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Let $X$ be a smooth projective curve over $\mathbb C$ and take a point $p \in X$. For some very large $k$, $\mathcal O(kp)$ is very ample, so gives an embedding $i: C \to \mathbb P^n$. How can I find a hyperplane $H \subset \mathbb P^n$ with hyperplane section divisor equal to $kp$? Or at the bare minimum, I just want a hyperplane divisor intersecting $X$ at only $p$. For reference, I am trying to understand why a smooth projective curve minus a finite number of points is affine, and there is a proof here. Thank you.

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I don't know if you are still interested, but this is an application of the Riemann-Roch theorem. Indeed, as you said the divisor $D = (2g+1)P$ is very ample by Riemann-Roch, so $|D|$ gives an embedding of $\phi : X \to \Bbb P^{2g}$.

On the other hand, the elements in the linear system $L(D)$ corresponds exactly to the set of hyperplane sections $H \cap \phi(X)$, so in particular there is an hyperplane $H$ with $H \cap X = \{P\}$ and in particular $X \backslash \{P\}$ is affine.

This correspondence holds because if $f_0, \dots, f_n$ is a basis of $L(D)$, an divisor is an element of the linear system $E = D + \text{div}(g)$, with $g =\sum a_i f_i$, will corresponds to $X$ intersected with the hyperplane $H : \sum a_i x_i = 0$. For more details see the book by Rick Miranda, Riemann surfaces and algebraic curves, page 159.

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