Measure of the roots of a multivariate polynomial.

**RaisinBread** · June 29th 2011, 06:40 AM

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.

**hatsoff** · June 29th 2011, 08:24 AM

Originally Posted by RaisinBread

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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