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Math Help - Measure of the roots of a multivariate polynomial.

  1. #1
    Junior Member RaisinBread's Avatar
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    Measure of the roots of a multivariate polynomial.

    Hey, I'm trying to show the following proposition, but I have no idea where to start with it;

    Let P(x) be a polynomial, where x \in R^n. Show that the set \{x \in R^n ; P(x)=0\} has Lebesgue measure zero.
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  2. #2
    Senior Member
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    Re: Measure of the roots of a multivariate polynomial.

    Quote Originally Posted by RaisinBread View Post
    Hey, I'm trying to show the following proposition, but I have no idea where to start with it;

    Let P(x) be a polynomial, where x \in R^n. Show that the set \{x \in R^n ; P(x)=0\} has Lebesgue measure zero.
    This only works for nonzero polynomials, where we can use induction. Suppose the result holds for n-1 and let A=\{x_n:P(x_1,\cdots,x_n)=0\forall x_1,\cdots,x_{n-1}\}. Then A has measure zero in \mathbb{R}. Furthermore, for each fixed u\in\mathbb{R}\setminus A, the polynomial P(x_1,\cdots,x_{n-1},u)\in\mathbb{R}[x_1,\cdots,x_{n-1}] is nonzero, which means the set B(u)=\{(x_1,\cdots,x_{n-1}):P(x_1,\cdots,x_{n-1},u)=0\} has measure zero in \mathbb{R}^{n-1} by the inductive hypothesis. Show that

    \mathbb{R}^{n-1}\times A and

    C:=\bigcup_{u\in\mathbb{R}\setminus A}[B(u)\times\{u\}]

    both have measure zero in \mathbb{R}^n, and it will follow that \mathcal{Z}(P)=C\cup(\mathbb{R}^{n-1}\times A) has measure zero, where \mathcal{Z}(P)\subset\mathbb{R}^n is the set of zeros of P.
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