Let $f(x_1,...,x_n)$ and $g(x_1,...,x_n)$ be polynomials in $\mathbb{C}[x_1,...,x_n]$ such that all roots of $f$ are roots of $g$ as well (but not necessarily viceversa). The question is: Does $f$ necessarily divide $g$? In the univariate case ($n=1$) it is clear that indeed $f$ divides $g$ by appealing to the fundamental theorem of algebra. Is this true in general for $n>1$? Any references would be appreciated.



1 Answer 1


Already in the univariate case $f$ needn't divide $g$, contrary to your claim: just consider $f=x^4(x-1), g=x(x-1)^2$ .
In the case of several complex variables Hilbert's celebrated Nullstellensatz answers your question: it says that the roots of $f$ are included in the roots of $g$ if and only if $f$ divides some power $g^k$ of $g$ .

An important point to keep in mind is that you may not assume that $k=1$ : in my counterexample above, $f$ divides $g^4$ but does not divide any smaller power of $g$.

  • $\begingroup$ Thanks for your answer. I forgot to mention that the degrees of the multivariate polynomials $f$ and $g$ need to be the same. Would in that case then Hilbert's Nullstellensatz imply that $f=cg$, for a constant $c$? $\endgroup$ Aug 12, 2013 at 15:48
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    $\begingroup$ Dear Raul, the answer is still no: consider $x_1^2x_2$ and $x_1x_2^2$ $\endgroup$ Aug 12, 2013 at 17:20

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