4
$\begingroup$

As the definition referred from Silverman's book:

An elliptic curve is a pair $(E,O)$, where $E$ is a nonsingular curve of genus one and $O\in E$. (We generally denote the elliptic curve by $E$, the point $O$ being understand.) The elliptic curve $E$ is defined over $K$, written $E/K$, if $E$ is defined over $K$ as a curve and $O\in E(K)$.

We use Riemann-Roch theorem to prove $E$ has a Weierstrass form(This need us go into the deep part of algebraic geometry). There are some other nice properties, like the dual isogeny, elliptic have CM(there are two simple forms of $End(E)\neq \mathbb{Z}$)...

I wonder why it is so beautiful and what may be the most important part that influenced elliptic curve has this properties.

Thank you for sharing your mind.

$\endgroup$
  • 1
    $\begingroup$ "Why" is a difficult question when it comes to math. One could say that one answer is due to the axioms we use. But Im guessing thats not the answer you are looking for... $\endgroup$ – Jakob May 19 '14 at 14:41
  • $\begingroup$ Indeed,@ Jakob the theory of elliptic curves is so rich, and my question some what large.But I'd love to get some interesting idea from others. Even only one part may helpful. I believe the beautiful Weierstrass form may influence a lot. $\endgroup$ – sherry May 19 '14 at 14:55
1
$\begingroup$

As Jakob notes in the comments, it's difficult to say "why" to such a "philosophical" question. However, the remarks below may provide a starting point. Briefly, a smooth plane curve admits a "natural" group structure if and only if the curve has degree three.

In more detail, a smooth plane curve $E$ of genus one is a cubic by the genus-degree formula, and $E$ turns out to be isomorphic to its own Jacobian variety under identification of a point $p$ in $E$ with the divisor $p - O$ in $J(E)$, with $O$ denoting the identity element of $E$.

To express these facts more geometrically:

  • A smooth plane cubic (over an algebraically closed field) intersects each line exactly three times (counting multiplicity).

  • If $p_{1}$, $p_{2}$, $p_{3}$ are $q_{1}$, $q_{2}$, $q_{3}$ are triples of collinear points of $E$, then $\sum p_{i}$ and $\sum q_{i}$ are linearly equivalent: If $P$ and $Q$ are linear forms on $\mathbf{P}^{2}$ (a.k.a., sections of the hyperplane bundle $\mathcal{O}_{\mathbf{P}^{2}}(1)$) such that $P(p_{i}) = 0$ and $Q(q_{i}) = 0$ for each $i$, then $P/Q$ defines a rational function on $E$ such that $(P/Q) = \sum p_{i} - \sum q_{i}$.

  • A divisor $p_{1} + p_{2} + p_{3} - 3O$ is principal if and only if the $p_{i}$ are collinear. (The "only if" direction amounts to the claim that $E$ and $J(E)$ are isomorphic, and is not as obvious as the "if" direction above. Over the complex numbers, the "only if" direction amounts to an addition formula for the Weierstrass $\wp$-function.)

  • Consequently the group law for $E$ has a (beautiful and well-known) geometric interpretation.

    "Numerologically", it's significant that each line intersects $E$ in three points because addition is a binary operation: A condition $p + q = r$, i.e., $p + q - r = O$, involves triples of points.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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