components of E[p], E universal in char p.

Question

I have just realised that a group scheme I've known and loved for years is probably a bit wackier than I'd realised.

In this question, in Charles Rezk's answer, I erroneously claim that his construction of the space representing Drinfeld $\Gamma_1(p)$ structures on elliptic curves must be flawed, because the global properties of $Y_1(p)$ that I know from Katz-Mazur seemed to contradict global properties that his construction appeared to me to have. We took the conversation to email and I also started writing down my thoughts more carefully to check there were no problems with them. I found a problem with them---hence this question.

Let $p$ be prime, let $N\geq4$ be an integer prime to $p$, and consider the fine moduli space $Y_1(N)$ over an algebraically closed field $k$ of characteristic $p$. The $N$ isn't important, it just saves me having to use the language of stacks. Let $Y^o$ denote the open affine of $Y_1(N)$ obtained by removing the supersingular points. Over $Y^o$ we have an elliptic curve $E$ (obtained from the universal family over $Y_1(N)$).

In brief: here's the question. The $p$-torsion $E[p]$ of $E$---it's a group scheme and its identity component is non-reduced. But (regarded as an abstract scheme) does it have a component which is reduced? I think it might! This goes against my intuition.

Now let me go more carefully. Let's consider the scheme $E[p]$ of $p$-torsion points. This is finite flat over $Y^o$ and hence as an an abstract scheme over $k$ it's going to be some sort of 1-dimensional gadget. It also sits in the middle of an exact sequence of group schemes over $Y^o$:

$0\to K\to E[p]\to H\to 0$

with $K=ker(F)$, $F$ the relative Frobenius map (an isogeny of degree $p$). Now at every point in $Y^o$, the fibre of $K$ is isomorphic to $\mu_p$ and the fibre of $E[p]$ is isomorphic to $\mu_p\times\mathbf{Z}/p\mathbf{Z}$. In particular all components of all fibres are isomorphic and non-reduced. Now here is where my argument in the thread in the question linked to above must become incorrect. I wanted to furthermore claim that

(a) $K$ (as an abstract curve) is non-reduced, and then

(b) hence (because $K$ is the identity component of $E[p]$ and "all components of a group are isomorphic as sets") all components of $E[p]$ are non-reduced.

I now think that (b) is nonsense. In fact I know (b) is nonsense in the sense that $\mu_p$ over $\mathbf{Q}$ has only two components and they look rather different when $p$ is odd, but in some sense I feel here that the difference is more striking. In fact I now strongly suspect that $E[p]$ as an abstract scheme has two components, one being $K$ and the other being a regular scheme (an Igusa curve) mapping down in an inseparable way onto $Y^o$ (so the component isn't smooth over $Y^o$ but abstractly it's a smooth curve).

If someone wants a proper question, then there is one: am I right? The identity component of $E[p]$ is surely non-reduced---but does $E[p]$ have any regular components? I know how to prove this but it will be a deformation theory argument and I've got to go to bed :-/ If so then I think it's the first example I've seen, or at least internalised, of a group scheme where the behaviour of a non-identity component is in some sense a lot better than the behaviour of the identity component. I say "in some sense" because somehow it's the map down to $k$ that is better-behaved, rather than the map down to $Y^o$. Someone please tell me I'm not talking nonsense ;-)

Answer 1

Speaking of "connected components" is a delicate thing since you really mean in a relative sense, and more specifically the etale quotient $H$ can have its open and closed non-identity part with very nontrivial $\pi_1$-action (so more subtle than on geometric fibers over the base). But even if the $\pi_1$-action is trivial over whatever base, there are generally no no nontrivial sections through the non-identity part, so you can't do translation arguments, so there's no reason to expect intuition about "homogeneity" to have any relevance. Likewise for any property which isn't local for whatever topology the thing admits local sections. (In this case the fppf topology, for which regularity is not a local property, and ditto for reducedness.)

In this case Katz-Mazur (or better: Kummer!) did all of the deformation theory work. If we pass to the complete local ring at a geometric point of the base curve then (by the Serre-Tate deformation theorem) you're really asking a question about something over the universal deformation ring $R = k[[x]]$ of the $p$-divisible group

$$\mu_{p^{\infty}} \times (\mathbf{Q}_ p)/\mathbf{Z}_ p$$

over which the universal $p$-divisible group $\Gamma$ has finite flat $p$-torsion $G = \Gamma[p]$ with a connected etale sequence that is described explicitly in Katz-Mazur. So there you can stare at the non-identity factors, and if those are regular then you're done. And if not regular somewhere then likewise for the global case over the modular curve.

If you look at (8.7.1.1) in KM (with $i \ne 0$) and then the proof of Prop. 8.10.5 there, or instead think about Kummer theory for group schemes, you'll see that there is a unit $q$ unique up to $p$-power unit multiple which "classifies" (up to isomorphism) the extension structure on $G$ (as $p$-torsion extension of $\mathbf{Z}/p \mathbf{Z}$ by $\mu_p$). We can scale so $q$ is a 1-unit.

Now I claim that $q-1$ has ord equal to 1 in the deformation ring $k[[x]]$. Indeed, otherwise it would say that every first-order deformation of the $p$-divisible group has split $p$-torsion, which we know is nonsense (since we can use the unit $1+x$ to build a deformation violating that).

So then we can change $x$ so $q = 1 + x$ and the equations of the non-identity components are $T^p - (1+x)^i$ for $1 \le i \le p-1$ (from K-M, or thinking on our own). I claim the quotient in each case is a discrete valuation ring. Since $(1+x)^i = 1 + ix + x^2(\dots)$, by change of $x$ it is always the same as $T^p - (1+y)$ over $k[[y]]$, and writing it as $(T-1)^p - y$ since in characteristic $p$ we see it is Eisenstein, so we're done.

Since function fields of the modular curves you had in mind are not perfect, perhaps a more amusing example for you of funny behavior is to give a reduced group scheme over a field which is not smooth, and a non-reduced group scheme whose underlying reduced scheme is not a subgroup scheme (affine groups of finite type, but ground field must be imperfect of course).