On page 89 of The Arithmetic of Elliptic Curves (second edition), Silverman says:

Let $\phi:E_1\rightarrow E_2$ be an isogeny of elliptic curves. Then $\phi$ induces maps $\phi:E_1[l^n]\rightarrow E_2[l^n]$, and hence induces a $\mathbb{Z}_l$-linear map $\phi_l:T_l(E_1)\rightarrow T_l(E_2)$.

I'm only just starting to understand the Tate module from the same chapter. I'm still trying to come to grips with what the Tate module actually is from the idea of the 'inverse limit'.

My questions are:

  1. How is a $\mathbb{Z}_l$-linear map different from a linear map?
  2. How does $\phi$ induce the $\mathbb{Z}_l$-linear map?
  3. Where can I see an example of such an isogeny and the maps that it induces?

1 Answer 1

  1. It depends on what you mean by "linear map." If $A$ is an abelian group and $M$, $M'$ are $A$-modules, then the term "linear map" is sometimes used as shorthand to mean "morphism of $A$-modules" which is equivalent to "$A$-linear map." All of these things are just different ways of saying that it is a morphism that respects the module structure which is just a linear action of $A$.

    That is a long way of saying that those terms do not differ if you use them in the usual way.

  2. On the one hand $\phi: E_1 \to E_2$ is an algebraic morphism of curves, so you can think of it as mapping points of $E_1$ to points of $E_2$. This isn't good enough to conclude that it maps $\ell^n$-torsion to $\ell^n$-torsion. Luckily, an isogeny is also a group homomorphism. Since it preserves the group structure of the elliptic curve we see that $\ell^n \phi(P) = \phi(\ell^n P) = \phi(0)=0$.

    Once you have that $\phi$ induces a map $E_1[\ell^n]\to E_2[\ell^n]$ it is basically "general nonsense" about inverse limits that you get a map $T_\ell (E_1)\to T_\ell (E_2)$. As an exercise, try the easier case of just $\mathbb{Z}_\ell$ to see how knowing a map on the finite levels of the limit $\mathbb{Z}/\ell^n\mathbb{Z}\to \mathbb{Z}/\ell^n\mathbb{Z}$ gives you a map on the limit.

    Update: The comment below points out that in order to apply the general nonsense we have to check that the induced maps are actually compatible with the inverse system. This means we have to check that all the squares in the sequence commute: $$ \begin{array}{ccccc} \cdots & \longrightarrow & E_1[\ell^{n+1}] & \overset{\cdot \ell}{\longrightarrow} & E_1[\ell^n] & \longrightarrow & \cdots \\ & & \phi \downarrow & & \phi\downarrow \\ \cdots & \longrightarrow & E_2[\ell^{n+1}] & \overset{\cdot\ell}{\longrightarrow} & E_2[\ell^n] & \longrightarrow & \cdots \end{array}$$ This can be checked by a similar argument to the earlier one.

  3. You should play with the multiplication-by-$m$ map $E\to E$ for various $m$ since this is always an isogeny. This will give you feel for it in an explicit case (e.g. is it always an isomorphism on Tate modules?).

You also write that you are coming to grips with the Tate module as an inverse limit. As long as $\ell$ is not the characteristic of your base field, then $T_\ell(E)$ is abstractly isomorphic to $\mathbb{Z}_\ell \oplus \mathbb{Z}_\ell$ which might give you something more concrete to think about.

  • 3
    $\begingroup$ This is a very nice answer, but it sort of hides under "general nonsense" the part that is fundamental to the whole notion of inverse limit, which is the fact that the elements of the inverse limit are formed by coherent (or compatible) sequences of elements, and it does not give any indication of why an isogeny respects the coherent nature of the elements in a Tate module. I'd suggest adding a hint about this. As it stands, the reader may think that it is enough to have any random collection of maps at each level to obtain a map at the $\ell$-adic level, which is not true. $\endgroup$ Commented Apr 28, 2014 at 2:44

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