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Is it easy to find an example of a complete linear system on a smooth projective curve (say over $\mathbb C$) which separates points but which is not an embedding?

(for just a linear system, one can take the linear system induced by a linear projection (in an embedding) from a point which is not on any secant line of the curve but lies on a tangent line of the curve).

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Take for $X$ a trigonal curve of genus $\geq 5$; that is, $X$ carries a unique degree 3 pencil $P$. Suppose some divisor of $P$ is of the form $p+2q$. Consider the linear system $|K-p|$. If $r,s$ are two distinct points of $X$, we have $h^0(p+r+s)=1$ by unicity of the $g^1_3$, hence $h^0(K-p-r-s)=h^0(K-p)-2$ by Riemann-Roch; thus $|K-p|$ separates $r$ and $s$. But $h^0(K-p-2q)=h^0(K-p)-1$, which means that the associated map is not an embedding at $q$.

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I believe that on any non-rational and non-hyperelliptic curve , the complete linear system $L = K_X(2p)$ will work. In such a case $ 2p - p_1 -p_2$ will always be a non-trivial divisor of degree zero unless $p = p_1 = p_2$ and hence $h^0(L) = g+1$ and $ h^0(L(-p_1 -p_2) ) = g-1 $ unless as stated $p = p_1 = p_2$.

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  • $\begingroup$ @abx your answer appeared after I started mine, but before I put mine in. Same idea I think. $\endgroup$ – meh Jun 13 '19 at 15:22
  • $\begingroup$ Yours is certainly simpler, I just didn't think of it. $\endgroup$ – abx Jun 13 '19 at 18:42

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