17
$\begingroup$

I know that the answer is $\mu_p \times \mathbb{Z}/p\mathbb{Z}$ if $E$ is ordinary, and $\alpha_p$ if $E$ is supersingular, where $\mu_p$ and $\alpha_p$ are the kernels of Frobenius on $\mathbb{G}_m$ and $\mathbb{G}_a$ respectively. But why is it this true?

Suppose $E'$ is a lift of $E$ to characteristic 0. Then $E'[p] = (\mathbb{Z}/p\mathbb{Z})^2$. If $E$ is ordinary then we have $E[p] =\mathbb{Z}/p\mathbb{Z}$, and one way to reconcile these two facts is to have $E[p] = \mu_p \times \mathbb{Z}/p\mathbb{Z}$, since the group of closed points of $\mu_p$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$ in characteristic 0, whereas in characteristic p, it is just a single (nonreduced) point. I'm not sure why this is the only possibility though--I think it has to do with the height of the formal group, but I just can't nut out the details. In Katz-Mazur "Arithmeticic moduli of elliptic curves" (proof of theorem 2.9.3) they say that "any p-divisible group over an algebraically closed field is the product of a p-divisible commutative formal Lie group and finite number of copies of $\mathbb{Q}_p/\mathbb{Z}_p$," but I don't see why this is true.

For the supersingular case, I'm even hazier. Is $\alpha_p$ the unique one-paramater formal group of height 2, and if so, how can you see this? For an affine scheme Spec(R), we have $\alpha_p(R) = \mathrm{Spec}(R[x]/(x^p))$. Is it true that in characteristic 0 we have (for example) $\alpha_p(\mathbb{C}_p) = (\mathbb{Z}/p\mathbb{Z})^2$?

Sorry if this is a mess, i'm really confused, and I haven't been able to find a sufficiently dumbed down explanation of this stuff anywhere.

Improve this question
$\endgroup$
8
  • $\begingroup$ What you say seems a bit confused to me. I think your claim in the ordinary case is only true if, for example, the base is a separably closed field. Is this what you mean by "characteristic p"? In the supersingular case the answer can't be $\alpha_p$ because $\alpha_p$, assuming you're using the same notation as the standard notation, is a group scheme of order $p$, and $E[p]$ has kernel of order $p^2$. In the supersingular case, over a sep closed field, $E[p]$ is a non-split extension of $\alpha_p$ by itself. I don't know what you mean by "$E[p]=Z/pZ$"; this is only true "on points"... $\endgroup$
    – Kevin Buzzard
    Commented Aug 18, 2010 at 6:14
  • $\begingroup$ ...and ignores the non-reduced structure on $E[p]$ which presumably is precisely what you're asking about (I'm talking about your statement in the 2nd para). In the 3rd para you talk about $\alpha_p(C_p)$ but this doesn't make any sense because $\alpha_p$ is a group scheme in characteristic $p$ so you can't evaluate it at a field of characteristic zero (assuming $C_p$ is what I think it means). Unfortunately I can't remember how I learnt this stuff myself---perhaps from talking to my advisor :-/ Did you try Conrad's paper in the Boston Fermat proceedings? LEARN DIEUDONNE MODULES they're easy! $\endgroup$
    – Kevin Buzzard
    Commented Aug 18, 2010 at 6:18
  • $\begingroup$ PS I think this isn't a real question. I don't think you have a precise question; I think your question is really just asking for a basic reference, and I don't know the answer to this. $\endgroup$
    – Kevin Buzzard
    Commented Aug 18, 2010 at 6:20
  • 1
    $\begingroup$ Kevin's advice is right (as usual). The question has the same flavor as trying to prove general theorems about Lie groups without exploiting Lie algebras, just bare hands. Bad idea; the reason Dieudonne modules were invented was exactly to render these kinds of questions straightforward to figure out by a little computation. You also seem to have a shaky grasp of finite flat group schemes; read Tate's article in the big FLT book to get a better grip on that. $p$-torsion in char. $p$ & $p$-div. groups are subtle things, don't expect to find "dumbed down" explanations: need some real theory. $\endgroup$
    – BCnrd
    Commented Aug 18, 2010 at 6:26
  • 3
    $\begingroup$ Kevin: one small correction is that $p$-torsion commutative extensions of $\mathbf{Z}/(p)$ by $\mu_p$ over a field $k$ of char. $p > 0$ are classified by ${\rm{H}}^1_ {\rm{fppf}}(k, \mu_ p) = k^{\times}/(k^{\times})^p$, so if $k$ is imperfect and separably closed then this is non-trivial (and so such non-split extension structures exist: nice example is $p$-torsion on Tate curve over sep. closure of $\mathbf{F}_p((q))$, classified by $1 + q$, which is not a $p$-power). This is also discussed within Example A.8.3 in the book "Pseudo-reductive groups"... $\endgroup$
    – BCnrd
    Commented Aug 18, 2010 at 6:32

1 Answer 1

Reset to default
33
$\begingroup$

Dear Maxmoo,

Just to offer a slightly different perspective than that given by Kevin and Brian:

While their advice is certainly correct, when I was learning this I also found it very helpful to make a couple of "bare hands" computations, as a kind of reality check.

For this, begin with an elliptic curve in char. $2$, in fact with two, of the form:

$$y^2 + y = x^3$$

and

$$y^2 + x y = x^3 + x $$

One of these is supersingular, the other ordinary. (I won't tell you which here!)

Now try computing the $2$-torsion concretely, using lines passing through three points and so on.

Remember that in the end you are looking for a degree $4$ equation (you may need to change variables to see the point at infinity; this won't show up in the affine equations I've given you). By general theory, you know this equation won't be separable: non-reduced group scheme structure will show up concretely as inseparability in this polynomial.

In one case (the s.s. case) it will be entirely inseparable; in the other (ordinary) case it will have inseparability degree $2$ (so "half" inseparable, "half" separable).

Once you've done the case of char. $2$, you might want to try char. $3$ as well (since computing the equation for the 3-torsion is also just about in reach by hand).

The reason I suggest this is that I remember, when I was learning this stuff, that all these group schemes (especially the non-reduced ones) seemed fairly ephemeral, but after I had made these kind of explicit computations, I had a much more concrete mental model for what the general theory was talking about, which gave me a lot more confidence in reading and making arguments about these kinds of things.

Best wishes,

Matt

Improve this answer
$\endgroup$
2
  • 6
    $\begingroup$ Matt, great examples; now that you mention it, I remember doing the same calculations for myself early in life (so to speak). Another nice reality check I did was to see by hand that $[p]$ factors through the Frobenius map (as it must since the kernel of the latter isogeny sits inside that of the former, as there's only one closed subscheme of any given length supported at the identity) by explicitly working out what $[p](x,y)$ is in coordinates for $p = 2, 3$ in such examples: nifty to see how the algebra makes these be entirely in terms of $x^p$ and $y^p$ since in characteristic $p$. $\endgroup$
    – BCnrd
    Commented Aug 18, 2010 at 15:28
  • 5
    $\begingroup$ I'll disagree strongly with Kevin this time and agree with Matt to say that in this case, the route to understanding is not via Dieudonn\'e modules, but by direct computation of simple cases. You'll see very clearly, when you express the addition on $y^2+y=x^3$, that the kernel of $[2]$ here is not contained in $G_a$, and of course is not equal to $\alpha_2$, as others have pointed out. $\endgroup$
    – Lubin
    Commented Sep 15, 2011 at 3:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .