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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 $f,g$?

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    1. Might depend on $F()$. 2. Do you mean $f(x,y)$? 3. What do you expect: $\frac{F'}{F}=?$2012-03-12
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    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.2012-03-12
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    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.2012-03-12
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    Thanks for the response. I was wondering if there is an expression for $\frac{F'}{F}$ in terms of the variety of $F$.2012-03-14
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    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?2012-03-14

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