Let $\Omega$ be a connected open set in the complex plane. What is the closure of the polynomials in $\mathcal{H}(\Omega)$ the set of holomorphic functions on $\Omega$? The topology is the usual compact convergence topology. Take, for instance, an annulus such as $D(r,R)$, the set of all complex $z$ such that $r<|z|< R$, you cannot recover the function $z\mapsto \frac{1}{z}$ because of the residue at $0$, so what holomorphic functions are limits of polynomials?
[Math] What holomorphic functions are limits of polynomials
cv.complex-variablesfa.functional-analysis
Related Solutions
We have that there are no non-constant bounded functions on $\mathbb C^*=\mathbb C\setminus\{0\}$. The easiest way to see that is to notice that such a function has a removable singularity at the origin and hence comes from a bounded function on $\mathbb C$ (which incidentally has a removable singularity at $\infty$ and hence extends to the Riemann sphere and therefore is constant).
As for the higher-dimensional problem it is no doubt hopeless to get anything like a complete description: There are as you say domains in $\mathbb C^n$ (and there are many of those), the compact manifolds but also compact manifolds minus a codimension $2$ closed analytic subspace, the blowing up of some space that has the property, any product of two manifolds with the property and so on. Complex manifolds in higher dimension are simply too varied.
Finally, there are non-compact manifolds with only constant holomorphic functions. If $L\rightarrow X$ is an analytic line bundle over a complex manifold $X$ and $f\colon L\rightarrow\mathbb C$ is a holomorphic function, then we may Taylorexpand $f$ along the zero section of $L$: First we just look at the restriction of $f$ to the zero section which gives a function on $X$. Then we may take any local section of $L$, think of that as a tangent vector at the zero section and take the derivative of $f$ along this tangent vector. This glues together to give the first derivative as a global section of $L^{-1}$ and similarly the $n$'th derivative of $f$ along the zero section will be a section of $L^{-n}$. If $X$ is compact and $L^{-n}$ for $n>0$ only has the zero section as global section, then $f$ is constant along the zero section and all its higher derivatives along it are zero so that $f$ is constant in a neighbourhood of the zero section and hence constant. There are lots of such examples $(X,L)$.
$\def\CC{\mathbb{C}}\def\cO{\mathcal{O}}$
Here is a candidate counterexample for $M= \CC$: Is $e^{-z}$ in the closure of the algebra generated by $e^z$ and $e^{\sqrt{2}z}$? My current guess is "no", but I need to move on to actual work.
I will show that separating points and separating tangents is not enough for $M = \CC^2$.
Let $A \subset \cO(\CC^2)$ be those holomorphic functions $f$ such that $f(z,z^{-1})$ extends holomorphically to $z=0$. We observe:
$A$ is a subalgebra: This is obvious.
$A$ is closed: Proof We have $f \in A$ if and only if $\oint f(z,z^{-1}) z^n dz=0$ for all $n \geq 0$, where the integral is on a circle around $0$. This fact is preserved by uniform limits on compact sets. (Specifically, by uniform limits on that circle.)
$A$ separates points: Note that the functions $f(x,y) = x$, $g(x,y) = xy$ and $h(x,y) = y (xy-1)$ are all in $A$. The functions $f$ and $g$ alone separate $(x_1, y_1)$ and $(x_2, y_2)$ unless $x_1=x_2=0$. In that case, $h$ separates them.
$A$ separates tangent vectors: Again, $df$ and $dg$ are linearly independent at all points where $x \neq 0$, and $df$ and $dh$ are linearly independent at $x=0$.
$A \neq \cO(\CC^2)$ Clearly, $y \not \in A$.
I remembered this counterexample from an old blog post of mine.
Note that we could replace $z \mapsto (z, z^{-1})$ with any map $\phi$ from the punctured disc $D^{\ast}$ to $\CC^2$. There are many such $\phi$'s, and they all appear to impose independent conditions. This makes me pessimistic about any simple criterion for equality when $M = \CC^2$.
Best Answer
Let $\Sigma\supset\Omega$ be the union of $\Omega$ and all bounded components of ${\mathbb C}\setminus \Omega$. The algebra you get is the algebra of all holomorphic functions on $\Sigma$.
First, every $f\in{\cal H}(\Sigma)$ is a locally uniform limit of polynomials as a consequence of Runge's Theorem, see Corollary 1.15 in John Conway's "Functions of one complex variable I", 2nd Ed.
Second, if $p_n$ is a sequence of polynomials converging locally uniformly on $\Omega$ and if $K$ is a bounded component of ${\mathbb C}\setminus \Omega$, then $p_n$ also converges uniformly on a neighborhood of $K$ by Cauchy's integral Theorem, as there exists a path around $K$ with winding number 1. (Take the positively oriented boundary of an $\varepsilon$-neighborhood of $K$, where $\varepsilon$ is chosen so small that it does not hit any other components.)