Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $q$ and $q'$ be complex numbers with $0<|q|,|q'|<1$, and let $m$ and $n$ be positive integers.
Suppose that $q^m={q'}^n$. Then the map $$ f:\mathbb{C}^\times/q^{\mathbb{Z}} \to \mathbb{C}^\times/{q'}^{\mathbb{Z}}\qquad \text{defined by}\qquad u\mapsto u^m $$ gives an isogeny of (analytic) elliptic curves over $\mathbb{C}$.

The Tate curve $\mathrm{Tate}(q)$ is an (algebraic) elliptic curve over the Laurent series ring $\mathbb{Z}((q))$ which can be used to give a uniformization of the curve $\mathbb{C}^\times/q^\mathbb{Z}$ by means of certain well known explicit formulae.

My question is:

Does there exist an isogeny $\mathrm{Tate}(q)\to \mathrm{Tate}(q')$ of elliptic curves defined over $\mathbb{Z}((q,q'))/(q^m-{q'}^n)$ which "lifts" the map $f$ above, and if so, how do you prove it exists?

It should suffice to construct such an isogeny for $(m,n)=(m,1)$, and use dual isogenies and composition to get the general case.

(I'm being vague about "lifts", because one has to worry about convergence somewhere. Probably you want to say that the isogeny is defined over some subring of $\mathbb{Z}((q,q'))/(q^m-{q'}^n)$ of power series which are analytically convergent near $q=0$, or something like that.)

I presume (though I probably can't prove) that the existence of the analytic isogenies means that such a map of schemes is defined over $\mathbb{C}((q,q'))/(q^m-{q'}^n)$, so that this is just a question about integrality.

This is very closely related to exercise 5.10 in Silverman, Advanced Topics in the Arithmetic of Elliptic Curves. There, he apparently asks us to show that for a $p$-adic field $K$, if $q,q'\in K$, $0<|q|,|q'|<1$, and $q^m={q'}^n$, then the function $\overline{K}^\times/q^\mathbb{Z}\to \overline{K}^\times/{q'}^{\mathbb{Z}}$ defined by $u\mapsto u^m$ lifts to an isogeny $E_q\to E_{q'}$ of elliptic curves over $K$, where $E_q$ and $E_{q'}$ are defined by the Tate curve equations. (An answer to my question solves his exercise, right?)

Unfortunately, I have no idea how to do Silverman's exercise either (he marks it as difficult). Any hints?

share|improve this question
    
The answer is "yes", but only way I understand such things is via formal schemes & generalized ell. curves over rings (such as $\mathbf{Z}[[q,q']]/(q^n - {q'}^m)$, & then invert $q$ and $q'$ to get to ell. curves). Methods in AECII & rigid-analytic geometry are useless when base ring is not field. I asked Tate if he knew how to rigorously understand structure of $N$-torsion on Tate curve over $\mathbf{Z}((q))$ without hard alg. geom; he said "no". Saying it's "just" integral refinement on analytic theory misses too many subtleties. "Black box" approach in Katz-Mazur hides difficulties too. –  BCnrd Aug 20 '10 at 0:19
    
To do the exercise, a natural method is to use rigid-analytic geometry and GAGA, much as one would do over the complex numbers. In effect, one is trying to show that an "analytic" viewpoint allows one to construct "algebraic" maps. Think how one would do the complex-analytic version, and the ideal would be to carry out the same reasoning in the non-archimedean case. However, if you don't know rigid-analytic geometry, it might inspire you to try to learn about that topic (that sure worked for Tate!). There might be a brute force argument with power series, but is that illuminating? –  BCnrd Aug 20 '10 at 2:59
    
Thanks, BCnrd. Can you say a little more about the approach via "formal schemes & generalized elliptic curves"? (Is something like this in Deligne-Rappaport?) That sounds closer to things I know than rigid-analytic things (of which my ignorance is total). (I mentioned AEC2 because it's the only place I could find a statement along the lines I want, not because I care about its methods so much.) –  Charles Rezk Aug 20 '10 at 4:32
    
Yes, in D-R. For some highlights from D-R with extras tossed in, see sec. 2.1, 2.2 (especially 2.2.4), and 2.5 in my paper "Arith. moduli of gen. ell. curves" (on my webpage), which includes reference to section of D-R which discusses Raynaud construction of Tate curve via formal schemes. (Raynaud procedure is in more concrete terms in sec. 2.5.) Must be careful about rigor when relating things over $\mathbf{Z}((q))$ and over $\mathbf{C}$ (since can't evaluate formal series). Simpler for non-arch. fields, but algebraization of formal schemes and then generic fiber will give what you want. –  BCnrd Aug 20 '10 at 6:27

1 Answer 1

up vote 6 down vote accepted

No matter how you define Tate(q), it should have the following properties:

(a) for any $n$ it contains a subgroup $M_n$ canonically isomorphic to $\mu_n$ (which corresponds tho $\mu_n\subset\mathbb{C}^\times$ in the complex model),

(b) the (co)tangent space along the unit section is canonically trivialized (by $d\log u$ in the complex model).

