)S

N=5___Let R=S^2+(G/S). Then R^2+R=S^2+S^4+(FG/S^2)=S^2. So S^2=(1,2), and once again the (k)S lie in Z/2[F,G] and are a Z/2 basis of M(odd). Since G=RS+S^3=(1)S+(1,2)S=(2)S, it follows that (G^2)S=(5,6)S. Then (0)S, (1)S, (2)S, (3)S, (4)S, and (5)S are a Z/2[G^2] basis. The traces of these elements are 0,0,0,G,G, and G. The C_j are:

C1=F=(0,1)S

C3=(S^2)F=(1,2,3,4)S

C5=G=(2)S

C7=(S^2)G=(3,4)S

C9=(S^4)G+G^3=(4,5)S

N=7___Let R=S^2+S^4+P. Then R^2+R=S^2+S^8+P^2+P=S^2 once again, and R^2=S^4+S^8+P^2=S^4+P. Set J=G+RS. Then J lies in M(odd), and we find that the (k)J and the (k)S are a Z/2 basis of M(odd).

___Now J^2+JS=(G+RS)(G+(R+1)S)=G(G+S)+S^4=P+S^4=(2). Also since G=(0)J+(1)S, G^2=(0)(J^2)+(2)(S^2). Combining these equations with those of the last paragraph we find that:

(*)__(G^2)J=((1,3,4)J+(2)S, and (G^2)S=(1,2)J+(2,3,4)S.

---From (*) we deduce that (0)J, (1)J, (2)J, (3)J, (0)S, (1)S, (2)S, and (3)S are a Z/2[G^2] basis of M(odd). Their traces are 0,0,G,0,0,0,0, and G. The C_j are given by:

C1=F=(0)J+(0,1)S

C3=(0,1)J

C5=(1)J

C7=G=(0)J+(1)S

C9=(S^2)G=(1,2)J+(2,3)S

C11=(S^4)*G=(2)J+(3)S+(G^2)C_1

C13=(5)J=(1,2,3)J+(2,3)S+(G^2)C_3

Fix an odd prime N. Let F in Z/2[[x]] be sum (x^n) where n runs over the odd squares. Set G=F(x^N). There is a degree N+1 irreducible polynomial relation between F and G over Z/2, (The relation has the form (F+G)^(N+1)+(lower degree terms) =0, and is symmetric).

Examples: Let S=F+G and P=FG. When N=3, S^4=P. When N=5, S^6=P. When N=7, S^8=P^2+P. When N=11, (S^4+P)^3=P.

Now let M be the integral closure of Z/2[G] in the degree N+1 extension field of Z/2(G) generated by F. View M as a subring of Z/2[[x]], and let M(odd) consist of those g in M for which each exponent n appearing in g is odd. The trace map Z/2(F,G)-->Z/2(G) maps the Z/2[G^2] module M(odd) into the cyclic module generated by G. Let M_0 consist of all elements of M(odd) of trace 0. M_0 is a free rank N module over Z/2[G^2].

For small N there are very nice bases of M(odd) and M_0 over Z/2[G^2].

N=3... Let C_1=F, C_3=G, C_5=(F^2)(G). Then the C_j are a basis of M_0. There is an element of M(odd) whose trace is G; it follows that this element and the C_j form a basis of M(odd). Furthermore for each exponent n appearing in C_j, n=j mod 8 and the Legendre symbol (n/3) is either (j/3) or 0.

N=5... Let C_1=F, C_3=(F^3)+(G^2)(F), C_5=G, C_7=(F^2)(G), C_9=(F^4)(G). Then the above results continue to hold, with 3 replaced by 5 in the final sentence.

When N=7 or 11 one can write down C_j where j is odd and <2N, and prove the corresponding facts. But when N=11 it's not possible for all the C_j to be in Z/2[F,G]; nevertheless one can arrange that the product of each C_j by (1+G^8) is in Z/2[F,G].

To what extent do the above results generalize to larger N? More precisely, let C(plus), (resp. C(minus)), consist of those g in M(odd) in which each exponent n that appears has (n/N) equal to 0 or 1 (resp. -1). Let C be C(plus)+C(minus).

Question 1--- Does M_0=C?

Remark 1... M is indeed the Serre Swinnerton-Dyer ring of characteristic 2 modular power series for Gamma_0 (N), though this isn't obvious. This allows one to introduce Hecke operators.

