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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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@paul-Monsky, it's probably worth editing in the $LaTeX$ by putting $-marks around your math LaTeX terms. That'll really simplify reading this. –  sleepless in beantown Oct 30 '10 at 23:17
    
@sleepless: I set it into tex to make it more readable...was hoping I'd have an answer once I knew what the question was, but sadly, not so much. –  Charles Siegel Nov 9 '10 at 18:12
    
Michael Stillman tells me that when l=11 P has degree 55 and J=P. He also confirms my belief that when l=7 P has degree 14 and J=P and that when l=9 P has degree 27 and J is the intersection of P and the K above. (So it's conceivable that J=P whenever l is prime). He says that the primary decompositions of J when l is 7,9 or 11 are a challenge for Macaulay2. –  paul Monsky Nov 11 '10 at 14:38
    
When l=7 I understand the curve attached to P=J better now. Look at the model in affine 3 space defined by the 3 quintic relations. This curve admits a group of 7 automorphisms (u_1,u_2,u_3)-->(ru_1,(r^4)u_2,(r^2)u_3),r^7=1, as well as automorphisms by cyclic permutation of coordinates. At (0,0,0) there are 3 linear branches,with the branch tangents being the coordinate axes. An order 7 automorphism fixes the 3 branches and one gets a degree 7 Galois cover (presumably of the projective line) ramified at 3 points. So the genus is 3, and the curve surely is birational to the Klein quartic. –  paul Monsky Nov 11 '10 at 22:28
    
Here's a proof of the relation between a,b,c,d,e and f that I give in my question. If u is in Z/2[[x]] then u is uniquely v+w with v and w/x in Z/2[[x^2]]. Let p and q be the maps u-->v and u-->w. Evidently p(c) and p(d) are a^4 and b^4.So p(cd)=p(c)p(d)+q(c)q(d)=d(a^4)+c(b^4)+cd. Thus it suffices to show that p(cd)=(ef)^2. But this follows from the fact that if 2n is r^2+s^2, then n is the sum of the squares of (r+s)/2 and (r-s)/2. –  paul Monsky Nov 23 '10 at 14:57
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2 Answers

In my question I remarked that when l=2m+1 a modular form argument might be used to show that the field generated by [1], ... ,[m] over Z/2 has transcendence degree 1. Below I give an elementary proof that the transcendence degree is 1 when l is prime, based on the quintic relations. The proof is too long to be a comment or edit, so I'm posting it as an answer.

Lemma----Let F be a field of characteristic 2 and n-->a_n be a function Z/l-->F satisfying:

(1) a_0 =1

(2) a_i =a_-i

(3) The sum of (a_2i)(a_j)^4, (a_2j)(a_i)^4, (a_2i)(a_2j) and (a_(i+j)a_(i-j))^2 is 0.

Then if a_1 =0, each of a_2, ... ,a_(l-1) is 0.

The proof proceeds in 3 steps: First I claim that if a_2 is 0 then a_1 is 0. For suppose the contrary. Since a_2l is 1, there is an odd positive r with a_2r non-zero. Take such an r as small as possible; since a_2 =0,r>1 and so (r+1)/2 is less than r. Taking i=r and j=1 in (3) we find that (a_2r)(a_1)^4 is the square of (a_(r+1))(a_(r-1)). So a_(r+1) and a_(r-1) are non-zero. But one of (r+1)/2, (r-1)/2 is odd. This contradicts the minimality of r.

Observe that if a_2s is 0 then a_s is 0. To see this note that s is not 0 in Z/l, and apply the result of the paragraph above to the function (s)(i)-->a_i.

Suppose finally that a_r and a_s are non-zero while a_(r+s) is 0. Then a_((r+s)/2) is 0. Applying (3) with i=(r+s)/2 and j=(r-s)/2 we find that the square of (a_r)(a_s) is 0, a contradiction. So the n in Z/l with a_n non-zero form a subgroup of the additive group, completing the proof.

Theorem---Let K be an algebraic closure of Z/2 and T be the subring of K[[x]] generated over K by all the [i]. Then the only prime ideal of the affine domain T that contains [1] is the maximal ideal ([1], ... ,[m]).

Note first that T is generated by [1], ... ,[m]. So the ideal of T generated by these elements is indeed maximal. Let I be a prime ideal that contains [1], and F be the field of fractions of T/I. Consider the function Z/l-->F taking the congruence class i+lZ to the image of [i] in T/I. This function clearly satisfies (1) and (2) of the Lemma. The quintic relations show that it satisfies (3) as well. As the function takes 1+lZ to 0, we find by the Lemma that it takes 2+lZ, ... , m+lZ to 0 as well. So I contains each of [1], ... ,[m].

