MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $p$ be a prime number, let $K$ denote a finite extension $\mathbb{Q}_{p}$ and let $\overline{K}$ be an algebraic closure of $K$. Let $A$ be an ellitpic curve over $K$ and denote by $T_{p}A$ its Tate module.

When $A$ has ordinary reduction then it is known that there is an exact sequence of $Gal(\overline{K}:K)$-modules: $$ 0 \rightarrow \mathbb{Z}_{p}(-1) \rightarrow T_{p}A \rightarrow \mathbb{Z}_{p} \rightarrow 0. $$ I would be interested in a characterization of the quotient $\mathbb{Z}_{p}$ and its applicability in the case of supersingular reduction. When $$ HT: H^{1}(A,{\cal O})\otimes_{K} \mathbb{C} _{p}\oplus H^{0}(A,\Omega ^{1})\otimes_{K} \mathbb{C} _{p}(-1) \simeq Hom_{\mathbb{Z} _{p}}(T_{p}A ,\mathbb{C} _{p}). $$ denotes the Hodge-Tate isomorphism, let $M \subset T_{p}A$ be the intersection of all kernels of homomorphisms in the image of $H^{1}(A,{\cal O})\otimes_{K} \mathbb{C} _{p}$ in $Hom_{\mathbb{Z} _{p}}(T_{p}A ,\mathbb{C} _{p})$ under $HT$. It is easy to see that the quotient $T^{Q}_{p}A:= T_{p}A/M$ is a free $\mathbb{Z} _{p}$-module of rank $\leq 2$, and that it gives rise to an exact sequence: $$0 \rightarrow M \rightarrow T_{p}A \rightarrow T^{Q}_{p}A \rightarrow 0. $$ If $A$ has ordinary reduction, then $T^{Q}_{p}A$ may be identified with $\mathbb{Z}_{p}$ from the first diagram.

If $A$ has supersingular reduction, is then $T^{Q}_{p}A$ also a free $\mathbb{Z}_{p}$-module of rank one? More generally, the quotient $T_{p}^{Q}A$ could be also defined when $A$ is an abelian variety of dimension $g$. Will $T_{p}^{Q}A$ be a free $\mathbb{Z}_{p}$-module of rank $g$?

share|cite|improve this question
If $A$ is non-ordinary, then $T_p A$ is irreducible as a Galois representation. The submodule you are defining is Galois stable, isn't it? So it must be either zero or everything. – David Loeffler May 17 '12 at 21:45
Thank you, this indeed seems to be the case and should answer the question. My original goal was define a free $\mathbb{Z}_{p}$-module $T^{Q}_{p}A$ of rank $g$, together with a canonical surjection of $\mathbb{Z}_{p}$-modules: $f:T_{p}A \rightarrow T^{Q}_{p}A$. The canonical inclusion $Hom_{\mathbb{Z}_{p}}(T^{Q}_{p}A, \mathbb{\C}_{p}) \subset $Hom_{\mathbb{Z}_{p}}(T_{p}A, \mathbb{\C}_{p})$ should be equal to the image of $ H^{1}(A,O) \otimes _{K} \mathbb{C}_{p}$. Can such a module be constructed in the non-ordinary case? (The morphism f does not need to compatible with Galois) – user23778 May 17 '12 at 23:24
Sorry, have I misread or seriously misunderstood the question? I don't see that there is such an exact sequence of ${\mathrm{Gal}}(\overline K\colon K)$-modules, because the kernel of reduction becomes isomorphic to the multiplicative (formal) group only after extension of the base to the complete maximal unramified extension of $K$. – Lubin May 19 '12 at 4:32

Although this question isn't really well-defined (you'd surely need to be more precise about the word "canonical" in the comment under the question) let me make two comments which hopefully put this to bed. Firstly it's hard to see how a "canonical" map from $T_pA$ would not have Galois-stable kernel -- because if the kernel were not stable then just hit it with Galois and you've got a different, but equally "canonical", map, so, as David Loeffler pointed out in the comments, if the Tate module is irreducible (which it certainly sometimes is: for example with a supersingular elliptic curve over $\mathbf{Q}_p$ I guess even the $p$-torsion is irreducible) then you're already in trouble. "Galois-stable" is just a more concrete say of saying "only depends on things up to isomorphism", which surely you want any "canonical" construction to have.

But here's another way of looking at it (which is probably just the same way in disguise). Let me argue that it is unreasonable even to hope for a "canonical" map mod $p$, i.e. a map from $A[p]$ to a 1-dimensional space over $\mathbf{F}_p$. This time I'll interpret "canonical" as "behaves well in families" -- because if $A$ is a family of ordinary elliptic curves then there is a family of subgroups of order $p$ which at each fibre is the so-called "canonical subgroup", and there's your kernel. However such a family cannot possibly exist for a general family which includes non-ordinary curves, because if it did then you would be constructing a section of the forgetful functor $Y_0(p)\to Y_0(1)$ over $\mathbf{Q}_p$, and such a section does not exist [to make this completely rigorous you could add a point of order $N$ everywhere to make the functors representable, suppress the point from the notation, and then just note that $Y_0(p)$ is now a connected curve with bigger genus than $Y_0(1)$]. The fact that you can write down such a section on the ordinary locus, first noticed by Katz I guess, is exactly the mod $p$ version of your question, and so even mod $p$ it's hard to see you moving far into the non-ordinary locus. You can move a little way in -- this is the theory of the canonical subgroup -- but the moment you're a little way in there are only canonical subgroups of order $p^n$ for finitely many $n$ so you'll never glue to get a map on the whole Tate module.

share|cite|improve this answer

Thank you for your answers.

After thinking some time about it, it seems that such a map as in my comment above cannot even exist in the case where $A$ has super singular reduction.

If there were a surjection $f:T_{p}A \rightarrow T_{p}^{Q}A$ of $\mathbb{Z}_{p}$-modules as described in my comment above, then the kernel $M'$ of $f$ would be a non-zero by dimension considerations.

Since by assumtion the image of $Hom_{\mathbb{Z}_{p}}(T_{p}^{Q}A,\mathbb{C}_{p})$ in $Hom_{\mathbb{Z}_{p}}(T_{p}A,\mathbb{C}_{p})$ coincides with $HT(H^{1}(A,{\mathcal O})\otimes _{K}\mathbb{C}_{p})$, the module $M'$ must be a submodule of $M$, which was defined as the intersection of all kernels of homomorphisms in $HT(H^{1}(A,{\mathcal O})\otimes _{K}\mathbb{C}_{p}) \subset Hom_{\mathbb{Z}_{p}}(T_{p}A,\mathbb{C}_{p})$.

As $M$ must be irreducible as a Galois module, as remaked by David, and since it has a nontrivial submodule $M'$, the module $M$ must be equal to $T_{p}A$. Therefore, all homomorphisms in $HT(H^{1}(A,{\mathcal O})\otimes _{K}\mathbb{C}_{p})$ must be trivial. And this cannot be since it a $\mathbb{C}_{p}$-vector space of dimension one.

Is this correct or am I missing something?

share|cite|improve this answer

Your Answer


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.