4
$\begingroup$

Suppose ell is prime and (N,ell)=1. Consider those power series over Z that are expansions at infinity of modular forms for gamma_0 (N) of weight a multiple of ell-1. I'll say that an element of (Z/ell)[[x]] is in M#(N) (or more briefly in M#) if it is the mod ell reduction of such a power series.

When ell is 2 or 3 there is no weight restriction, and so M# is just the Serre Swinnerton-Dyer ring, M, of question 93059. When ell>3 we can use the fact that the expansion of E_(ell-1) is 1 mod ell to see that M# is closed under addition, and is a subring of M.

I'd like to know the structure of M#(N), or more geometrically what the affine curve C(N) over Z/ell attached to M#(N) looks like. My guess is that there's a simple answer in terms of the characteristic ell non-singular projective modular curve X_0 (N) constructed by Igusa.

Explicitly let J be the mod ell reduction of the "Laurent expansion" (1/x)+744+... of j(z). Let K(1) be the extension field of Z/ell generated by J(x), and let K(N) be the field of Laurent series generated over Z/ell by the J(x^d) with d dividing N. If my understanding is correct, X_0 (1) is the projective j-line over Z/ell, there is a branched covering phi: X_0(N)-->X_0(1) defined over Z/ell, and the function fields of X_0(1) and X_0(N) identify with K(1) and K(N).

Now let P be the monic separable element of Z/ell[x] whose roots are the supersingular j-values. Then it can be shown that M(1) is the subring of (Z/ell)(J) generated by the (J^k)/P(J) with k< deg P. In other words, C(1) is the "ordinary part" of X_0(1); it is the projective j-line with the supersingular j omitted.

QUESTION: Is it true that C(N) identifies with the inverse image of C(1) under phi: X_0(N)-->X_0(1)? Alternatively (assuming my description of the function field of X_0 (N) is correct), is it true that M#(N) is the integral closure of M#(1) in the extension field generated by the J(x^d) with d dividing N?

EDIT: I've corrected some typos, where I wrote X(N) when I meant X_0(N). Kevin Buzzard, in response to an inquiry, has kindly told me that the answer to my question is yes, and indicated a proof.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.