The result I mentioned is an immediate consequence. For let X be the irreducible algebraic set in affine m-space over K corresponding to T. The Theorem tells us that the intersection of X with one of the coordinate hyperplanes consists of the origin alone. So X has dimension 1 and T has transcendence degree 1 over K.

EDIT: Let X be as in the paragraph above. In my question I showed that X is contained in the zero-locus of a set of "quintic relations". In my comment below I showed that this zero-locus imbeds in projective m-space in such a way that it has only finitely many "points at infinity". So all its irreducible components are of dimension 0 or 1. Let Y_i be the 1-dimensional irreducible components and Y their union, so that X is one of the Y_i. All this goes through even when l is composite. But when l is prime I now think I can prove more:

A)... Each Y_i passes through the origin and is the image of X under a certain permutation of coordinates. (The permutation corresponds to the permutation of the theta-series given by [i]-->[ri] for some r prime to l).

B)... deg(Y)= l(l-1)(l+1)/24. So the degree of X divides this number.

I believe that X is the only component of Y. To prove this amounts to showing that if S is the subring of Z/2[[x]] generated by the theta-series, then for each r prime to l there is an automorphism of S taking [i] to [ri] for each i. (This conjecture has a modular forms feel. Can anyone provide a proof of it?)

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Here's another argument which works for composite l as well. Add a new variable T and homogenize the quintics. Let U, contained in projective m-space over K, be the zero locus of these quintics. If we can show that the intersection of U with the hyperplane T=0 is finite, then every irreducible component of U has dimension 0 or 1; since X is contained in U, X is a curve. Let (a_1...a_r 0) be the projective coordinates of a point of U "at infinity". Looking at the quintics we see that (a_2j)(a_i)^4 = (a_2i)(a_j)^4. Fix i with a_i non-zero. Taking j=i/2 (to be continued) –  paul Monsky Jun 30 '11 at 16:17
    
we see a_2i is non-zero. Also, (a_2j)/(a_2i) is the fourth power of (a_j)/(a_i). So if 2^r =1 mod l, then (a_j)/(a_i) is a root of z^(4^r)=z. Since there are only finitely many roots we're done. –  paul Monsky Jun 30 '11 at 16:21
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Felipe Voloch referred me to two 1959 papers of Igusa in v. 81 of Amer. J. of Math. pages 453-475 and 561-577. Results from these papers and techniques I've developed on MO give an answer to my question when l is prime.

As I suggested in the edit to the question, the genus is (l-3)(l-5)(l+2)/24. More is true. In the first of the above-cited papers Igusa constructs, for each prime p and each N prime to p, a "field of modular functions of level N", finite and Galois over k(j) where k is the algebraic closure of Z/p. When N is a prime, l, he shows that this field has Galois group PSL_2(Z/l) over k(j) and is the splitting field of the "invariant transformation equation" Phi(X,j). In Lemma 2 of the second paper cited above he shows that this symmetric 2 variable Phi is the mod p reduction of the classical modular equation. I use these results to show that the field of my question, generated over the algebraic closure, k, of Z/2 by the theta series, identifies with Igusa's field of modular functions of level l and characteristic 2.

A key observation(the key observation according to Kevin Buzzard--it allows me to pass from modular functions that appear to be of even level to ones evidently of level l) is that Phi(1/G,1/F)=0 where F=x+x^9+x^25+... and G=F(x^l). To see this recall Jacobi's identity (1-q)(1-q^2)(1-q^3)...=1-3q+5q^3-7q^6+9q^10... where the exponents are the triangular numbers. Raising to the power 8, multiplying by q, and reducing mod 2 shows that the mod 2 reduction of the Fourier expansion of Delta(z) is F(q). Since j(z) is the quotient of (E_4(z))^3 by Delta(z), the mod 2 reduction of the Fourier expansion of j(z) is 1/F(q), while that of j(lz) is 1/G(q). This together with the result from Igusa's second paper gives the observation.

It now suffices to show that my field is the field generated over the algebraic closure, k, of Z/2 by G together with the l+1 conjugates of F over k(G). In various answers to other MO questions I've sketched a proof that my field admits PSl_2(Z/l) as an automorphism group, that it contains all of the above elements, and that the elements of PSL_2 all fix G. I now look at G sitting inside the fixed field of PSL_2, and show that it has exactly one zero (counted with multiplicity) in that field. So the fixed field is precisely k(G). It follows that my field is generated over k(G) by F and its k(G)-conjugates concluding the proof.

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