Question

Suppose $p$ is a point in $\mathbb{R}^n$ so that among the set $S$ of polynomials in $\mathbb{Z}[x_1,\ldots,x_n]$ which equal zero at $p$, $p$ is the only point in some neighborhood of $p$ at which all of them equal zero.

Is there necessarily a finite set $S_2\subseteq S$ of polynomials so that $p$ is the only point in a neighborhood of $p$ at which every polynomial in $S_2$ is zero?

This fact appears obvious to me, mostly because the coordinates of $p$ should be algebraic, but every attempt I make at showing the coordinates are algebraic numbers requires this finitely-many-polynomials condition.

Auxilliary questions: When $p$ is viewed as a point in $\mathbb{C}^n$, is it obvious that $p$ is still the only point in a neighborhood of $p$ at which these polynomials all equal zero?

Is the set $S$ finitely generated, as a ring? That is, are there finitely many ($n$?) polynomials $g_1,\ldots,g_n$ so that $S$ is $[g_1,\ldots,g_n]$? When $n=1$, I think this is called Gauss' lemma, and is non-trivially stronger than the fact that $\mathbb{Z}[x]$ is a principle ideal domain.

Answer 1

The existence of a finite $S_2$ follows from Hilbert's basis theorem, as $\mathbb{R}$ is obviously Noetherian. Just take a finite generating system of the ideal generated by $S$.

About your auxilliary questions: a) This is wrong. Just take $f = x^2 + y^2$; then ${ f }$ has the unique zero $(0, 0)$ in $\mathbb{R}^2$, while it has no isolated zero in $\mathbb{C}^2$.

b) The ideal generated by $S$ is finitely generated. The set $S$ itself does not needs to have any algebraic structure, unless you take the set of all polynomials which vanish in $p$ (then it is a maximal ideal in $\mathbb{R}[x_1, \dots, x_n]$). In any case, replacing $S$ by the ideal generated by $S$ does not change the zero set (variety) of $S$.