Isogenies between Tate curves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T09:10:33Z http://mathoverflow.net/feeds/question/36128 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36128/isogenies-between-tate-curves Isogenies between Tate curves Charles Rezk 2010-08-19T21:27:34Z 2010-08-20T10:33:25Z <p>Let $q$ and $q'$ be complex numbers with $0&lt;|q|,|q'|&lt;1$, and let $m$ and $n$ be positive integers.<br> 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 $\mathbb{C}$. </p> <p>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. </p> <p>My question is: </p> <blockquote> <p>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?</p> </blockquote> <p>It should suffice to construct such an isogeny for $(m,n)=(m,1)$, and use dual isogenies and composition to get the general case.</p> <p>(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.)</p> <p>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.</p> <p>This is very closely related to exercise 5.10 in Silverman, <em>Advanced Topics in the Arithmetic of Elliptic Curves</em>. There, he apparently asks us to show that for a $p$-adic field $K$, if $q,q'\in K$, $0&lt;|q|,|q'|&lt;1$, 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?)</p> <p>Unfortunately, I have no idea how to do Silverman's exercise either (he marks it as difficult). Any hints?</p> http://mathoverflow.net/questions/36128/isogenies-between-tate-curves/36179#36179 Answer by Laurent Moret-Bailly for Isogenies between Tate curves Laurent Moret-Bailly 2010-08-20T10:33:25Z 2010-08-20T10:33:25Z <p>No matter how you define Tate(q), it should have the following properties:</p> <p>(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),</p> <p>(b) the (co)tangent space along the unit section is canonically trivialized (by $d\log u$ in the complex model).</p> <p>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:</p> <p>(a') its kernel is $M_m$ (i.e. it induces an isomorphism Tate($q$)$/M_m\to$Tate($q^m$)),</p> <p>(b') it induces multiplication by $m$ on the tangent space, modulo the identification (b).</p> <p>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.</p> <p>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.</p> <p>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.</p>