Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $\mathrm{div}(f) = 2 [\alpha] - 3 [1] + 1 [\infty]$.

share|cite|improve this answer
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)

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.