5

For any smooth manifold $M$, the collection $F = C^\infty(M, \mathbb{R})$ of smooth real valued functions on $M$ can be made into a ring, and every point $x \in M$ determines a ring homomorphism $F \to \mathbb{R}$ in $F$. If $M$ is compact, how do I see that every maximal ideal in $F$ arises in this way from a point of $M$?

Alan
  • 16,582
Student
  • 1,032

2 Answers2

6

Let $L$ be a maximal ideal, suppose that $\bigcap_{f\in L}\{x:f(x)=0\}=\phi$. Then for every $x\in M$, there exists $f_x\in L$ such that $f_x(x)\neq 0$. There exists a neighborhood $U_x$ of $x$ such that the restriction of $f_x$ to $U_x$ does not vanish, we suppose that $f_x>0$ on $U_x$ and by multiplying $f_x$ by a cut off function, you obtain $g_x$ whose support is in $U_x$ and the restriction of $g_x$ to a neighborhood $V_x\subset U_x$ of $x$ is strictly positive. Since $M$ is compact, you can cover $M$ with a finite number of $V_{x_i}$ and $\sum_i g_{x_i}$ does not vanish. thus it is invertible and $L$ contains a unit, contradiction. So there exists $x$ such that for every $f\in L$, $f(x)=0$, so $L$ is contained in the maximal ideal $L_x=\{f:f(x)=0\}$, since $L$ is maximal, $L=L_x$.

Michael Hardy
  • 1
Tsemo Aristide
  • 87,475
  • Is it just me, or is the opposite trivially true? "For every point $x\in M$ there a maximal ideal $L_x={f\in F|f(x)=0}$." I guess you are't actually sure that ideal is maximal? Hence the need to show anything not in $L_x$ is strictly non-vanishing when $f(x)=0$. – levitopher Nov 13 '16 at 02:17
0

There is actually a very easy proof that works for this ring, and also if you relax the ring to just be the ring of continuous functions.

Denote the set of elements that $f$ maps to zero as $z(f)$.

Lemma: For a proper ideal $I$, if $f,g\in I$, then $z(f)\cap z(g)\neq \emptyset$. By induction, the same can be said for a finite number of functions in $I$.

Lemma: A space is compact iff, given any family $F$ of closed subsets with the property that all finite intersections of members of $F$ are nonempty, the intersection of all members of $F$ is nonempty.

By the first lemma, you can see the family of zero sets for elements in $I$ form a collection which the compactness criterion in the second lemma applies to. Therefore the intersection of the zero sets of all elements in $I$ is nonempty. By maximality, $I$ is therefore a set of functions which are zero at a single point.

You can find more details here where I gave the solution for a less general question.

Community
  • 1
rschwieb
  • 153,510