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Let $X/k$ be a smooth projective curve over an algebraically closed field $k$ of genus $g$, then when is it that the canonical sheaf $\omega_X$ is very ample, i.e. $\omega_X = i^*\mathcal{O}_{\mathbb{P}^n}(1)$ for some closed immersion $i:X\hookrightarrow\mathbb{P}^n$? My intuition is that this is true for any $g\gg0$, and probably for something like $g\ge 2$ or $g\ge 3$, but I don't immediately see how to prove this. Is Riemann-Roch the correct approach?

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    $\begingroup$ Have you heard of Hyperelliptic curves? $\endgroup$
    – Mohan
    Commented Apr 25, 2017 at 22:24
  • $\begingroup$ Yes. What are you implying? $\endgroup$ Commented Apr 25, 2017 at 22:25
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    $\begingroup$ They are typical cases when the canonical bundle is not very ample. $\endgroup$
    – Mohan
    Commented Apr 26, 2017 at 2:17
  • $\begingroup$ Oh, ok. Is there any sufficient condition for the canonical divisor being very ample? $\endgroup$ Commented Apr 26, 2017 at 2:18
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    $\begingroup$ I don't understand why so much hate in the comments. @Sasha, basically every question can be answered by looking into the appropriate book, so your comment es irrelevant and inappropriate. Plus, this is a site for comments of all levels. Having a constructive answer would be great. $\endgroup$
    – Bilateral
    Commented Jul 9, 2018 at 22:15

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Hartshorne, Proposition IV.5.2 has an answer, and I guess this is what Mohan hinted at in the comments:

Proposition 5.2 Let $X$ be a curve of genus $g \geq 2$. Then $|K|$ is very ample if and only if $X$ is not hyperelliptic.

As by Exercise IV 1.7. there are hyperelliptic curves of any genus $g$, the very-ampleness of $K$ is independent of the genus (save for the cases $g = 0$ and $g = 1$, in which $K$ has degree $\leq 0$, and hence can never be (very) ample).

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