# Ring of functions over M

1. May 4, 2008

### neworder1

Let $$M$$be a compact manifold and $$C(M), C^{\infty}(M)$$ denote rings of continuous (resp. smooth) real functions on $$M$$. Let m be a maximal ideal of functions vanishing at some point $$x_{0} \in M$$. Prove that $$m$$ is finitely generated over $$C^{\infty}(M)$$, but is not finitely generated over $$C(M)$$.

2. May 4, 2008

### mathwonk

isnt that what taylor polynomials are for? [n.j. hicks, notes on differential geometry, lemma, page 6.]

oops, that is just the local solution. i presume you can globalize it without too much trouble, using partition of unity.

on second i am not so clear on this. at least the ideal is generated by dimM elements in the local ring of the point.

but it is not immediately clear to me that this is even true except at the level of germs.

but compactness is a very strong property.

Last edited: May 4, 2008
3. May 4, 2008

### Hurkyl

Staff Emeritus
neworder1: What ideas have you had on this problem? What have you tried? Have you considered any simpler problems to search for ideas?

4. May 4, 2008

### neworder1

Well, I tried to solve the problem locally first. My guess was that you can approximate a function by some polynomials, perhaps, since it's smooth, something involving its derivatives (i.e. the ring is generated by some simple functions, and the coefficients are something like derivatives), and then somehow you can choose a finite set of these polynomials, but that's all. My other approach was to try to find a similar generating set for smooth vector fields (i.e. every smooth function defines a gradient vector field, and with some assumptions you can recover a function from its gradient field, so if you could finitely generate vector fields...), but I don't think that's much simpler.

5. May 5, 2008

### Doodle Bob

Since $C^\infty(M)\subset C(M)$, it would seem to me that, if m is finitely generated over $C^\infty(M)$, then it is automatically finitely generated over $C(M)$. Otherwise I am misunderstanding the problem.

6. May 5, 2008

### Hurkyl

Staff Emeritus
Just to be clear -- you're currently stuck at proving it in the local case, correct? (As opposed to being stuck at passing from local to global)

Well, try considering the case where we are only interested in polynomial functions on some coordinate chart -- IMHO that case should be fairly easy and quite suggestive. (If you have trouble seeing it, look at the one-dimensional case of polynomials over R that vanish at 0)

That you can do this locally is almost built into the definition of a (finite-dimensional) vector bundle -- there is be an obvious spanning set (in fact, a basis!) for the module of vector fields on n-dimensional real space. Remember that the coefficients can be functions; you're not limited to constants! (coefficients in $C(\mathbb{R}^n)$, $C^{\infty}(\mathbb{R}^n)$, or other coefficient ring, depending on how smooth the vector fields are supposed to be)

7. May 5, 2008

### Hurkyl

Staff Emeritus
The different problems have different m's. In particular, if I let m and $\mathfrak{m}_\infty$ be the maximal ideals of $C(M)$ and $C^\infty(M)$ respectively of functions vanishing at a particular point P, then:

$$\mathfrak{m} \neq \mathfrak{m}_\infty \cdot C(M)$$

(the right hand side is, of course, a subset of m)

Last edited: May 5, 2008
8. May 5, 2008

### mathwonk

i knew hurkyl was the man who would take this **** seriously.

just kidding!!!

9. May 8, 2008

### neworder1

I think this makes sense:

I cover $$M$$ with finite number of charts $$U_{k}$$ with maps $$\phi_{k}$$. In $$\mathbb{R}^{n}$$, functions $$f_{i}=x_{i}-x_{0i}$$ generate my ideal (from Taylor series expansion, right?), so in $$U_{n}$$ functions $$\phi_{k}f_{i}$$ generate my ideal locally. So I have a finite family of generators $$f_{i,k}$$ and I want to make it global, so I take a partition of unity $$g_{k}$$ and put $$f_{n}=\sum{g_{i}f_{i,n}$$. Is this correct?