Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Suppose $f$ is a degree $n$ univariate polynomial with roots $\alpha_1, \ldots, \alpha_n$. Then we know that $$\frac{f'(x)}{f(x)} = \sum_{i=1}^n \frac{1}{(x-\alpha_i)}.$$ Can we say something similar for a system $F=(f,g)$, where $f,g$ are bivariate polynomials with only finitely many common zeros? That is, can we express $F'/F$ in a form dependent upon the common roots of $f,g$?

share|cite|improve this question
1. Might depend on $F()$. 2. Do you mean $f(x,y)$? 3. What do you expect: $\frac{F'}{F}=?$ – draks ... Mar 12 '12 at 9:13
Actually, what exactly do you want to know? Systems of bivariate polynomials have many interesting properties, but without a motivation for your question, it has hard to find out which kind of result would help you. – Johannes Kloos Mar 12 '12 at 9:18
If you are just interested in a result on the common zeroes, you could use some computer algebra system to calculate them (e.g., by calculating the primary decomposition of the system $F$). I doubt there will be a nice expression of the form you are looking for, since multivariate systems like your tend to be far more complex than univariate systems - if you wish to know more, read up on (classic) algebraic geometry and commutative algebra. – Johannes Kloos Mar 12 '12 at 10:55
Thanks for the response. I was wondering if there is an expression for $\frac{F'}{F}$ in terms of the variety of $F$. – Vikram Mar 14 '12 at 10:05
You could probably work out something per coordinate (calculate $F \cap K[x]$ and $F \cap K[y]$), but I am not aware of any formula that gives direct results. Why exactly do you need such a correspondence? – Johannes Kloos Mar 14 '12 at 10:10

Your Answer


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

Browse other questions tagged or ask your own question.