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Can someone give a really concrete example of such a sequence? I am looking at several notes related with such things, but haven't seen any well-calculated example. And I'm really confused at this point.

Besides asking for a good example, I am also wondering about the following two things:

  1. There is an exact sequence for elliptic curves defined over a local field $K$, $0 \rightarrow \hat E(m) \rightarrow E(K) \rightarrow \tilde E(k) \rightarrow 0$, where $\hat E(m)$ is the formal group associated to $E$ and $\tilde E(k)$ is the reduction. (See Silverman AEC I, page 118), is this sequence related with connected-etale sequence?

2.Take the p-torsion kernel $E[p]$ of $[p]: E \rightarrow E$ for $E$ defined over $K$ a local field.Is $E[p]$ a finite flat group scheme over $R$ the valuation ring? And if so, what is its connected-etale sequence? (maybe I should change $p$ to an $n$, but I'm also curious what will happen if $p$ is the characteristic of $k$?)

Thank you.

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2 Answers 2

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Regarding 1. : If you pass to the $n$-torsion parts of the members of this exact sequence, you will get the $R$-valued points of the connected-etale sequence for $E[n]$.

Regarding 2. : If $E$ has good reduction, then $E[n]$ is a finite flat group scheme. If the residue char. $p$ of $R$ does not divide $n$ then it is even etale.

In general, by the Chinese remainder theorem, when analyzing $E[n]$ we may assume $n$ is a prime power, so suppose now that $n$ is a power $p^r$, where $p$ is the residue char. of $R$.

If $E$ has supersingular reduction, then $E[p^r]$ is connected.

If $E$ has ordinary reduction, then the connected and etale parts of $E[p^r]$ each have order $p^r$. The special fibre of the connected part is the kernel of $Frob^r$.

Note that if $E$ has bad reduction, then $E[n]$ can still have a finite flat extension over $R$ for some finite number of $n$. (E.g. $X_0(11)$ has bad reduction at 11, but its 5-torsion is still finite flat at 11.)

To compute an example, I suggest considering elliptic curves in char. 2, and computing the 2-torsion.

Consider the two examples: $y^2 + y = x^3 $ and $y^2 + x y = x^3 + x$. If you compute the 2-division equation it will have degree 4. In the first case, it will be purely inseparable: this case is supersingular. In the second case it will have inseparability degree 2; this case is ordinary. In the first case the 2-torsion group scheme is connected; in the second, it has a connected subgroup scheme of order 2.

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Thank you very much! It's really helpful! –  natura Feb 23 '10 at 6:18
    
If you read french and want the pedestrian computation of the 2-torsion of ell. curves in char. 2 you may have a look at people.math.jussieu.fr/~romagny/notes/points_de_torsion.pdf –  Matthieu Romagny Jun 18 '11 at 12:52

For concepts related to algebraic geometry when the base is not a field, it can be difficult for a beginner to reconcile the approach in Silverman with the approach via schemes. I wasted a lot of time as a student trying to relate the "3 points through a line" definition of the group law over fields with the concept of "reduction mod $p$" on points. Likewise, the approach with formal groups tends to make things confusing, despite their apparent "concreteness".

This sort of stuff drove me crazy when I was a student, until I realized that the best way to understand such topics is to give up working over fields and with equations, and to work over the valuation ring and with functorial viewpoints (only translating into field language at the very end). The relevant schemes in the question are really torsion schemes over the valuation ring, not torsion in the separate fibers (so you mean to assume $E$ has good reduction in both questions). I address this below. The following answer is way too long, since I do not know of a suitable reference not involving EGA/SGA. Tate's article on finite flat group schemes probably explains some aspects, but I doubt it addresses the link with the concrete stuff for elliptic curves.

