# Basic Projective Geometry Question

Can someone help me to see why any two points in $\mathbb P^1$ are linearly equivalent as divisors? If this is true, how come two points on a smooth projective cubic curve are not linearly equivalent?

Linear equivalence means there is a rational function such that one of the points is a pole (of order 1) and the other is a zero (of order 1). For a smooth projective curve, we may think of a rational function as a map to $\mathbb{P}^1$. So the question is: if I give you two points $z_0, z_1$ of $\mathbb{P}^1$, can you find a map with $z_0 \mapsto \infty$ and $z_1 \mapsto 0$, such that these are the only preimages of $\infty$ and $0$? The answer is yes. Send $$z \mapsto \frac{z - z_1}{z-z_0}.$$
Now why can't you do this on a higher genus curve (like an elliptic curve)? Well, for many reasons. The key point is that such a map would have to be an isomorphism; indeed, its degree is 1 (as there is only one point over $0$ and it is unramified there, by definition of multiplicity), so the extension of function fields is trivial. But it can't be an isomorphism because isomorphism preserves genus.
If $a$ and $b$ are points corresponding to the divisors $[a]$ and $[b]$, the rational function $f(z) = (z - a)/(z - b)$ has divisor $(f) = [a] - [b]$.
If $a \neq b$ were linearly equivalent points of a cubic, and if $(f) = [a] - [b]$, there would exist a rational function $f$ with a simple zero at $a$ and a simple pole at $b$. This is impossible by the residue theorem; the sum of the residues of a meromorphic function on a compact curve is zero.