Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
LaTeX tip: < and > are not the same as \langle and \rangle. – Zev Chonoles Feb 1 '13 at 7:33
you'r right, thanks! I overlooked it. – Bruce Wayne Feb 1 '13 at 16:59
It is good that Batman studies stability theory. After all, we need Gotham to be a stable city. – user60578 Feb 1 '13 at 17:00
by the way, thanks for the edit. looks nicer now. – Bruce Wayne Feb 1 '13 at 17:09
@JohnJKN: hahaha got to be prepared, you never know what's out there... – Bruce Wayne Feb 1 '13 at 17:11

Finite linear combinations of polynomials are still polynomials ...

share|cite|improve this answer
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. – Bruce Wayne Feb 1 '13 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.