Do the polynomial germs generate all the ring of germs?

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I'm trying to understand some equality that comes up in stability theory involving sets of germs and I think I need a result like the next one, so if anyone knows anything about this and helps me it would be wonderful!

Let $C^{\infty}_0(\mathbb{R}^n)$ be the ring of germs at $0\in \mathbb{R}^n$ of smooth maps from $\mathbb{R}^n \to \mathbb{R}$. We write $[f]\in C^{\infty}_0$ to denote the germ of a smooth function $f \colon \mathbb{R}^n \to \mathbb{R}$, and define the set $$\mathrm{Pol}_k(x_1, \cdots,x_n)=\{x_1^{i_1} x_2^{i_2}\dots x_n^{i_n}\in K[X]\;| \;i_1+i_2+\dots+ i_n=k \},$$ where $K[X]$ is the polynomial ring in $n$ variables. Now, we can think of $C^{\infty}_0(\mathbb{R}^n)$ as a module over itself so the question is, Does the class of all polynomials generate $C^{\infty}_0(\mathbb{R}^n)$? Specifically, is this true?

$$C^{\infty}_0(\mathbb{R}^n) = \langle 1,[\mathrm{Pol}_1(x_1,\cdots,x_n)],[\mathrm{Pol}_2(x_1,\cdots,x_n)],\cdots\rangle_{\mathbb{R}}$$

where by $\langle [f_1],\cdots,[f_n]\rangle_{\mathbb{R}} \subset C^{\infty}_0(\mathbb{R}^n)$ we mean the $\mathbb{R}$-submodule generated by $[f_1],\cdots,[f_n]\in C^{\infty}_0(\mathbb{R}^n)$.

--I'm sorry I don't think I expressed correctly what I had in mind. Actually what I was trying to ask is something like this:

If we denote by $\mathfrak{m}(n)$ the maximal ideal in $C^{\infty}_0(\mathbb{R}^n)$ consisting of elements in $[f]\in C^{\infty}_0(\mathbb{R}^n)$ such that their representatives have $f(0)=0$. Is this equality (or a similar result) correct? $$C^{\infty}_0(\mathbb{R}^n) = \langle 1,[\mathrm{Pol}_1(x_1,\cdots,x_n)],\cdots,[\mathrm{Pol}_n(x_1,\cdots,x_n)]\rangle_{\mathbb{R}} +\mathfrak{m}(n)^n$$ It looks to me that it could be right. I mean at least if $f$, a representative of $[f]\in C^{\infty}_0(\mathbb{R}^n)$, is equal to it's Taylor series it looks like true, but then again there are functions like $e^{-1/x^2}$...

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