The Faltings theorem states that the number of rationals over an algebraic curve is finite if the genus is greater than 1. The genus decreases by increasing the number of singularities. My question is this. Should one count only the singularities that are rational points or all the singularities over the complex field?
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$\begingroup$ What do you mean by a non-rational singularity for a curve ? What do you mean by a rational singularity for a curve ? In any event, all singularities count. $\endgroup$– mehCommented Jul 11, 2018 at 21:38
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2$\begingroup$ A singular point of a curve f(x,y) is one such that the first derivatives of f(x,y) are zero. This can occur in a point where x and y are real or, more generally, complex. Do these singularities count? in other words, does the geometric genus depend on the field over which the "curve" is defined? $\endgroup$– AlmCommented Jul 11, 2018 at 22:04
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1$\begingroup$ You have an accepted answer, but I was merely trying to make the point that rational singularity is a pointless concept for curves. (pun is unintentional). $\endgroup$– mehCommented Jul 12, 2018 at 15:41
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1$\begingroup$ @aginensky: I think the OP was using the phrase "rational singularity" to mean "rational point which is a singular point" rather than the usual meaning from higher-dimensional geometry. $\endgroup$– PopCommented Jul 12, 2018 at 21:40
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$\begingroup$ Indeed, this is what I was meaning. Sorry for my imprecise language. I have a background as physicist, I am not a mathematician. $\endgroup$– AlmCommented Jul 13, 2018 at 7:37
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The definition of the geometric genus in terms of (d-1)(d-2)/2 minus the contributions of the singularities is not a great one. It's better to give a more intrinsic definition, as the dimension of the space of global section of the canonical line bundle of the normalization (or the first sheaf cohomology of the normalization). In particular, this definition makes it possible to compute what the contributions of different types of singularities are.
In particular, this definition is straightforwardly invariant under change of base field. So it's possible to define and compute it just over an algebraically closed field. This is what most references do (as most introductions are focused more on the pure algebraic geometry than in the arithmetic applications) which is probably the source of your contribution.
So because singular points not defined over the base field correspond to multiple points over an algebraically closed field, they actually have a larger contribution than singular points defined over the base field.
answered Jul 11, 2018 at 23:03
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$\begingroup$ @AlbertoMontina One needs to count the singularities over an algebraically closed field. Also, one needs the singularities to be nodes for that formula to hold. $\endgroup$ Commented Jul 12, 2018 at 1:25