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 ;-)