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Let us fix a projective curve X over a field k. With nt, I mean a variety with all irreducible components of dimension 1. Let us suppose that there is a smooth rational point $x \in X$. My question is:

Is it possible to find a hyperplane such that the intersection of X and this hyperplane is a point? If so, how?

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If you are looking for some projective immersion for your curve such that this happens, then if your curve is irreducible just consider the divisor $(2g+1)p$, where $p$ is the point you were talking about and $g$ is the genus of the curve. This divisor is very ample, and so if we immerse the curve in projective space with this divisor, there exists a hyperplane such that the set theoretical intersection of the hyperplane with the curve is just $p$.

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  • $\begingroup$ Dear Robert: Thank you. $\endgroup$ – Riemannrock Jul 16 '13 at 16:18
  • $\begingroup$ Dear Riemannrock: You're welcome. PS: nice name, I always thought Riemann Rock would make a cool band name $\endgroup$ – rfauffar Jul 16 '13 at 16:20
  • $\begingroup$ I must however bother you a bit - what if we don't assume that the curve is irreducible? (Actually, I am trying to use this to show that such a curve must be irreducible) $\endgroup$ – Riemannrock Jul 16 '13 at 16:21
  • $\begingroup$ Then it can't happen unless your point is in the intersection of all the irreducible components. This is because a hyperplane is ample and so must intersect any positive dimensional subvariety of projective space in at least one point. In particular, it must intersect all the irreducible components of your curve. $\endgroup$ – rfauffar Jul 16 '13 at 16:24
  • $\begingroup$ Robert Auffarth: Thank you! I will accept your answer now, if you have the time, however: Is it true that a curve such as this one should be integral if it has a smooth rational point? $\endgroup$ – Riemannrock Jul 16 '13 at 16:52

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