Thursday, 5 May 2011

ag.algebraic geometry - How do you see the genus of a curve, just looking at its function field?

Let k be the ground field.



The Kahler differentials OmegaK/k are the K vector space generated by formal symbols dg subject to d(f+g)=df+dg, d(fg)=fdg+gdf and da=0 for aink. This is a one dimensional K vector space.



Let omega be a differential. For any valuation v on K, let t be such that v(t)=1. We say that omega is regular at v if v(omega/dt)geq0. We say that omega is regular if it is regular at every valuation of K/k. Then the space of regular differentials is a k vector space of dimension g.



FURTHER THOUGHTS



Both Ben's answer and mine used the set of valuations of K/k. This essentially means that we used the ground field k. A valuation of K/k is defined as a valuation of K which is trivial on k; conversely, the ground field can be recovered from the valuations that respect it by the formula k=bigcapvv1(mathbbRgeq0).



Here is a cautionary example to show that there can not be any solution which only uses properties of the field K, without reference to k. Let C and D be two irreducible curves of different genuses. Let K be the field of meromorphic functions on CtimesD, let k and ell be the fields of functions on C and D. Then K is a transcendence degree 1 extension of both k and ell, but has different genuses when considered in these two ways.

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