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 $e^{-1/x^2}$...

  • 2
    $\begingroup$ LaTeX tip: < and > are not the same as \langle and \rangle. $\endgroup$ Feb 1, 2013 at 7:33
  • $\begingroup$ you'r right, thanks! I overlooked it. $\endgroup$ Feb 1, 2013 at 16:59
  • $\begingroup$ It is good that Batman studies stability theory. After all, we need Gotham to be a stable city. $\endgroup$
    – user60578
    Feb 1, 2013 at 17:00
  • $\begingroup$ by the way, thanks for the edit. looks nicer now. $\endgroup$ Feb 1, 2013 at 17:09
  • $\begingroup$ @JohnJKN: hahaha got to be prepared, you never know what's out there... $\endgroup$ Feb 1, 2013 at 17:11

2 Answers 2


Finite linear combinations of polynomials are still polynomials ...

  • $\begingroup$ ok of course, thank you... but I think I didn´t express myself correctly so I edited the question to what I actually wanted to mean. $\endgroup$ Feb 1, 2013 at 17:01

Let $m$ be the ideal of all such germs, that are represented by functions $f$ such as $f(0)=0$. There is Hadamard's theorem which states that $m$ is generated by the so called coordinate functions $x_i:(t_1,t_2,\dots,t_n)\rightarrow t_i$. I think it may help.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .