# Eichler-Shimura congruence

I'm trying to understand the Eichler-Shimura congruence which relates the Hecke operator $T_p$ to Frobenius at $p$ in characteristic $p$.

Two possible ways to compute $T_p$ mod $p$ seem to be:

A) Look at the map $Div(X_0(N)) \to Div(X_0(N))$ induced by the correspondence $X_0(N) \leftarrow X_0(Np) \to X_0(N)$ where the first map forgets the subgroup at $p$ and the second mods out by it. And then just take this map and look at the induced map $Div(X_0(N)_{F_p}) \to Div(X_0(N)_{F_p})$ in characteristic $p$.

B) Look directly at the correspondence $X_0(N)_{F_p} \leftarrow X_0(Np)_{F_p} \to X_0(N)_{F_p}$ in characteristic $p$ and try to compute directly the induced map $Div(X_0(N)_{F_p}) \to Div(X_0(N)_{F_p})$.

In case (A), starting with a point in $X_0(N)$, it has $p+1$ lifts to $X_0(Np)$ and thus the induced map on divisors will result in divisors of degree $p+1$. Explicitly, I'm getting: $$(E,C) \to (E,(C+\ker(F))/\ker(F)) + p (E,(C+\ker(V))/\ker(V))$$ which looks at lot like the standard Eichler-Shimura relation.

But in case (B), starting with an ordinary point in $X_0(N)_{F_p}$, it has $2$ lifts to $X_0(Np)$ --- one where we pick $\ker(F)$ as our group scheme of order $p$, and another for $\ker(V)$. Thus the induced map on divisors will result in divisors of degree $2$, and it seems to yield $$(E,C) \to (E,(C+\ker(F))/\ker(F)) + (E,(C+\ker(V))/\ker(V)).$$

What's wrong with the argument in case B?

• In case (B), one of the two lifts has multiplicity $p$. – ACL Nov 21 '14 at 20:27
• – Filippo Alberto Edoardo Nov 21 '14 at 23:51

## 1 Answer

There's a little more geometry here that should be accounted for in characteristic $p$. Namely, the curve $X_0(Np)_{\mathbb{F}_p}$ is reducible -- its two components are isomorphic to $X_0(N)_{\mathbb{F}_p}$ and they intersect transversally at the supersingular points (see e.g. Ribet and Stein's online notes). So the components are disjoint on the ordinary locus and account for the two subgroups you wrote down.

The left pointing map $\pi: X_0(Np)_{\mathbb{F}_p} \to X_0(N)_{\mathbb{F}_p}$ in the correspondence is an isomorphism on the component corresponding to $\ker (F)$, but it is an inseparable degree $p$ map on the second component. Roughly speaking, that this is degree $p$ and inseparable comes from the fact that there are $p$ "non-canonical subgroups" in characteristic 0 which all reduce to $\ker (V)$ in characteristic $p$.

To compute the effect of the correspondence on divisors, you need to pull back divisors along an inseparable map. So your argument in Case B is forgetting to multiply by the ramification index (which is $p$) for the unique preimage of $(E,\ker(V))$ under $\pi$ in the "$\ker (V)$ component" of $X_0(Np)_{\mathbb{F}_p}$

Edit: The OP asks how to check the inseparability rigorously and without using characteristic 0. As Eric says in the comments, this is a somewhat subtle issue which can be found in Katz-Mazur (see section 13.5), probably also in Deligne-Rapoport. The Ribet-Stein book has a more approachable discussion (read the proof of Theorem 12.6.4 carefully), but they implicitly use the fact that the Frobenius map $X_0(N) \to X_0(N)$ corresponds under the moduli interpretation to the Frobenius map $(E, C) \mapsto (E^{(p)}, C^{(p)})$, which is really the heart of the matter.

But if you're willing to ignore issues of representability of functors, then there's an elementary way to see this last fact: first note that base change reduces us to the case where $N = 1$. Then use the parameterization of $X_0(1)$ by the $j$-invariant to see that both the Frobenius $X_0(1) \to X_0(1)$ and the map $E\mapsto E^{(p)}$ can be identified with $j \mapsto j^p$.

• Thanks. But I'm still confused as to why the map on the second component of X_(Np)_{F_p} is inseparable. On the level of moduli, the map from X_(Np)_{F_p} to X_(N)_{F_p} is defined by forgetting the p-isogeny --- which as far as I understand can be made sense of completely in characteristic p (without thinking of it as reducing the corresponding characteristic 0 map modulo p). And for a fixed elliptic curve in characteristic p, there are exactly two p-isogenies given by F and V. Can you explain how I see the inseparability of this map without referring to characteristic 0? – guest Nov 23 '14 at 3:31
• You're not going to see inseparability by looking at geometric points (which is what you're doing -- saying things like "there are exactly two $p$-isogenies" is only true if you're over e.g. an alg closed field); inseparability is something else. Think of a simpler example -- think about raising to the $p$th power map on the affine line in char p; the fact that "every element has one $p$th root" does not give you any insight into the fact that the map is inseparable. In my next comment I'll say something about another way of thinking about it. – eric Nov 23 '14 at 21:21
• Looking at points you've seen that the pre-image of one point in $X_0(N)$ is (usually) two points in $X_0(Np)$. But now consider instead what the degree of the forgetful map is. This is a delicate calculation, and you can't just think about points. Instead of just counting the points, fix an elliptic curve and now consider the problem of representing the functor of cyclic subgroups of degree $p$. When you think about it that way, and all the details in all their glory are written down in Katz-Mazur, you realise that the representing object is not just a point because it deforms differently – eric Nov 23 '14 at 21:23
• to a point, and then when you do the explicit calculation you find it's a non-reduced point, and now suddenly you can see the inseparability. But if you don't want to lift to char 0 you'll have to get your hands dirty with questions like Grassmannians representing degree $p$ subgroups of $\mu_p\times (Z/pZ)$. – eric Nov 23 '14 at 21:25
• @eric: Thank you. I think I see the idea now. – guest Nov 24 '14 at 4:06