Is there an example of an integral projective curve $X$ over an algebraically closed field $k$ with nodal singularities whose normalization is a (smooth) genus 1 curve?

More generally, given any $g,n$, can we always find an integral projective curve $X$ over $k$ with $n$ nodes whose normalization is genus $g$?

  • $\begingroup$ A quartic with 2 double points has genus $1$. Is this what you are looking for ? $\endgroup$ – Rene Schipperus May 1 '17 at 20:52
  • $\begingroup$ @ReneSchipperus How would you construct such a quartic? Is there a general procedure for constructing curves with $n$ nodes and whose normalization is genus $g$? $\endgroup$ – oxeimon May 1 '17 at 21:28
  • $\begingroup$ Doesnt such a curve automatically have genus 1 by the genus formula ? $\endgroup$ – Rene Schipperus May 1 '17 at 21:40

The answer to your question is yes. Let $X$ be a smooth genus $g$ curve and let $L$ be a very large degree line bundle ($\deg L>2g+2n+2$ should do). Let $P_i, Q_i\in X, 1\leq i\leq n$ be $2n$ distinct points. Then you have a surjection $H^0(L)\to H^0(L_{|{\cup P_i\cup Q_i}})=\prod k(P_i)\times\prod k(Q_i)$. Take $\prod_{i=1}^n k\to \prod k(P_i)\prod k(Q_i)$ given by each $k$ mapping to $k(P_i)\times k(Q_i)$ as the diagonal embedding. Let $V\subset H^0(L)$ the sections mapping to this $\prod k\subset \prod k(P_i)\times \prod k(Q_i)$. Then, you can consider the the map given by this linear system $V$ and check that the image is precisely a curve where the $P_i,Q_i$ for all $i$ are identified into nodal points. The checking is tiresome, but follows the same principle as you would find in Hartshorne's book.

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  • $\begingroup$ Aren't you just saying that any curve is birational to a plane curve with ordinary singularities ? $\endgroup$ – Rene Schipperus May 2 '17 at 3:04
  • $\begingroup$ @ReneSchipperus Not really, since if genus is fixed, the number of double points for a plane curve birational to it is fixed too, so not all $n$ would work. $\endgroup$ – Mohan May 2 '17 at 13:39
  • $\begingroup$ @ReneSchipperus You are right. The only minor extra in the above construction is that you can choose the points you want to identify, a bit more difficult when you are dealing with generic projections to plane curves, at least in my mind. $\endgroup$ – Mohan May 2 '17 at 14:09

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