If $R$ is a local ring, then an "elliptic curve over $R$" can be defined in two ways: the concrete way is as a Weierstrass plane cubic with unit discriminant, and the right way is as a smooth proper $R$-scheme with geometrically connected fibers of dimension 1 and genus 1 and a distinguished section. As usual, the concrete way is hard to use to actually prove anything interesting (and it is the "wrong" notion when the base is non-local and especially has non-trivial line bundles; e.g., over suitable number fields with class number $> 1$ one can make CM elliptic curves with "everywhere good reduction" which do not admit a global planar model with unit discriminant). How to prove there is a unique $R$-group structure with the indicated section as the identity? A real nightmare with the concrete definition, and elegantly explained in Chapter 2 of Katz-Mazur with the right definition. Likewise, that $E$ is functorial in its generic fiber when $R$ is a discrete valuation ring is a mess to prove by hand (which affine opens to use?), but has an elegant proof when approached through the "smooth and proper" viewpoint. Of course it is important and interesting that these concrete and abstract notions agree, and that is explained in Katz-Mazur Chapter 2.

That being said, if $E$ is an elliptic curve over any noetherian (say) scheme $S$ and $[n]_E:E \rightarrow E$ is multiplication by a positive integer $n$, then on geometric fibers this is a finite flat map, so $[n]$ is quasi-finite. Now proper and quasi-finite maps are finite (by Zariski's Main Theorem), so $[n]_E$ is a finite map, and the fibral flatness criterion implies that it is also flat. Being a finite flat map between noetherian schemes, it has a "degree" which is locally constant on the target and yet is $n^2$ on fibers over $S$. Hence, we conclude that $E[n] := {\rm{ker}}([n]_E)$ is a finite flat commutative $S$-group with constant fiber rank $n^2$. Honestly, I do not know any way to prove this which avoids the serious results that I just cited. But that's why the theorems are useful: because we can use them to make our intuition over fields carry over to cases when the base is not a field. (The noetherian condition can be dropped if we are more careful with the phrase "finite flat". I won't dwell on it here.) This answers the first part of the 2nd question (taking the base to be spectrum of the valuation ring there). It the notation there, the $p$-torsion of the elliptic curve over $K$ is not a finite $R$-scheme, and in general it may extend to a finite flat $R$-group in many ways. But the elliptic curve over $K$ uniquely extends to one over $R$ by the theory of Neron models, and its torsion levels provide the "right" finite flat groups you want to use over the valuation ring.

OK, now assume $R$ is a complete local noetherian ring (e.g., a complete discrete valuation ring). Could even assume it is a henselian local ring, but the complete case is easier to deal with and covers the case in the question. Let $G$ be a finite flat $R$-group, a case of interest being $E[n]$ for an elliptic curve $E$ over $R$. Let $k$ be the residue field, and consider $G_k$. Being a finite $k$-scheme, it has an open and closed identity component $G_k^0$ which is cut out by an idempotent. By 8.15 (or thereabouts) in Matsumura's Commutative Ring Theory, every idempotent in the special fiber of a finite $R$-algebra uniquely lifts. In particular, if $X$ is a finite $R$-scheme then its connected component decomposition uniquely lifts that of $X_k$. If $X$ is $R$-flat then so is each of its connected components. This is all compatible with products, so if $X$ has a structure of $R$-group then the open and closed connected component $X^0$ containing the identity section is an $R$-subgroup. Returning to our friend $G$, we get the so-called "relative identity component" $G^0$, an open and closed (hence finite flat) $R$-subgroup.

Remark: The formation of $G^0$ commutes with any flat local extension on $R$, as follows from the uniqueness! It doesn't usually commute with non-local extension, such as inclusion of a complete dvr into its fraction field.

Example: $G = E[n]$. Suppose $R$ is a complete discrete valuation ring with fraction field $K$, and $n \in K^{\times}$. What is $(G^0)_K$? Well, each "point" occurs over a finite extension $K'/K$, say with valuation ring $R'$, and $G(K') = G(R')$ by elementary integrality considerations (or in fancy terms, valuative criterion, which is killing a fly with a sledgehammer). Since the spectrum of $R'$ is connected, a point in $G(R')$ lies in $G^0(R')$ if and only if its specialization into $G_k(k')$ vanishes ($k'$ the residue field of $R'$). In other words, $(G^0)(\overline{K})$ consists of the $n$-torsion geometric points of $E_K$ whose specialization into geometric points of $E_k$ by valuative criterion for the $R$-proper $E$ ($E_K(K') = E(R') \rightarrow E(k') = E_k(k')$!) is 0.

