Ring of germs of smooth functions on $\mathbb{R}^{n}$ in $0$.

Question

First of all, I'm quite new to this theory, so it may be very dumb questions. Sorry for that.

Let $R$ be the ring of germs of $C^{\infty}$ functions on $\mathbb{R}^{n}$ in $0$. Let $K$ be the intersection of all powers of maximal ideal. I would like to prove that $K$ is generated by all nonnegative $\varphi\in R$.

I've been told that there exists only one maximal ideal:

$$M=\{f\in R\mid f(0)=0\}\tag{1}$$

It is easy, I think, to prove it is an ideal. However, I've no clue how to prove it is maximal and it is the only maximal ideal... Next, it is said that

$$M^{k}=\left\{f\in R\,\Big\vert\, \frac{\partial^{\vert\alpha\vert}f}{\partial x^{\alpha}}(0)=0,\,\forall\alpha\in\mathbb{N}^{n}:\vert\alpha\vert=0,1,\dots,k-1\right\}\tag{2}$$

How can one find that? I mean, the definition of the $k$-th power of an ideal is the ideal generated by finite sums of product of its elements. For example, $M^{2}=\langle \{f\cdot g\mid f,g\in M\}\rangle$, no? I do not understand how one can get $(2)$ with such a definition. I do understand that

$$f\in M^{k}\implies\frac{\partial^{\alpha}f}{\partial x^{\alpha}}(0)=0,\,\forall\alpha\in\mathbb{N}^{n}:\vert\alpha\vert=0,1,\dots,k-1$$

since it follows from the Leibniz product rule. But I don't see how we get the other direction. I'm kind of lost here and far from the result I want to show...

Answer 1

To your first question:

Given an ideal $I \subset R$ and the desire to proof that $I$ is the only maximal ideal of $R$, you just have to show that $R \setminus I$ contains only units. Because this shows, that any proper ideal is contained in $I$, hence $I$ is the unique maximal ideal.

In our situation, this is easy to show: If $f \notin M$, we have $f(0) \neq 0$, hence $f(x) \neq 0$ in some neighborhood of $0$ by continuity. Hence $1/f$ is well defined on this neighborhood (and by some calculus, again smooth) and this shows that $f$ is invertible in $R$.

To your second question: The fact, that any $f$, whose partial derivatives vanish up to order $k-1$, is contained in $M^k$, is precisely the Taylor formular.

Answer 2

Note that if $f \in R$ with $f(0) \neq 0$ then $f$ is invertible in $R$ (as $\frac{1}{f} \in R$). If an ideal $I$ of $R$ contains a function $f$ with $f(0) \neq 0$ then it contains $\frac{1}{f} \cdot f = 1$ and so we must have $I = R$. This shows that $M$ is maximal and that any proper ideal of $R$ is contained in $M$ so $M$ is the only maximal ideal of $R$.

For the interesting direction regarding $M^k$, you need to show Hadamard's lemma. Using Hadamard's lemma, any smooth germ can be written as $f(x) = f(0) + \sum_{i=1}^n x^i g_i(x)$ where $g_i$ are smooth. If $f(0) = 0$ and $\frac{\partial f}{\partial x^i}(0) = 0$ then using Leibniz rule you can check that $g_i(0) = 0$ and so $f \in M^2$. By induction, you can similarly show the characterization of $M^k$.