Genus of a curve

Question

My question is:

How to show that for all $g>0$ there exists an algebraic curve of genus $g$? The intuition of this is simple, but I don't know how to show it.

Answer 1

The Wikipedia article Hyperelliptic curve states:

In algebraic geometry, a hyperelliptic curve is an algebraic curve of genus $g > 1$, given by an equation of the form $$ y^{2}+h(x)y=f(x) $$ where $f(x)$ is a polynomial of degree $n = 2g + 1 > 4$ or $n = 2g + 2 > 4$ with $n$ distinct roots, and $h(x)$ is a polynomial of degree $< g + 2$ (if the characteristic of the ground field is not $2$, one can take $h(x) = 0$).

The article goes on to refer to the Riemann-Hurwitz formula which can be used to give the proof you want.

Answer 2

Here is a perspective from working over the field of complex numbers. I'll assume you want to talk about nonsingular curves, in which case the genus unambiguously means the geometric genus $h^0(X,\Omega^1)$, i.e. dimension of the global sections of the holomorphic cotangent bundle. Let $X$ denote a Riemann surface. It is well known that the geometric genus of $X$ defined as above equals the topological genus of the underlying real surface.

Next, using Kodaira's Embedding Theorem (or simply any other way to prove that a Riemann surface holomorphically embeds in $\Bbb{P}^n$), you can holomorphically embed $X$ as an analytic subvariety of some $\Bbb{P}^n$ and by Chow's Theorem this realizes the image of $X$ as an algebraic variety in $\Bbb{P}^n$. The genus of this variety is then $g$.

Of course, a better construction in some sense is given by using hyperelliptic curves as mentioned in Somos' answer.