1
$\begingroup$

I read the definition of the group law on elliptic curves and there's one thing I don't understand. In the link above it is stated that

In the projective plane, each line will intersect a cubic at three points when accounting for multiplicity.

In an algebraically closed field this is a consenquence of Bézout's theorem, but is it true if the elliptic curve is defined over a field which is not necessarily algebraically closed? If not, how is the group law defined over such fields?

$\endgroup$
  • 3
    $\begingroup$ It is not true that any line will intersect the elliptic curve in 3 points with coordinates in the field. But any line which passes through 2 points with coordinates in the field will meet the curve in a third point with coordinates in a field, and that is what you need for the group law. $\endgroup$ – Art May 11 at 19:09
  • $\begingroup$ Can you please elaborate why a line can't intersect an elliptic curve at exactly two points (with coordinates in the field)? $\endgroup$ – MCL May 11 at 21:30
  • $\begingroup$ The third point can be $P_\infty$ $\endgroup$ – kelalaka May 11 at 21:42
  • 1
    $\begingroup$ $f(x,y) = y^2-x^3-ax-b$, given two points $(A,B),(A+C,B+D)$ on the curve with $C \ne 0$ there is a unique 3rd root $T$ of the polynomial $g(t)=f(A+tC,B+tD) = -C^3 t(t-1)(t-T)$ which is found from $g(-1) = C^3 2 (T+1)$ and the curve intersects the line in $(A,B),(A+C,B+D),(A+TC,B+TD)$ $\endgroup$ – reuns May 11 at 22:53
  • $\begingroup$ Now I get it. Since we have two roots, the polynomial must split into linear factors and there has to be a third root. And in the case that C=0 the third intersection point is at infinity. Thank you. $\endgroup$ – MCL May 12 at 17:01

Your Answer

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

Browse other questions tagged or ask your own question.