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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.

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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 $S$.

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Thanks! I may have poorly stated the first sentence: $S$ was (supposed to be) defined to be the set of all polynomials vanishing at $p$. The second question may have been poorly phrased: Among $S$ (which inlcudes all polynomials which are zero at $p$) is $p$ the only point near $p$ in $\mathbb{C}^n$ at which all vanish? For $p=(0,0)$, $S$ contians $x$ and $y$, so this holds here. It looks like Hilbert's basis theorem answers my third question also. –  Joe Mar 2 '11 at 8:46
    
If $S$ is defined as the set of all polynoms vanishing at $p$, then $S$ is a maximal ideal of $\mathbb{R}[x_1, \dots, x_n]$ and is in fact finitely generated by $x_1 - p_1, \dots, x_n - p_n$, where $p = (p_1, \dots, p_n)$. And in this case, the only zero in $\mathbb{C}^n$ is $p$ as well: this follows from $x_i - p_i \in S$ for all $i$. –  felix Mar 2 '11 at 22:17

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