How can i calculate the principal divisor $(f)$ where
$$f = \frac{(x^{3}-1)}{(x^{4}-1)}$$
with $f\in\mathbb{F}_2(x)$. I am recently reading about the subject, so i am looking for a simple solution (the more clear the better)
Anyone can give at least an idea?
The decomposition into irreducible factors is $$f = \frac{x^2+x+1}{(x-1)^3}.$$ This shows that $f$ has a pole of order $3$ in the rational point corresponding to the maximal ideal $(x-1) \subseteq \mathbb{F}_2[x]$, and that $f$ has a zero of order $1$ in the closed point of degree $2$ corresponding to the maximal ideal $(x^2+x+1) \subseteq \mathbb{F}_2[x]$. It follows $$\mathrm{div}(f)|_{\mathbb{A}^1_{\mathbb{F}_2}} = 2 [x^2+x+1] - 3 [x-1].$$ Since the whole principal divisor $\mathrm{div}(f)$ on $\mathbb{P}^1_{\mathbb{F}_2}$ has, as always, degree $0$, it follows $$\mathrm{div}(f) = 2 [x^2+x+1] - 3 [x-1] + 1 [\infty].$$ This can also be checked directly: For $t=\frac{1}{x}$ we have $f = t \cdot \frac{1 - t^3}{1 - t^4}$, so the order at $\infty$ is $1$.
If $\mathbb{F}_4=\{0,1,\alpha,\beta\}$, one may also write $\mathrm{div}(f) = 2 [\alpha] - 3 [1] + 1 [\infty]$.