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Suppose $M$ is a complex affine algebraic manifold in ${\mathbb C}^n$ (I mean, a set of common zeroes of a system of polynomials on ${\mathbb C}^n$, which is at the same time a smooth manifold). Consider the space ${\mathcal O}({\mathbb C}^n)$ of all holomoprphic functions on ${\mathbb C}^n$ with the usual topology of uniform convergence on compact sets. Alexei Pirkovskii in his paper Arens–Michael envelopes... (page 42, example 3.6) shows that

If a function $f\in {\mathcal O}({\mathbb C}^n)$ vanishes on $M$, $$ f\Big|_M=0, $$ then it can be approximated in ${\mathcal O}({\mathbb C}^n)$ by a sequence of polynomials $g_i\in {\mathbb C}[x_1,...,x_n])$ which also vanish on $M$: $$ g_i\Big|_M=0,\qquad g_i\overset{{\mathcal O}({\mathbb C}^n)}{\underset{i\to\infty}{\longrightarrow}}f $$

Pirkovskii uses Cartan's theorems for this, but I have a feeling that this is too complicated. Is it possible that there is a more or less simple way to prove this, without Cartan?

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  • $\begingroup$ An observation: Let $I$ be the ideal in $\mathcal O(\mathbb C^n)$ consisting of all $f$ such that $f|_{M}=0$. The statement seems to follow from (maybe to be equivalent to) the assertion that $I$ is generated by polynomials. $\endgroup$ Commented Aug 2, 2021 at 20:42
  • $\begingroup$ Tom, excuse me, what is meant by "generated by polynomials"? $\endgroup$ Commented Aug 2, 2021 at 21:41
  • $\begingroup$ I meant that, as an ideal in the ring of global holomorphic functions, $I$ is generated by elements of $I$ that belong to the subring consisting of polynomials. I would guess that this is true, and it would imply your statement. $\endgroup$ Commented Aug 2, 2021 at 23:22
  • $\begingroup$ Tom, $$ \forall f\in I\quad \exists p_1,...,p_k\in{\mathbb C}[x_1,...,x_n] \quad p_i\Big|_M=0\quad \&\quad \exists h_1,...,h_k\in{\mathcal O}(M)\quad f=\sum_{i=1}^k p_i\cdot h_i $$ -- if this is your point, then of course this implies what I need. But why is this true? $\endgroup$ Commented Aug 3, 2021 at 0:55
  • $\begingroup$ I don't know whether it is true, but I am guessing that it is true. $\endgroup$ Commented Aug 3, 2021 at 10:57

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