Remark 2... By using these operators one can show that C has rank N or N+1. The bases that I've exhibited when N=3,5,7 or 11 show that in each of these cases M_0 is contained in C. When N=3,5 or 7, I can show that M_0=C. This is surely also true when N=11, though I haven't proved it.

Remark 3... For each odd n prime to N there is a formal Hecke operator Z/2[[x]]-->Z/2[[x]]. As I've indicated, Remark 1 can be used to show that the T_n stabilize M, M(odd) and C. So one may ask the perhaps more accessible weakening of Question 1:

Question 2... Do the T_n stabilize M_0?

Fix an odd prime $N$. Let $$ F=\sum_{n\text{ an odd square}}x^n\in\mathbb{Z}/2[[x]]. $$ Set $G=F(x^N)$. There is a degree $N+1$ irreducible polynomial relation between $F$ and $G$ over $\mathbb{Z}/2$, (The relation has the form $(F+G)^{N+1}+(\text{lower degree terms}) =0$, and is symmetric).

Examples: Let $S=F+G$ and $P=FG$. When $N=3$, $S^4=P$. When $N=5$, $S^6=P$. When $N=7$, $S^8=P^2+P$. When $N=11$, $(S^4+P)^3=P$.

Now let $M$ be the integral closure of $\mathbb{Z}/2[G]$ in the degree $N+1$ extension field of $\mathbb{Z}/2(G)$ generated by $F$. View $M$ as a subring of $\mathbb{Z}/2[[x]]$, and let $M(\text{odd})$ consist of those $g$ in $M$ for which each exponent $n$ appearing in $g$ is odd. The trace map $\mathbb{Z}/2(F,G)\to \mathbb{Z}/2(G)$ maps the $\mathbb{Z}/2[G^2]$ module $M(\text{odd})$ into the cyclic module generated by $G$. Let $M_0$ consist of all elements of $M(\text{odd})$ of trace 0. $M_0$ is a free rank $N$ module over $\mathbb{Z}/2[G^2]$.

For small $N$ there are very nice bases of $M(\text{odd})$ and $M_0$ over $\mathbb{Z}/2[G^2]$.

$N=3$... Let $C_1=F$, $C_3=G$, $C_5=(F^2)(G)$. Then the $C_j$ are a basis of $M_0$. There is an element of $M(\text{odd})$ whose trace is $G$; it follows that this element and the $C_j$ form a basis of $M(\text{odd})$. Furthermore for each exponent $n$ appearing in $C_j$, $n\equiv j \mod{8}$ and the Legendre symbol $(n/3)$ is either $(j/3)$ or 0.

$N=5$... Let $C_1=F$, $C_3=(F^3)+(G^2)(F)$, $C_5=G$, $C_7=(F^2)(G)$, $C_9=(F^4)(G)$. Then the above results continue to hold, with 3 replaced by 5 in the final sentence.

When $N=7$ or 11 one can write down $C_j$ where $j$ is odd and $<2N$, and prove the corresponding facts. But when $N=11$ it's not possible for all the $C_j$ to be in $Z/2[F,G]$; nevertheless one can arrange that the product of each $C_j$ by $(1+G^8)$ is in $Z/2[F,G]$.

To what extent do the above results generalize to larger $N$? More precisely, let $C(\text{plus})$, (resp. $C(\text{minus})$), consist of those $g$ in $M(\text{odd})$ in which each exponent $n$ that appears has $(n/N)$ equal to 0 or 1 (resp. -1). Let $C$ be $C(\text{plus})+C(\text{minus})$.

Question 1--- Does $M_0=C$?

Remark 1... $M$ is indeed the Serre Swinnerton-Dyer ring of characteristic 2 modular power series for $\Gamma_0(N)$, though this isn't obvious. This allows one to introduce Hecke operators.

Remark 2... By using these operators one can show that $C$ has rank $N$ or $N+1$. The bases that I've exhibited when $N=3,5,7$ or 11 show that in each of these cases $M_0$ is contained in $C$. When $N=3,5$ or 7, I can show that $M_0=C$. This is surely also true when $N=11$, though I haven't proved it.

Remark 3... For each odd $n$ prime to $N$ there is a formal Hecke operator $\mathbb{Z}/2[[x]]\to\mathbb{Z}/2[[x]]$. As I've indicated, Remark 1 can be used to show that the $T_n$ stabilize $M$, $M(\text{odd})$ and $C$. So one may ask the perhaps more accessible weakening of Question 1:

Question 2... Do the $T_n$ stabilize $M_0$?