I suffer from lack of concrete examples in Algebraic Geometry, so I will appreciate it if somebody can help me in understanding a bit better this one:

Let $\mathcal{C}$ be a genus $2$ curve (projective, smooth and complex). Let $K$ be the canonical class and take a couple of different points $p,q\in\mathcal{C}$. Consider the divisor $K+p+q$.

If I'm correct, (using Riemann-Roch theorem) this divisor defines a map taking $\mathcal{C}$ into a plane projective curve with one node. Is this map birational? How can I visualize it in a concrete way?. Many thanks for any useful comments.


1 Answer 1


This is a very nice example! In fact, provided that $p+q$ is not a canonical divisor, the linear system $K+p+q$ separates "most" points, since $K$ is the only $g^1_2$ on the curve. Thus for any $r$ and $s$ not equal to $p$ and $q$, $\ell(K+p+q-r-s) = 1 = 3-2$, so $r$ and $s$ are separated by the linear system (the dimension drops twice, so you know that $s$ is not in every divisor of the linear system containing $r$, and thus $s$ maps to a different point). The only pair of points for which this doesn't happen is $p$ and $q$ themselves! So the image of $q$ is in every hyperplane ($=$line) containing the image of $p$ (note that imposing that a hyperplane section contain $p$ drops the dimension by $1$, from $3$ to $2$, but further imposing that it contain $q$ leaves the dimension at $2$). This shows that indeed the mapping is birational and that there is exactly one node, which is the image of both $p$ and $q$, in the image curve.


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