Let $X$ be a non-singular projective curve over $\mathbb C$ and let $P\in X$ be a closed point. Which is, in your opinion, the easiest way to prove that there exists a divisor $D$ of $X$ satisfying the following conditions:
- $D$ is linearly equivalent to $P$.
- $D$ doesn't contain $P$.
Thank you in advance.
I have it on good authority that Relapsarian is suffering an identity crisis at the moment, so let me post the comment above as an answer as per @GeorgesElencwajg's impassioned suggestion.
Let $H_1$ and $H_2$ be nonsingular hyperplane sections of $X$ such that $P$ lies in $H_1$ but not $H_2$. Then the divisor you want is $D=P−H_1+H_2$. Why? "Nonsingular" means that each hyperplanes is intersecting $X$ transversely; that implies in particular that the divisor $H_1$ contains $P$ with coefficient 1. On the other hand $P$ is not in $H_2$, so in the divisor $P-H_1+H_2$ we end up with no term involving $P$.
Finally, (exercise!) any two hyperplanes in projective space are linearly equivalent, and linear equivalence is preserved by restriction, so $-H_1+H_2$ is a principal divisor on $X$.