Isogenies between Tate curves - MathOverflow

Isogenies between Tate curves

2010-08-19T21:27:34Z

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?

Answer by Laurent Moret-Bailly for Isogenies between Tate curves

2010-08-20T10:33:25Z

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.