Now we need to explain the "etale quotient" in concrete terms. This is best understood as a generalization of the following procedure over a field.

Example: Let $k$ be a field and $A_0$ a finite $k$-algebra. There is a unique maximal \'etale $k$-subalgebra $A_0'$ in $A_0$: concretely, in each local factor ring of $A_0$ uniquely lift the separable closure of $k$ in the residue field up into the local factor ring via Hensel's Lemma and the primitive element theorem. Since it is uniquely characterized by lifting separable closures of $k$ in the residue fields of the factor rings, it is a good exercise to check the following crucial thing: if $B_0$ is another finite $k$-algebra then $(A_0 \otimes_k B_0)' = A_0' \otimes_k B_0'$, and $A_0'$ is functorial in $A_0$. Observe that $A_0' \rightarrow A_0$ is faithfully flat since at the level of factor rings of $A_0$ it is an inclusion of a field into a nonzero ring. Also observe that any \'etale $k$-algebra equipped with a map to $A_0$ uniquely factors through $A_0'$.

Exercise: The formation of $A_0'$ commutes with any field extension on $k$. (Hint: use Galois descent to reduce to the separate cases of separable algebraic extensions and the easy case $k = k_s$.)

In geometric terms, for a finite $k$-scheme $X_0$, the preceding Example constructs a finite \'etale $k$-scheme $X_0'$ and a faithfully flat $k$-map $f_0:X_0 \rightarrow X_0'$ which is initial among all $k$-maps from $X_0$ to finite \'etale $k$-schemes, and its formation is functorial in $X_0$ and commutes with products in $X_0$ and with any extension on $k$. In particular, if $X_0$ is a $k$-group then $X_0'$ has a unique $k$-group structure making $f_0:X_0 \rightarrow X_0'$ a $k$-homomorphism.

Example: Now let $R$ be a complete discrete valuation ring with residue field $k$, and let $X$ be a finite flat $R$-scheme. (Can relax the hypothesis on $R$ if familiar with finite \'etale maps in general.) In this setting, "finite 'etale" over $R$ just means "product of finitely many unramified finite extensions". By using Hensel's Lemma in finite local $R$-algebras, to give a map from a finite \'etale $R$-algebra $A$ to a finite $R$-algebra $B$ is the same as to give a map $A_0 \rightarrow B_0$ between their special fibers. In particular, finite \'etale $k$-algebras uniquely and functorially lift to finite \'etale $R$-algebras, and so $X_0'$ uniquely lifts to a finite \'etale $R$-scheme $X'$ and there is a unique $R$-map $f:X \rightarrow X'$ lifting $f_0:X_0 \rightarrow X_0'$. By fibral flatness (using $X$ is $R$-flat!), $f$ is faithfully flat since $f_0$ is. By uniqueness of everything in sight, the formation of $f$ commutes with products and local extension on $R$ and is also functorial in $X$. In particular, if $G$ is a finite flat $R$-group then $G'$ admits a unique $R$-group structure making $f$ an $R$-homomorphism. We call $G'$ the maximal \'etale quotient of $G$.

Now we can put it all together and obtain the connected-etale sequence:

Proposition: Let $G$ be a finite flat group scheme over a complete discrete valuation ring $R$. (Even ok for complete local noetherian $R$, or even henselian local $R$.) The faithfully flat $R$-homomorphism $f:G \rightarrow G'$ to the maximal \'etale quotient has scheme-theoretic kernel $G^0$.

Proof: The kernel $H = \ker f$ is a finite flat $R$-group. To show it contains $G^0$ we have to check that the composite map $G^0 \rightarrow G \rightarrow G'$ vanishes. Being a map from a finite $R$-scheme to a finite \'etale $R$-scheme, the map is determined by what it does on the special fiber, so it suffices to show that $G_k^0 \rightarrow G_0'$ vanishes. This is a map from a finite infinitsimal $k$-scheme to a finite \'etale $k$-scheme which carries the unique $k$-point to the identity point. Thus, it factors through the identity section of $G_0'$, which is open and closed since $G_0'$ is finite etale over $k$.

