# Simple proof that the arithmetic genus is non-negative

I take an irreducible and reduced closed curve $$C\subseteq \mathbb{P}^n$$, defined over an algebraically closed field $$k$$ and define the arithmetic genus $$p_a(C)$$ as the integer such that the Hilbert polynomial of $$C$$ (or of its ideal of polynomials vanishing on it) is $$h_C(X)=\deg(C) X+1-p_a(C).$$ I would then like to have a simple proof that $$p_a(C)\ge 0$$. Of course, one may say that it is because it is the dimension of a vector space obtain using cohomology, but I would like a proof not involving cohomology, as it is for a master work and in my course of algebraic geometry I (unfortunately you will say) did not define the cohomology. It would then take a lot of time for the student to understand the cohomology just to prove this.

In the course, we defined the Hilbert Polynomial of any ideal, computed local intersections, Bézout theorem in $$\mathbb{P}^n$$ and studied blow-ups of surfaces. We also proved that $$p_a(C)=g(C)$$ when $$C$$ is smooth, where $$g(C)$$ is given by Riemann-Roch (the smallest integer such that $$\ell(D)\ge \deg(D)+1-g$$ for each divisor $$D$$ on $$C$$). I would like the student to use the fact that $$p_a(C)\ge 0$$ to bound the type of singularities of a plane curve and to show that one can have a resolution by blowing-up the singular points and repeating this process finitely many times.

Thanks for your help.

## 2 Answers

Mumford (Complex Projective Varieties, section 7) has the following, reasonably simple proof. Let $$d$$ be the degree of $$\mathrm{C}$$, $$m$$ big enough such that $$h_{\mathrm{C}}(m)=\mathrm{dim}_{\mathbf{C}} (\mathbf{C}[\mathrm{T}_0,\ldots,\mathrm{T}_n]/\mathrm{I}(\mathrm{C}))_m$$ and $$md/2>p_a$$. Embed $$\mathrm{C}$$ into $$\mathbf{P}^N$$ by the degree $$m$$ Veronese embedding; let $$\mathrm{L}\subset\mathbf{P}^N$$ be a linear space containing $$\nu_m(\mathrm{C})$$ such that $$\nu_m(\mathrm{C})$$ is nondegenerate in $$\mathrm{L}$$. Then $$\dim(\mathrm{L})=h_\mathrm{C}(m)=md+1-p_a$$, in other words $$\mathrm{L}\simeq\mathbf{P}^{md-p_a}$$, and the degree of $$\nu_m(\mathrm{C})$$ is $$md=\mathrm{deg}(\nu_m(\mathrm{C}))\geqslant\mathrm{codim}(\nu_m(\mathrm{C}))+1\geqslant md-p_a$$ since $$\nu_m(\mathrm{C})$$ is nondegenerate in $$\mathrm{L}$$.

• Thanks. It is a nice proof that I did not know. – Jérémy Blanc Mar 26 at 16:35

Maybe what follows should be modified depending on what is exactly assumed in the question; nevertheless, let me try.

We fix $$m$$ positive integer large enough so that $$h_C(m)=dim H^0(C,\mathcal{O}_C(m))$$. We have to show that $$dim H^0(C,\mathcal{O}_C(m)) \leq deg(D)m +1$$.

Let $$H$$ be a degree $$m$$ hypersurface of $$\mathbb{P}^n$$ intersecting $$C$$ in finitely many points of $$C$$, all smooth (such $$H$$ exists as $$C$$ is reduced and so has finitely many singular points). One can view the intersection $$D:=H \cap C$$ as an effective Weil divisor on the smooth part of $$C$$, we have $$D=\sum_i a_i p_i$$ with $$p_i$$ smooth point of $$C$$ and $$a_i$$ positive integer multiplicity. We have $$\sum_i a_i =deg(C)m$$.

The equation defining $$H$$ restricted to $$C$$ defines a rational trivialization of the line bundle $$\mathcal{O}_C(m)$$, singular along $$D$$. Therefore, sections of $$C$$ can be identified with rational functions $$f$$ on $$C$$ such that $$div(f)+D \geq 0$$ (remark that this makes sense as $$D$$ is made of smooth points of $$C$$). Concretely, $$f$$ is allowed to have a pole of order at most $$a_i$$ at the point $$p_i$$. Near each point $$p_i$$, fix a local coordinate and consider the polar part of the Laurent expansion of $$p_i$$ (terms with negative power of the local coordinate): it consists of $$a_i$$ coefficients. If $$f$$ and $$g$$ have the same polar parts at every $$p_i$$, then $$f-g$$ is a regular function on $$C$$ and so a constant as $$C$$ is reduced and irreducible. Therefore the kernel of the map from $$H^0(C,\mathcal{O}_C(m))$$ to the space of polar parts is one-dimensional and so $$dim H^0(C,\mathcal{O}_C(m)) \leq (\sum_i a_i)+1=deg(C)m+1$$.

• Thanks for the proof. I have to try to see if I can use it with what I did in the course. – Jérémy Blanc Mar 26 at 16:38