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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?

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

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  • $\begingroup$ How is this related to the question? $\endgroup$ – abx Jan 13 '16 at 14:09
  • $\begingroup$ @abx, the relation is that the OP's question is based on an incorrect premise as I pointed out. $\endgroup$ – Mikhail Katz Jan 13 '16 at 14:29
  • $\begingroup$ There is no mention of principal ideals in the OP question. $\endgroup$ – abx Jan 13 '16 at 14:51
  • $\begingroup$ @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. $\endgroup$ – Mikhail Katz Jan 13 '16 at 14:53
  • $\begingroup$ You have a strange definition of a principal ideal! Look at "principal ideal" in Wikipedia. $\endgroup$ – abx Jan 13 '16 at 15:20
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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$.

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