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In section I.6 of Algebraic Geometry, Hartshorne establishes a that every curve is birationally equivalent to a nonsingular projective curve. To do this, he defines for any given curve $C$ with function field $K$ an abstract nonsingular curve $C_K$ whose points are just DVR's of $K$ with an appropriate topology. It can be shown that $C_K$ is isomorphic to a nonsingular projective curve.

Later on in ex II.3.8, Hartshorne defines the normalization of a scheme, which in the case of curves will be a nonsingular curve that is birational to the original curve.

In general, do these two constructions give the same curve? They give birational curves by II.6.12, but I would like to say that they are actually the same. If this is not true, can we at least say that the normalization has an open immersion into $C_K$? I have a hunch that the latter statement should be true since the points of a nonsingular curve just correspond to DVR's of $K.$

Thanks ahead of time.

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1 Answer 1

If $k$ is an algebraically closed field, then the procedure described in Hartshorne's text establishes an (anti)equivalence of categories between:

  1. The category whose objects are finitely generated field extensions of $k$ of transcendence degree 1, and whose morphisms are $k$-algebra homomorphisms.
  2. The category whose objects are nonsingular projective $k$-curves, and whose morphisms are morphisms of $k$-schemes.

In particular, if you have a nonsingular projective $k$-curve, the corresponding function field is unique up to isomorphism, and two such curves are isomorphic if and only if they are birational.

If you start with a curve that is not projective, then its normalization admits an open immersion to a projective curve, and that curve is unique up to isomorphism.

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So in particular we know that the abstract singular curve and the normalization are birational so isomorphic. Thanks! That does it. –  Lalit Jain Jul 13 '11 at 19:09

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