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Consider an irreducible polynomial $P \in \mathbb{R}[x_1,\ldots ,x_n]$ and define

$$V := P^{-1}(0) = \left\{ (x_1, \ldots ,x_n) \in \mathbb{R}^n \mid P(x_1,\ldots ,x_n)=0 \right\}$$

It's well known that $V$ can have many connected components (a hyperbole has two components, for example).

Suppose V contains a sphere

$$S = \left\{ (x_1,\ldots ,x_n) \in \mathbb{R}^n \mid (x_1-a_1)^2+\ldots +(x_n-a_n)^2=r^2 \right\}$$

Is there any relationship between $P(x_1,\ldots ,x_n)$ and the polynomial $Q(x_1,\ldots ,x_n)=(x_1-a_1)^2+\ldots +(x_n-a_n)^2-r^2$ ?

Is it true that $Q$ divide $P$ (and then $Q=P$ by irreducibility)?

Alexandre Paiva Barreto
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1 Answers1

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Yes, this is true when $r\neq0$. Then the polynomial $Q$ is square-free (when $n>1$ even irreducible) and the real part $S$ of the complex zero set $X\subset\mathbb{C}^n$ of $Q$ contains a smooth point of $X$. This implies that $S$ is Zariski dense in $X$ meaning that any polynomial vanishing on $S$ will also vanish on $X$. Since $Q$ is square-free, Hilbert's Nullstellensatz implies that the vanishing ideal of $X$ is generated by $Q$. In particular, any polynomial that vanishes on $S$ must be divisible by $Q$.

For $r=0$ this is no longer true.

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