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In Katz's article p-adic properties of modular schemes and modular forms in the Antwerp proceedings, the following definition of an elliptic curve over a base scheme $S$ is given:

By an elliptic curve over a scheme $S$, we mean a proper smooth morphism $p: E \to S$, whose geometric fibres are connected curves of genus one, together with a section $e : S \to E$.

Now this is a quite reasonable definition, which coincides with the usual notion of an elliptic curve when $S$ is the spectrum of a field. However, it does not seem (to me) to follow directly from the definition that such an elliptic curve over $S$ should be a group scheme over $S$ (an obviously desirable property which Katz seems to take as an obvious fact). When $S= \text{Spec }k$, I understand that this essentially follows from Riemann-Roch...

So, what principle allows one to come to this conclusion in the general case?

Thank you!

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    $\begingroup$ There is a result of Grothendieck, proven in Mumford's GIT book (Th. 6.14, p. 124 in the third edition), which says that a smooth proper morphism, which has one fibre, which is an elliptic curve, can be endowed with the structure of an abelian scheme. So if one fibre of your family has a rational point, the two notions will coincide... (in particular if there is a section). $\endgroup$
    – Damian Rössler
    Apr 9, 2013 at 11:50
  • $\begingroup$ Sorry I just noticed that pranavk made a similar comment... I am leaving my comment because it contains the corresponding bibliographical reference. $\endgroup$
    – Damian Rössler
    Apr 9, 2013 at 14:56
  • $\begingroup$ Strictly speaking, the argument in GIT has a projectivity hypothesis on the abelian scheme (due to the role of projectivity for the existence of Hilbert schemes), and there are non-projective abelian schemes over non-normal noetherian domains. If one uses algebraic spaces, which didn't exist at the time that GIT was written, then Hilbert functors are representable without projectivity hypotheses and so the proof of Grothendieck's theorem works without projectivity hypotheses. $\endgroup$
    – user30379
    Apr 10, 2013 at 5:12

3 Answers 3

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The argument that allows you to show that an elliptic curves defined as you say is a group scheme, and even a commutative one is the construction of a functorial and natural isomorphism $E(T) \rightarrow Pic_{E/S}^0(T)$ for every $S$-scheme $T$. This allows to see the functor $T \mapsto E(T)$ as a funtor in group, and since this functor is representable by $E$, this gives a structure of group scheme on $E$, which is the one you are looking for.

Essentially this map is defined as follows: one attached to a point in $E(T)$, that is a $T$-section of $E_T$, the invertible sheaf of divisor this section minus the trivial section $e_T$ (obtained by base change from the section $e$ which is part of the definition). To prove that this map is an isomorphism, one essentially reduces, using base change theorems for direct image in coherent cohomology, to the case of a field, where it becomes a consequence of Riemann-Roch. Note that by definition, the trivial section $e_T$ is send to to the trivial sheaf, so that $e$ is the neutral section of the group scheme structure on $E$, as desired.

Of course there are many details to deal with to make this argument a complete one, but this is done with great care in the beginning of the (unique) book by Katz and Mazur.

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    $\begingroup$ A summary as I understand it: "The same construction that makes an elliptic curve over a field, namely $x \mapsto [x] - [e]$, into a group scheme, works for a relative elliptic curve as well." Then there are the details, of course. $\endgroup$
    – Ryan Reich
    Apr 9, 2013 at 3:13
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This is N. Katz, B. Mazur, Arithmetic Moduli of Elliptic Curves, Ann. of Math. Studies 108, Princeton University Press, Theorem 2.1.2.

Note that the group scheme structure is unique by rigidity results.

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If I were allowed to comment, I would comment that you can also prove this in a more down-to-earth fashion by using that on each fiber, including the generic fiber, there is a unique group structure with the given zero.

Added: For some clues on how to do such things (without using Picard schemes and the like), see Artin's proof of "Weil's theorem" in section 2 of: Artin, M. Néron models. Arithmetic geometry (Storrs, Conn., 1984), 213--230, Springer, New York, 1986.

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  • $\begingroup$ Dear anon, would you mind expanding a little bit? Thank you! $\endgroup$
    – Bruno Joyal
    Apr 9, 2013 at 0:25
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    $\begingroup$ There is a general theorem of Grothendieck to the effect that a smooth proper morphism equipped with a section and geometrically connected fibers is an abelian scheme if it is so on a single geometric fiber, but the proof isn't in any sense down-to-earth. I am amazed to hear that there could be a "down-to-earth" proof of the existence of the group scheme structure even just when the base is an artin local ring (for which "generic fiber" is the special fiber). Anon, what method do you have in mind which isn't the deformation-theoretic proof of Grothendieck's version? $\endgroup$
    – user30379
    Apr 9, 2013 at 1:10
  • $\begingroup$ @anon: Did you mean to write that there is a down-to-earth proof that there is at most one group law with the given zero when the base is reduced (in which case the proof is immediate from consideration of the generic fibers over the base, due to flatness and separatedness considerations over the base)? To say "there is a unique" (which I read as including an existence assertion) over any base or to say "unique" over a non-reduced base both seem to lie beyond the reach of "more down-to-earth" arguments. $\endgroup$
    – user30379
    Apr 10, 2013 at 13:08

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