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Given a smooth algebraic curve $C$, say projective over an algebraically closed field $k$, is it always possible to identify two distinct closed points $x, y$ on $C$ to produce a curve with a single node?

In more precise terms, does there always exist a nodal curve $C'$ whose normalization is $C$ such that $x, y$ are the points above the node? Thanks in advance!

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    $\begingroup$ A geometric approach would be to embed the curve nondegenerately in $\mathbf P^3$, project away from a general point on the secant line $\overline{xy}$, and then resolve any nodes other than the one you want. It might take some work to make that rigorous, though. $\endgroup$ – Schemer Jul 23 '15 at 16:11
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The answer is yes. Here is the basic idea. Take a very large degree line bundle $L$ and consider the sections $s$ of $L$ such that $s(x)=s(y)$ for the two given points $x,y$. These sections define a morphism from the curve to a nodal curve identifying just the two points $x,y$ if the degree of $L$ is sufficiently large.

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  • $\begingroup$ Thank you. One thing is unclear to me though, and I would appreciate any help: $L$ will set up a projective embedding by the standard Riemann-Roch arguments (for a high enough degree), but how is it ensured that there are enough sections $s$ with $s(x)=s(y)$? Or can we simply take generating sections for $L$ and somehow alter them such that they take identical values on $x$ and $y$? $\endgroup$ – Paul Jul 27 '15 at 14:04
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    $\begingroup$ If degree of $L$ is large, the vector space of sections of $L$ with $s(x)=s(y)$ is codimension one in the space of all sections. $\endgroup$ – Mohan Jul 27 '15 at 16:34

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