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## relation between polynomial $f$ and $\delta (f)$

Given a polynomial $f \in \mathrm{C} [y_{0}, y_{1}, \cdots, y_{n-1} ]$, where $n$ is a natural number with $n >1$. Define a derivation $\delta$, s.t., it is linear and $\delta(y_{a}*y_{b})= y_{a+1}*y_{b}+ y_{a}*y_{b+1}, all a, b=1, \cdots, n-1$. For example, if $f=3*y_{1}*y_{4}-y_{5},$ then, we have $\delta (f)=3*y_{2}*y_{4}+3*y_{1}*y_{5}-y_{6}$. Then, the question is : given a point $z=(z_{0}, \cdots, z_{n})$, with $z\neq 0$, is there any relation between $|f(z)|$ and $|\delta (f)|$ ? (here, $|\ |$ denotes the absolute value defined by $|z|= \sum_{k=0}^{n} z_{k}$).\

One of my motivation of this problem is that I expect to use the result combining with W.Dale.Brownawell's result on a analytic description of Nullstellensatz Theorem to test a problem in differential algebra, however, I do not know how to get any useful result on the relation of $|f(z)|$ and $|\delta (f)|$ mentioned above. c.f.link text for the differential algebra problem.

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I don't understand the definition. What is, for example, $\delta(y_1\cdot y_2\cdot y_3)$? – Emil Jeřábek Mar 1 2011 at 12:32
$\delta(y_{1}*y_{2}*y_{3})=\delta(y_{1}*y_{2})*y_{3}+y_{1}*y_{2}*y_{4}=(y_{2}^2+y_{1}*y_{3})*y_{3}+y_{1}*y_{2}*y_{4}$ – Jiang Mar 1 2011 at 13:28
If the asterisks stand for multiplication, then please don't do that. It is immensely confusing. And do you really mean for the absolute value of an arbitrary point in $\mathbb{C}^n$ to be the sum of the components? That would be a first, for me. And also, what is $\delta(y_{n-1})$? Zero? – Harald Hanche-Olsen Mar 1 2011 at 14:35
I think $\delta$ maps $C_n:=\mathbb{C}[y_0,\dots,y_n]$ into $C_{n+1}$ so that it could be defined as the unique derivation on the algebra $A:=\oplus_n C_n$ such that $\delta(x_n)=x_{n+1}$ for all $n$ (I'm not saying that I am convinced of what I'm saying, though). – Pietro Majer Mar 1 2011 at 16:51
Pietro Majer's understanding is what I want to say. And the absolute value of an arbitrary point is the sum of the components. – Jiang Mar 2 2011 at 2:46