Let $V$ be an affine algebraic variety in ${\mathbb C}^n$, i.e. a set of common zeroes of some set $S$ of polynomials on ${\mathbb C}^n$: $$ V=\{z\in {\mathbb C}^n:\ \forall p\in S\quad p(z)=0\}. $$ Let $I$ be the ideal in the algebra ${\mathcal P}({\mathbb C}^n)$ of polynomials on ${\mathbb C}^n$ consisting of all polynomials that vanish on $V$: $$ I=\{f\in {\mathcal P}({\mathbb C}^n):\ \forall z\in V\quad f(z)=0\}. $$ By Hilbert's basis theorem, $I$ is generated by a finite set $q_1,...,q_k$ of polynomials in ${\mathcal P}({\mathbb C}^n)$, so we can think that $$ V=\{z\in {\mathbb C}^n:\ q_1(z)=...=q_k(z)=0\}, $$ and
each polynomial $f\in {\mathcal P}({\mathbb C}^n)$ that vanishes on $V$, $$ \forall z\in V\quad f(z)=0, $$ has the form $$ f(z)=f_1(z)\cdot q_1(z)+...+f_k(z)\cdot q_k(z) $$ for some $f_1,...,f_k\in {\mathcal P}({\mathbb C}^n)$.
A.Yu.Pirkovskii in his work of 2008 (Example 3.6, p.42) uses sheaves, the analitization functor and the Cartan theorem B for proving a similar statement for holomorphic functions:
each holomorphic function $f\in {\mathcal O}({\mathbb C}^n)$ that vanishes on $V$, $$ \forall z\in V\quad f(z)=0, $$ can be approximated in ${\mathcal O}({\mathbb C}^n)$ by the functions of the form $$ f_1(z)\cdot q_1(z)+...+f_k(z)\cdot q_k(z) $$ where $q_1,...,q_k$ are still the same polynomials in ${\mathcal P}({\mathbb C}^n)$ that generate $I$ (and they don't change), and $f_1,...,f_k$ are arbitrary holomorphic functions from ${\mathcal O}({\mathbb C}^n)$ (or polynomials from ${\mathcal P}({\mathbb C}^n)$, this is not important since ${\mathcal P}({\mathbb C}^n)$ is dense in ${\mathcal O}({\mathbb C}^n)$), and ${\mathcal O}({\mathbb C}^n)$ is endowed with the usual topology of uniform convergence on compact sets in ${\mathbb C}^n$.
I have a feeling that using analytization and the Cartan theorem, is extra in this situation, and one can prove this in a simpler way. Am I right? Can this have a more or less simple explanation?
