# Principal divisors

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?

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I´m not even sure what i need to do, at least anyine can say if i am going i the wrong direction? i factorize f (denominator and numerator) then i count the zeros and poles – Dimitri Feb 20 '13 at 23:57

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 $\mathrm{div}(f) = 2 [\alpha] - 3 [1] + 1 [\infty]$.

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Thanks that was the idea i was getting close, but i have seen this example in a book: $f=(x-i)^{4}$ in $\mathbb{C}(x)$, then $div(f)=4[i]-4[\infty]$ – Dimitri Feb 21 '13 at 1:15
Thanks very much, i understand now :) – Dimitri Feb 21 '13 at 1:53
I think that you don´t have to multiply by the degree of the maximal ideal (correct me if i am wrong) – Dimitri Feb 22 '13 at 0:10

The divisor on the affine line over $\mathbb{F}_2$? It's almost too simple: they're essentially just prime factorizations. (and ignore units)

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But i am not sure what the result should be, for example if $f=(x-1)^4$ entonces $(f)=4(x-1)$ or $(f)=(x-1)^4$ or none of them? i am pretty confuse about the subject – Dimitri Feb 21 '13 at 0:35