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If I understand correctly:

  • A scheme is a functor $\mathbf{CRing} \rightarrow \mathbf{Set}$ satisfying certain axioms.
  • A morphism of schemes is a natural transformation.
  • A group scheme is a group object in the category of schemes, so in particular it includes the data of a scheme $G$ together with a natural transformation $\mu : G \times G \rightarrow G$. This means, in particular, that given any commutative ring $R$, we get a corresponding function $\mu_R : G(R) \times G(R) \rightarrow G(R).$
  • Every elliptic curve can be viewed as a group scheme.
  • This means, in particular, that given an elliptic curve $E$ and a commutative ring $R$, we get a function $\mu_R : E(R) \times E(R) \rightarrow E(R)$.

However, this last statement contradicts something else that I thought was true; namely, I thought that $E(R)$ only carried a group structure in the special case where $R$ is a field. In particular, recall that the group structure is defined by considering lines through points on the curve $E$. AFAIK, these lines might "miss" the curve if we're not working over a field.

Question. What's going on here?

goblin GONE
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    If $E$ is an elliptic curve over a field $k$, then $E$ is a group scheme "over $k$", that is a functor from the category of commutative $k$-algebras to groups satisfying appropriate conditions. – Angina Seng Jul 04 '18 at 05:52

1 Answers1

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What's going on is that what you thought was true is wrong. The intersections of lines used to define the group law on an elliptic curve can be written in coordinates using division only by expressions which are units modulo the equation of the curve (this is not obvious and takes some work to prove, and this work is a crucial part of proving that an elliptic curve actually is a group scheme). As a result, they make sense over any algebra over your base ring (see below).

There's also a small but important error in your first set of statements. We normally talk about group schemes over some base ring $A$ (or more generally, over some base scheme). In that case, all our functors are on the category of $A$-algebras, not on the category of commutative rings. Of course, one case of this is $A=\mathbb{Z}$ where you really would get commutative rings, but there actually are no elliptic curves over $\mathbb{Z}$ by a theorem of Tate (in case this seems obviously wrong to you, keep in mind that the definition of an elliptic curve requires nonsingularity, and over $\mathbb{Z}$ that means that the reductions mod $p$ need to be nonsingular for every prime $p$). In any case, it is most common to talk about elliptic curves over some field $k$, so our functors would be on the category of $k$-algebras.

Eric Wofsey
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