Now that $H$ contains $G^0$, to prove the resulting closed immersion $G^0 \hookrightarrow H$ between finite flat $R$-schemes is an isomorphism it suffices to do so on special fibers. But that reduces us to the variant of our problem over the residue field. We can increase it to be algebraically closed, and so the problem is to show that if $G$ is a finite flat group scheme over an algebraically closed field $k$ then $G \rightarrow G'$ has kernel exactly $G^0$. But $G'$ is a constant $k$-scheme since it is etale and $k$ is algebraically closed, so by construction $G'$ is just the disjoint union of the $k$-points of the connected components of $G$. It is then physically obvious that the kernel is $G^0$. QED

Remark: If $X$ is any finite flat $R$-scheme, with $X \rightarrow X'$ the initial map to a finite \'etale $R$-scheme, then the induced map on $\overline{k}$-points is bijective. Indeed, we can pass to geometric special fibers and connected components to reduce to the case when $X$ is local finite over an algebraically closed field (in place of $R$), in which case the assertion is clear.

By this Remark, the geometric points of the $n$-torsion in $E_k$ are identified with the geometric points of the special fiber of the maximal etale quotient $E[n]'$. In particular, if $n$ is not divisible by the characteristic of $K$ and if $K'/K$ is a sufficiently big finite separable extension which splits $E_K[n]$ then the finite etale $R'$-scheme $E[n]'_{R'}$ is constant (as it may be checked on $K'$-fiber), so the map $$E _K[n] (\overline{K}) = E _K[n] (K') = E[n] (R') \rightarrow E[n]' (k') \hookrightarrow E _k[n]' (\overline{k}) = E[n]'(R') = E[n]' (\overline{K})$$ is identified with the naive map in question 1. In other words, that step computes the "quotient" part of the connected-etale sequence of $E[n]$ after passing to $\overline{K}$-points!

Example: If $E$ has supersingular reduction then $E[p] = E[p]^0$ and the etale part of the sequence for $E[p]$ vanishes.

Example: If $E$ has ordinary reduction then working over an algebraic closure of the residue field shows that $E[p]^0$ and $E[p]'$ each have rank $p$ as finite flat $R$-groups.

Finally, it remains to relate $E[n]^0$ to $n$-torsion in the so-called "formal group" of $E$ (not the formal group of $E_K$, which loses contact with the integral structure and for ${\rm{char}} (K) = 0$ is actually the formal additive group which has no nontrivial $n$-torsion!). A moment's reflection on the definition of the formal group in Silverman shows that its $R'$-points for any finite local valuation ring extension of $R$ are precisely the local $R'$-points of the complete local ring $\widehat{\mathcal{O}}_{E,0_k}$ at the origin of the special fiber (or the completion along the identity section, comes to the same since $R$ is complete). By the universal properties of local rings on schemes and completions of local noetherian rings, such $R'$-points of the latter type are simply points in $E(R')$ specializing to $0_k$ in $E_k (k')$. But we saw earlier that $E[n]^0 (R')$ is exactly the set of points in $E[n] (R')$ specializing to $0_k$ on $E_k$. So indeed $E[n]^0 (R')$ inside of $E[n] (R') = E_K[n] (K')$ is exactly the $n$-torsion in the $K'$-points of the "formal group" in the sense of Silverman's book.

Voila, so that answers the questions. The arguments used are designed to apply equally well to abelian varieties.

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+1. Ceci n'est pas une réponse. C'est un petit cours! –  Chandan Singh Dalawat Feb 23 '10 at 7:47
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Yeah---Brian should have just written all of that down on the board in his office, videoed it, uploaded it to youtube, and posted the link. –  Kevin Buzzard Feb 23 '10 at 10:32
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this is really why I get so overwhelmingly happy when a couple of days ago I saw Brian showing up on MO. He is such a great teacher and expositor! :) –  natura Feb 23 '10 at 10:55

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