I recently proved the following statement:

Let $M$ be a smooth manifold and let $I \subseteq C^\infty(M)$ be an ideal such that $C^\infty(M)/I \cong \mathbb{R}$ (such an ideal is clearly maximal, since $\mathbb{R}$ is a field). Then $I= \mathfrak{m}_p$ for some $p \in M$, where $\mathfrak{m}_p= \left\{f \in C^\infty(M): f(p)=0 \right\}$.

This is fairly easy to prove for compact $M$. The case for non-compact $M$ requires a bit more work. Now, take $M=\mathbb{R}$ and $I$ the ideal of functions with compact support. $I \nsubseteq \mathfrak{m}_p$ for any $p \in \mathbb{R}$, but there must exist a maximal ideal $I' \subseteq C^\infty(\mathbb{R})$ containing $I$. Clearly $I' \neq \mathfrak{m}_p$, so if $C^\infty(\mathbb{R})/I' \cong \mathbb{F}$ for a field $\mathbb{F}$, which must be the case since $I'$ is maximal, $\mathbb{F} \neq \mathbb{R}$.

I have two questions: Does anyone have an explicit example of such an $I'$? Also, does anyone know what $\mathbb{F}$ is or could be?

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    $\begingroup$ I don't know if this helps, but $I\subset \mathfrak m_\infty$, where $\infty$ is the point at infinity of Alexandrov's compactification. $\endgroup$ May 23, 2016 at 15:55
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    $\begingroup$ Actually there is no such $m_\infty$; the functions in $C^\infty(M)$ do not extend to the one-point compactification. $\endgroup$ May 23, 2016 at 16:00
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    $\begingroup$ (And the functions in $C^\infty(\Bbb R)$ that do happen to tend to zero at infinity do not form an ideal.) I seriously doubt that an explicit description of $I'$ or $\Bbb F$ exists; I suspect they are indescribably axiom-of-choice-ish things. $\endgroup$ May 23, 2016 at 16:06

1 Answer 1


Here is an "explicit" example. Let $U$ be a nonprincipal ultrafilter on $\mathbb{N}$. Define $I'$ to be the set of $f\in C^\infty(\mathbb{R})$ such that $\{n\in\mathbb{N}:f(n)=0\}\in U$. It is easy to see that $I'$ is an ideal, and it contains $I$ since $U$ is nonprincipal. To show $I'$ is maximal, note that if $f\not\in I'$, then $\{n\in\mathbb{N}:f(n)\neq 0\}\in U$ since $U$ is an ultrafilter. We can find a smooth function $g$ such that $g(n)=1/f(n)$ for each $n\in\mathbb{N}$ such that $f(n)\neq 0$, and then $fg-1\in I$. Thus $f$ is a unit mod $I'$, so the quotient $C^\infty(\mathbb{R})/I'$ is a field.

To understand this quotient field $C^\infty(\mathbb{R})/I'$, note that any sequence of real numbers can be realized as the values $f(n)$ for $n\in\mathbb{N}$ for some smooth function $f$. Moreover, two functions which agree on $\mathbb{N}$ are equal mod $I$. So in taking the quotient, we can replace smooth functions by just their sequences of values on $\mathbb{N}$, and we find that $C^\infty(\mathbb{R})/I'$ is the ring of sequences $(x_n)$ of real numbers mod the ideal of sequences such that $\{n:x_n=0\}\in U$. This field is better-known as the ultrapower of $\mathbb{R}$ by the ultrafilter $U$, sometimes written $\prod_U\mathbb{R}$. This field is also sometimes called a field of "hyperreal numbers" and is studied in nonstandard analysis as a field with very deep similarities to $\mathbb{R}$ but which has infinitesimal elements.

Of course, all of this is only as "explicit" as our choice of nonprincipal ultrafilter $U$ is, and there's not really any way to explicitly write down a nonprincipal ultrafilter.

  • $\begingroup$ Thanks for your answer, after I asked this question I read a few papers relating maximal ideals to something called Z-ultrafilters, a kind of ultrafilter on the closed sets of a topological space (actually, this was for ideals in $C(X)$, but I believe things carry over nicely to $C^\infty$). I had never seen the construction from an ultrafilter on $\mathbb{N}$ before though, very nice. $\endgroup$
    – M10687
    Jul 10, 2017 at 19:35

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