# If $C \subseteq \mathbb{P}^2$ is a plane curve, then $genus(C)=\frac{1}{2}(d-1)(d-2)$. Compare with example in the notes

In my Algebraic Geometry notes (see http://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2002/main.pdf) there is the following exercise:

If $C \subseteq \mathbb{P}^2$ is a plane curve of degree $d$, then its arithmetic genus $g(C)$ is equal to $\frac{1}{2}(d-1)(d-2)$. Compare this to example $0.1.3$ (of the notes of course).

In fact, the first part of the exercise is easy to solve using the Hilbert polynomial of $C$.

The part of comparing with the example $0.1.3$ is what I don't understand.

What do you think the author wants here as an answer?

• where is the exercise? – Alex Fok Jun 7 '15 at 14:38
• Exercise 6.7.3 of page 118 – Leafar Jun 7 '15 at 14:52
• It seems that the point is to confront the geometric or topological genus with the arithmetic genus. I think so since in example 0.1.3 the discussion is related with the topological genus i.e. the number of holes in your surface. – Holonomia Jun 7 '15 at 14:57
• Here I would say that arithmetic and topological genus coincide, but only if the curve $C$ is non-singular. Is this true? – Leafar Jun 7 '15 at 16:22

The topological (=geometric) genus and the arithmetic genus indeed coincide for a smooth curve $C\subset\mathbb P^2(\mathbb C)$, and the common genus is $\frac{1}{2}(d-1)(d-2)$ if the degree of $C$ is $d$.
For example the curve $y^2z=x^3$ has geometric genus $g(C)=0$, since $\mathbb P^1(\mathbb C)$ is homeomorphic to $C$ under the map $(t:v)\mapsto (t^2v:t^3:v^3)$ .
However its arithmetic genus is $p_a(C)=\frac{1}{2}(3-1)(3-2)=1$, exactly as for any plane projective curve of degree $d=3$.
We always have $g(C)\leq p_a(C)$.
The difference between these numbers is a subtle invariant of the singularities of $C$ and there are formulas for the difference $p_a(C)-g(C)$: see Hartshorne Chapter IV, Exercise 1.8, page 298