I'm studying basics of elliptic curves. I'm reading An Elementary Introduction to Elliptic Curves by Leonard Charlap and David Robbins. It is stated there that the divisor of a line (i.e. a polynomial of the form $ax + by + c$) can have only few forms, among them is $3\langle P \rangle - 3\langle \mathcal{O}\rangle$. I tried to find an example of a curve and a line on it that has such divisor, but to no avail. Can anyone provide an example? If it helps, they suggest that $P$ is an inflection point.

  • $\begingroup$ That's actually an "if and only if". More precisely, assuming that $\mathcal O$ is an inflection point (which it is if you're using a standard Weierstrass equation), then the divisor of $ax+by+c$ has the form $3\langle P\rangle-3\langle\mathcal O\rangle$ if and only if $P$ is an inflection point and the line $ax+by+c$ is the tangent line at $P$. $\endgroup$ – Joe Silverman Nov 15 '16 at 18:54

Let the base field be $F_2$ (hopefully you're fine with a finite base field). Let the curve be $y^2+y=x^3$ and the line $y=0$. The function $y$ has a pole of order 3 at the point of infinity ${\mathcal O}$ and a triple zero at origin ${\mathcal P}=(0,0)$, so the divisor of $y$ is $3{\mathcal P}-3{\mathcal O}$ as prescribed.

Edit: D'oh. The OP asked for examples in other characteristics. I'm apparently at a my dullest. Doesn't the same example work in any characteristic? (Except at char 3, because then the curve has a singular point). See a figure of the real points below. real points of the curve $y^2+y=x^3$

The origin looks like an inflection point to me :-)

  • $\begingroup$ I'm perfectly fine with finite fields, as I'm doing this for cryptography purposes ;) I just want to make sure: it is impossible to give such example for field with characteristic other than 2? $\endgroup$ – Jasiu Sep 12 '11 at 8:17
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    $\begingroup$ @Jasiu: This kind of examples abound in all characteristics. The condition is equivalent to finding a point ${\mathcal P}$ of order 3. I would just have to think a bit harder to find one, as my experience is mostly with char 2 (for reasons similar to yours:-) $\endgroup$ – Jyrki Lahtonen Sep 12 '11 at 8:25
  • $\begingroup$ @Jyrki: When I find a point with order 3, how do I create the line? $\endgroup$ – Jasiu Sep 12 '11 at 8:36
  • $\begingroup$ @Jasiu: The same curve and point work unless we have char 3. $\endgroup$ – Jyrki Lahtonen Sep 12 '11 at 9:31
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    $\begingroup$ @Jasiu: The line is always the tangent. $\endgroup$ – Jyrki Lahtonen Sep 12 '11 at 14:02

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