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

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

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

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