# Genus over finite fields

Is there a way of computing the genus of a parametrized curve over a finite field? For instance I am interested in the genus of the following space curve in the m-dimensional space over $F_{q^k}$ given parametrically as: $(x, x^h, x^{h^2}, \dots , x^{h^{m-1}})$ Here h and m are some fixed constants, say.

I basically work in theoretical computer science, more specifically in algebraic coding theory, so am not that familiar with advanced algebraic geometry, so I apologize in advance if the question is vague. Any help or references in this regard would be appreciated.

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I came to find on further searching that a curve is parametrizable iff it has genus 0. But the results were stated for curves over complex numbers. Is it true over finite fields too, and are there caveats? –  BharatRam Jun 10 '13 at 3:35
Defining a curve by parametrization with the parameter ranging over a finite field is a bit problematic. After all, you are just listing a finite set of points. If you consider the points on the curve over an extension field, then what are they? You can fit many curves through this set of points, so you haven't really defined a curve yet. You might even get a surface or something else entirely! Presumably you want to use the same formula with the parameter $x$ ranging over all the extension fields of $\mathbb{F}_q$. Then the argument from the complex case goes through and you get genus zero. –  Jyrki Lahtonen Jun 10 '13 at 7:21
BTW, are you working your way towards Goppa codes aka algebraic geometry codes? –  Jyrki Lahtonen Jun 10 '13 at 7:32

The map you describe is also defined over the algebraic closure of $\mathbb F_q$ and is a morphism of algebraic varieties from the affine line to the affine space of dimension $m$. The projection to the first coordinate, when restricted to the image, is the inverse map. So the image is isomorphic to the affine line. You can say it has genus $0$ (usually, the genus is defined for projective curves).