Polynomial that forces a polynomial to be $0$

Asked Sep 02 '21 at 08:56

Active Sep 02 '21 at 09:02

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Suppose that $f(x_0,x_1,\dots,x_n)\in \mathbb C[x_0,\dots,x_n]$ is a polynomial. Then what are the conditions that $f$ must satisfy (preferably necessary and sufficient) such that there exists a $g\in \mathbb C[x_1,\dots,x_n]$ for which $f(g(x_1,\dots,x_n),x_1,\dots,x_n)$ is a polynomial identical to $0$?

I am sure that there is a theory based on this question, if someone could point me to the right direction I would appreciate it.

edited Sep 02 '21 at 09:02

asked Sep 02 '21 at 08:56

quantum

1,645

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What is $\Bbb{C}[\cdots]$? – FShrike Sep 02 '21 at 08:56
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@FShrike This is the standard notation for the set of all polynomial with $n+1$ indeterminates with coefficients in $\mathbb{C}$. – TheSilverDoe Sep 02 '21 at 08:58
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This just means $f$ Isa multiple of $x_0-g(x_1,x_2,\cdots,x_n)$. – Mohan Sep 02 '21 at 15:16
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Related $;-;$ Application of remainder theorem and factor theorems to multivariable polynomials. – dxiv Sep 02 '21 at 17:15
@dxiv: Oh wonderful! This is not only related, its more or less the answer. Two questions though:
1. Is there a proper reference that discusses this?
2. Is there a generalization of this e.g. $f(g_1,g_2,x_2,\dots,x_n)$ (or with more $g_i$) is identical to zero? I suspect this is not possible as one is behind the scenes using factorization in one variable.
– quantum Sep 03 '21 at 05:14
There is no such direct generalization. Consider for example the irreducible polynomial $,f(x_1,x_2,x_3)=x_1-x_2+x_3,$, then $,f\left(g(x_3), g(x_3)+x_3, x_3\right) \equiv 0,$ for any $,g,$. – dxiv Sep 03 '21 at 05:31

Polynomial that forces a polynomial to be $0$

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