3

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?

Riemannrock
  • 43

1 Answers1

2

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 $p$.

rfauffar
  • 7,509
  • Dear Robert: Thank you. – Riemannrock Jul 16 '13 at 16:18
  • Dear Riemannrock: You're welcome. PS: nice name, I always thought Riemann Rock would make a cool band name – rfauffar Jul 16 '13 at 16:20
  • 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) – Riemannrock Jul 16 '13 at 16:21
  • 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. – rfauffar Jul 16 '13 at 16:24
  • 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? – Riemannrock Jul 16 '13 at 16:52
  • Yes, since if the point is smooth, then it can only be in one irreducible component. Therefore, if there is an immersion of your curve such that the set theoretical intersection of a hyperplane and your curve is one point, then the curve must be irreducible. – rfauffar Jul 16 '13 at 17:21
  • OK - but do I need to know that O_x(2g+1p) is very ample for this? How do I know that the hyperplane section doesn't intersect another point, just from it being smooth? – Riemannrock Jul 16 '13 at 17:32
  • Well, a hyperplane section could very well intersect another point, and actually there are definitely hyperplane sections that do intersect other points as well (just take a line through your point and through any other point; this line is contained in some hyperplane). But the point is that if we can find a hyperplane section that intersects the curve in only one point and this point is smooth, then the curve must be irreducible. This is because if it is not irreducible then the hyperplane must intersect the other components. – rfauffar Jul 16 '13 at 17:38
  • The ampleness of $(2g+1)p$ just shows that if your curve is irreducible, then such a hyperplane section exists. – rfauffar Jul 16 '13 at 17:38
  • Right : I am however interested in the implication the other way around (sorry for not asking this at the start). Namely: IF we have a projective curve C with a smooth rational point, is it true that it must be irreducible? I do agree that if we could find a hyperplane divisor such as the one required, we are done. But I don't know that it is irreducible, or that $(2g+1)p$ is very ample. – Riemannrock Jul 16 '13 at 17:42
  • Not necessarily, if you take any reducible curve (say the union of two curves), then you can just take a smooth point on one component that doesn't belong to the other. – rfauffar Jul 16 '13 at 17:44
  • Hmm, OK, then I might have to rethink the problem. If the curve is of genus 1, maybe it should hold? – Riemannrock Jul 16 '13 at 17:45
  • What's your definition for the genus of a reducible curve? – rfauffar Jul 16 '13 at 17:46
  • Also careful, many times authors include irreducible in their definition of curve. – rfauffar Jul 16 '13 at 17:47
  • Robert Auffarth: I use the arithmetic genus here, so that it is $1-\psi_k(O_C)$ where $\psi$ is the Euler characteristic. – Riemannrock Jul 16 '13 at 17:50