Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know that for a given elliptic curve $E$ we can define a group $G$ with the points on this curve. However, can we define a ring on it? That is, can we define a multiplication on the curve, where we take two points $P$ and $Q$ and produce another point $R$?

Note: I am not talking about the point multiplication, where a point $P$ is added to itself repeatedly.

share|improve this question
6  
Sure: any group can be made into a ring by defining $ab=0$ for all $a$ and $b$; or you can use the fact that the group of points is a finitely generated abelian group, so isomorphic to $\mathbb{Z}^r\oplus(\mathbb{Z}/n_1\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}/n_k \mathbb{Z})$, fix an isomorphism, and define a ring structure by using the natural ring structure of the latter. But the question is: What would that be good for? The group of points has some very nice geometric and arithmetic properties; defining a ring structure that is not derived from similar properties is a bit pointless. –  Arturo Magidin Oct 10 '11 at 21:31
    
@Arturo: the group of rational points is f. g., or something. But the curve can well not be. –  Mariano Suárez-Alvarez Oct 10 '11 at 22:00
    
@Mariano: Yes, good point. The group of points over $\mathbb{C}$ is not a finitely generated abelian group... –  Arturo Magidin Oct 10 '11 at 22:20
    
Thanks for the comments. I am asking for something more natural. I guess the question would be better phrased if I asked for a "natural" way to define a ring structure. –  Aleks Vlasev Oct 10 '11 at 23:14
    
Might you share the incentive for this desire of a ring structure? Thanks very much. –  awllower Dec 14 '11 at 14:10

1 Answer 1

up vote 2 down vote accepted

Elliptic curves are used in cryptography because they do not have ring structures. See reference [12] in this paper.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.