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.



A unique zero of a system of polynomials is a zero of a finite system.
The existence of a finite $S\_2$ follows from [Hilbert's basis theorem](http://en.wikipedia.org/wiki/Hilbert%27s_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](http://en.wikipedia.org/wiki/Algebraic_variety)) of $S$.


