Example of curve whose coefficients are in field K but has no points with coordinates in K?

89 Views Asked by At

I'm reading Elliptic Tales.

It defines $E$ as an equation whose coefficients are in the field $K$.

It then assumes that $E$ has at least one point with coordinates in $K$.

Example:

An elliptic curve defined over Q is assumed to possess at least one point with rational projective coordinates.

The book then states:

This is a less innocuous assumption than it sounds. If you have a homogeneous polynomial in three variables with rational coefficients, defining the projective curve C, it can be difficult to tell whether C has any rational points at all.

I think I'm missing the obvious, but for some reason I'm surprised this assumption needs to be well... assumed.

I thought it would be necessarily true that if you have a polynomial with rational projective coordinates then all points on the polynomial would also have rational projective coordinates.

Of course, roots may not be rational. But roots are far different from a random point on a polynomial, no?

Is there a simple counter-example so I can see where my intuition is flawed?

1

There are 1 best solutions below

0
On BEST ANSWER

The curve $x^3 + y^3 + z^3$ has no rational point, except the obvious ones (one coordinate is $0$). But what about curves of the form $$x^3 + 2 y^3 + n z^3 = 0$$ There may be no "obvious" rational points, and perhaps for some $n$, there may be none ( see the paper by E. S. Selmer -- "The diophantine equation $a x^3+b y^3+c z^3=0$")

$\bf{Added:}$ If the elliptic curve is a cubic of the form $y^2 = f(x,1)$, $f(x)$ of degree $3$ with simple roots, the homogenized equation $y^2 z = f(x,z)$ has a rational point at $\infty$, $[x\colon y\colon z] = [0\colon 1\colon 0]$, so there is no need to require the existence of a rational point.