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Let $P:\mathbb{R}^n\rightarrow \mathbb{R}$ be a non-zero $n$-dimentional polynomial.

Let $C=\left \{ x\in \mathbb{R}^n: P(x) \neq 0 \right \} $. How can I show that $C$ is dense?

I don't even know where to start.

35T41
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2 Answers2

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Let $a\in \mathbb R^n, r>0.$ We need to show $P(x)\ne 0$ for some $x\in B(a,r).$ Assume to the contrary that $P\equiv 0$ in $B(a,r).$ Then for every unit vector $u,$ the function of $t\in \mathbb R$ given by $q_u(t) = P(a+ tu)$ is $0$ for $t\in (-r,r).$ But $q_u(t)$ is a one-variable polynomial, and a one-variable polynomial that vanishes in a full interval vanishes identically. Thus each $q_u$ is identically $0.$ That is the same as saying $P\equiv 0$ on each line through $a.$ That implies $P\equiv 0$ on $\mathbb R^n,$ contradiction

zhw.
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  • Thank you for your answer. Could you elaborate on the reasoning from "$P \equiv 0$ on each line through $a$" to "$P \equiv 0 $ on $\mathbb{R}^n$"? I do not fully understand this step. Thank you in advance. – Shi James Jun 16 '21 at 09:47
  • Consider that $x =[x_1 \quad x_2]$ and let $P(x) = x_1 x_2$. In addition, pick $a = [0 \quad 0]$. In this case, no matter how we perturb a single entry in $a$, $P \equiv 0$. However, $P$ is not identitically zero in this case. This shows that "$P \equiv 0$ on each line through $a$" is not sufficient for $P \equiv 0$, right? Does this falsify your proof? – Shi James Jun 16 '21 at 10:11
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    First question: Let $b\in \mathbb R^n, b\ne a.$ Consider the line through $a$ and $b.$ I've shown $P=0$ on this line. Therefore $P(b)=0.$ Since $b$ was any point in $\mathbb R^n,$ $P\equiv 0$ on $\mathbb R^n.$ – zhw. Jun 16 '21 at 14:11
  • Second question: No it doesn't falsify my proof. You can't reach all points by perturbing a single entry of $a.$ – zhw. Jun 16 '21 at 14:16
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Assume for the moment $n=2$.

Suppose the contrary. That is, assume that there is a disk $D(\mathbf x,r)$ where $P$ vanishes. Say that $\mathbf x=(a_0,b_0)$. Then the polynomial $P^2_{b_0}(x)=P(x,b_0)\in\Bbb R[x]$ has infinitely many roots, and then $P^2_{b_0}=0$. (The $2$ here is not a power, but just a superindex, and it points to what indeterminate is substituted by the subindex $b_0$).

That means that $b_0$ is a root of the polynomial $P^1_a(y)=P(a,y)\in\Bbb R[y]$ for every real $a$. That is, $P(a,y)=(y-b_0)Q^1_a(y)$ for every $a\in\Bbb R$, or $$P(x,y)=(y-b_0)Q(x,y)$$

Since the degree of $Q$ is lesser than of $P$ we can use induction on the degree.

After this, we can use induction now on the number $n$ of indeterminates.

ajotatxe
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  • Let $P\in\mathbb{R}[x_{1}, \dots, x_{n}]$ be identically zero on $(a_{1}, b_{1})\times \cdots \times (a_{n}, b_{n})$. Then for a fixed $x\in (a_{n}, b_{n})$, $P(y_{1}, \dots, y_{n-1}, x) = 0$ on an open rectangle in $\mathbb{R}^{n-1}$ and thus $P(y_{1}, \dots, y_{n-1}, x) = 0$ for all $(y_{1}, \dots, y_{n-1})\in\mathbb{R}^{n-1}$ (by the induction hypothesis). Now fix $(y_{1}, \dots, y_{n-1})$ in $\mathbb{R}^{n-1}$ and show that the polynomial in one variable $x\mapsto P(y_{1}, \dots, y_{n-1}, x)$ is identically zero. – Karthik Kannan Sep 11 '24 at 01:22
  • I think this is how your idea can be generalized to higher dimensions. – Karthik Kannan Sep 11 '24 at 01:23