# Are all ideals I in the ring of smooth functions on a compact manifold M equal to a set of smooth functions that vanish in $Z \subset M$?

Since $M$ is compact, we know that maximal ideals are $m_x$, the set of functions vanishing in $x \in M$. Thus by Zorn's Lemma we also have that $I$ must sit inside such a $m_x$ for some $x \in M$.

It seems logical that $I$ should be a $m_Z$ where $Z$ is the intersection of kernels over all functions from $I$. The set $Z$ is non-empty as a consequence of Zorn's Lemma.

Then clearly by definition of $Z$ we have $I \subset m_Z$. I just have trouble seeing that $M_Z \subset I$. Any ideas?

• As a counterexample take the ideal of all functions vanishing to order $\ge 2$ in a point. – user83633 Jan 13 '16 at 12:01

The compactness of $M$ does not guarantee that every maximal ideal is principal without additional assumptions. If $M$ has positive dimension then it contains infinitely many points and therefore there are always maximal nonprincipal ideals.

• How is this related to the question? – abx Jan 13 '16 at 14:09
• @abx, the relation is that the OP's question is based on an incorrect premise as I pointed out. – Mikhail Katz Jan 13 '16 at 14:29
• There is no mention of principal ideals in the OP question. – abx Jan 13 '16 at 14:51
• @abx, the set of functions vanishing at a point $x\in M$ is by definition a principal ideal in the ring of functions. However, there are other maximal ones. – Mikhail Katz Jan 13 '16 at 14:53
• You have a strange definition of a principal ideal! Look at "principal ideal" in Wikipedia. – abx Jan 13 '16 at 15:20

I realize that this is an old question, but it was never answered here. So I'll take a crack at it: no, not every ideal is of the form $$\frak{m}\!_Z$$.

Let $$M$$ be a compact Hausdorff manifold. Then (1) every ideal is contained in some $$\frak{m}\!_x$$, and (2) there are ideals strictly contained in a unique $$\frak{m}\!_x$$.

$$\textbf{Proof (1):}$$ Let $$I$$ be a proper ideal of $$\mathcal{C}$$, the smooth $$\mathbb{R}$$ valued functions on $$M$$. Define $$\mathcal{Z}[f]:=\{x\in M|f(x)=0\}$$, and suppose that $$\bigcap_{f\in I}\mathcal{Z}[f]=\emptyset$$. Then $$M=\bigcup_{f\in I}\mathcal{Z}[f]^c$$ and because $$M$$ is compact there is some finite collection of functions $$\{f_i\}_{i=1}^n$$ such that $$M=\bigcup_{i=1}^n\mathcal{Z}[f_i]^c$$. But that would mean that $$\bigcap_{i=1}^n\mathcal{Z}[f_i]=\emptyset$$, so $$\sum_{i=1}^nf^2_i\in I$$ would be a unit, so $$I=\mathcal{C}$$. Then it must be that, $$\exists x\in\bigcap_{f\in I}\mathcal{Z}[f]$$ and so $$I\subseteq \frak{m}\!_x$$. So, if $$I$$ is maximal then we have that $$I=\frak{m}\!_x$$.

$$\textbf{Proof (2):}$$ If we define $$\frak{o}\!_x:=\{f\in\mathcal{C}|f \text{ vanishes on a neighborhood of }x\}$$, then this is an ideal of $$\mathcal{C}$$. Clearly $$\frak{o}\!_x\subsetneq\frak{m}\!_x$$, but if we take $$y\neq x$$ there must be a neighborhood $$U$$ of $$x$$ such that $$y\notin\overline U$$ because $$M$$ is Hausdorff. Because compact and Hausdorff implies Tikhonov, we know that there must be $$f\in\mathcal{C}$$ such that $$U\subseteq\mathcal{Z}[f]$$ and $$f(y)=1$$. Then we may conclude that $$\frak{o}\!_x\not\subseteq\frak{m}\!_y$$.