Let me first treat the case $n=1$, as Charles suggests. The sought-for isogeny Tate($q$)$\to$Tate($q^m$) is characterized by two conditions:

(a') its kernel is $M_m$ (i.e. it induces an isomorphism Tate($q$)$/M_m\to$Tate($q^m$)),

(b') it induces multiplication by $m$ on the tangent space, modulo the identification (b).

Consider the scheme $X\to S:=\mathrm{Spec}\:\mathbb{Z}((q))$ parametrizing isomorphisms Tate($q$)$/M_m\to$Tate($q^m$). This is an unramified $S$-scheme, and in fact it is finite because Tate($q$) has no complex multiplication in any fiber over $S$ (I guess this has to be checked). Since it has a section over $\mathrm{Spec}\:\mathbb{C}((q))$ it is dominant, hence surjective over $S$. Since $S$ is normal it suffices to find a section at the generic point. But by flat descent, condition (b') guarantees that the above section over $\mathrm{Spec}\:\mathbb{C}((q))$ descends to the fraction field of $\mathbb{Z}((q))$. QED.

Remark: I am a bit uncomfortable about the "no CM" stuff, but we can probably avoid it by noting that $X\to S$ satisfies the valuative criterion of properness, even when it is not of finite type. This (together with unramifiedness) is enough to imply that a section at the generic point extends over a normal base.

For arbitrary $n$, observe that we have just constructed $\alpha_m:\text{Tate}(q)\to \text{Tate}(q^m)$ with kernel killed by $m$, so multiplication by $m$ factors as $\beta_m\circ\alpha_m$ for some $\beta_m:\text{Tate}(q^m)\to \text{Tate}(q)$. You can now treat the general case by taking the composition $$\text{Tate}(q)\to \text{Tate}(q^m)=\text{Tate}(q'^n)\to \text{Tate}(q')$$ where the two maps are $\alpha_m$ and $\beta_n$ respectively.

share|improve this answer
    
Very nice. The verification of (a) and (b) is the sort of "black box" as in Katz-Mazur. Ironically,even though K-M basically avoid ever alluding to generalized elliptic curves or moduli interpretation along cusps, to justify stuff like (a) and (b) (and formal group analogue) over $\mathbf{Z}((q))$ which underlies their analysis of structure along the cusps it seems to has to haul out the D-R approach to things anyway. –  BCnrd Aug 20 '10 at 15:47
    
My feeling is that we lack a good a priori definition of Tate curves (e.g. by a universal property, or a list of axioms such as (a) and (b)). All we have, it seems, is "constructions". Remember, for instance how Grothendieck's definition of the Picard functor made it easier to use Jacobians, which previously were just constructed. The difficulty in doing this, I think, is the "scheme vs formal scheme" bit, i.e. the algebraization process. In terms of functors, it makes it hard to guess on what category the represented functor should live. –  Laurent Moret-Bailly Aug 21 '10 at 10:54
    
Dear Laurent: The only nice viewpt I know is as univ. deformation of standard 1-gon over a field. Doesn't get around formal vs. algebraic issue, but suggests the formal aspect shouldn't be hidden. Deformation viewpt also loses contact with $\mathbf{Z}$, & more importantly with element $q$ in base ring (which is part of what makes it useful, as in algebraic theory of $q$-expansions). So in the end I think of it as the version for standard 1-gon over $\mathbf{Z}$ of what deformation theory provides over a field, but the formal construction is needed to do anything serious with it. Sad, but true. –  BCnrd Aug 22 '10 at 3:38
    
Dear Brian: You are probably right, but I should add that the picture is not complete without the classical analytic side (this is the convergence question raised by Charles). Specifically, the construction should prove that the Tate curve over $\mathbb{Z}[[q]]$ descends to the subring $R$ of series converging on the unit disk. Then so do our isogenies becouse $R$ is an integrally closed (albeit non-noetherian) domain. –  Laurent Moret-Bailly Aug 22 '10 at 7:45
    
Thanks for your help, both of you. I had not realized how opaque Katz-Mazur's "black box" (as you call it, BCnrd) is. –  Charles Rezk Aug 22 '10 at 16:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.