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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?

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Warning: there are a billion different sources that introduce algebraic curves, and they will all formalize things in a slightly different way (depending on what they will use later). So it's possible that my answer will be nonsense to you, or use machinery that you don't have yet, or be a tautology. If any of these occur, add more definitions to your question so it is clear what you are working with and what you are trying to prove.

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.

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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.

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