Let $P_1(x_1,y_1),P_2(x_2,y_2)...P_n(x_n,y_n)$ be $n$ rational points on given Elliptic curve. How do we prove they are independent? Are there any theorems/results/algorithms/softwares to prove their independence?
2026-03-27 13:19:25.1774617565
Independence of points on Elliptic curve
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This has already somewhat been answered in the comments but I'd like to add a little example of a Sage session playing with this.
The point is that the canonical height pairing is a symmetric bilinear map $$ E(\mathbf Q ) \times E(\mathbf Q) \to \mathbf R $$ so if there is a linear relationship between some points, there should be the same relationship between all of their heights.
You should be able to do all of this in a Sage session (Sage is free!) (there was a bug with the rational_points command in some recent versions of sage so that might not work depending on your version)
Let's start with a not-so-randomly-chosen elliptic curve (I picked this one to have rank 4 so that this would be interesting, see http://www.lmfdb.org/EllipticCurve/Q/?rank=4)
Lis the list of a whole bunch of points we found nowWe take the pairing matrix of the first two points, looks rank 1 so determinant 0!
and it is! we should expect this though, points 0 and 1 have the same x-coord so are inverses of each other
what about points 1,2?
doesn't look rank $\lt 2$ at all! though it is always symmetric
Similar for points 7,8
We can try 3 other points now
this is clearly non-zero so assuming correctness of the software these three points are independent.
What about non-independence, in general it is notoriously hard to prove real numbers are zero on a computer, leading to problems when trying to prove dependence in general. With rational points of elliptic curves though we are fundamentally in a finitely generated abelian group though, so we can do more
Here are 4 points, which don't have an obvious relationship by glancing at!
Looks like the determinant is zero:
So are they dependent? Lets give the matrix we think has some kernel a name.
Sage will complain if you ask it for the kernel as we are over the reals to some finite precision, so we use a little trick:
So it looks like $(1,1,0,-1)$ is a kernel vector (i.e. $L[4] + L[6] = L[10]$)
indeed this is a relation.