Ideals of the ring of smooth functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T07:23:17Z http://mathoverflow.net/feeds/question/77273 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77273/ideals-of-the-ring-of-smooth-functions Ideals of the ring of smooth functions balan-antoine 2011-10-05T18:19:05Z 2013-05-04T16:25:42Z <p>The ring $C^\infty(M)$ of smooth functions on a smooth manifold $M$ is a topological ring with respect to the Whitney topology and the usual ring operations. Is it possible to describe, maybe under some conditions on $M$, the ideals and the closed ideals of $C^\infty(M)$?</p> http://mathoverflow.net/questions/77273/ideals-of-the-ring-of-smooth-functions/77314#77314 Answer by Neil Strickland for Ideals of the ring of smooth functions Neil Strickland 2011-10-05T23:58:32Z 2011-10-06T13:41:50Z <p>Here are some examples, for the case $M=\mathbb{R}$.</p> <p>For each $n\geq 0$ we have a closed ideal $$I_n=\{f: f^{(i)}(0)=0 \text{ for } 0\leq i &lt; n\}$$ We can write $I_\infty$ for the intersection of these, which is again closed. We can also put $$J = \{ f : f(x)=0 \text{ for all } x \leq 0\}$$ and note that this is closed and contained in $I_\infty$.</p> <p>Next, for $n,a>0$ with $n\in\mathbb{Z}$ we can let $K_{n,a}$ be the principal ideal generated by the function $\exp(-a/x^{2n})$. These are all different and contained in $I_\infty$. I am not sure whether they are closed.</p> <p>For another kind of example, let $\mathcal{U}$ be a free ultrafilter on $\mathbb{R}$ and put $$L = \{f : f^{-1}\{0\} \in \mathcal{U} \}.$$ This is a non-closed maximal ideal.</p> <p>UPDATE:</p> <p>Now let $A$ be an arbitrary closed ideal in $C^\infty(\mathbb{R})$. Put $$X_n = \{ x\in\mathbb{R} : f^{(i)}(x)=0 \text{ for all } i \leq n \text{ and } f\in A\}.$$ Specialising Reimundo's answer to the case $M=\mathbb{R}$, we see that $$A = \{ f : f^{(i)}=0 \text{ on } X_n \text{ for all } i\leq n \}.$$ The sets $X_n$ are closed, with $X_n\supseteq X_{n+1}$. Moreover, if $x$ is a non-isolated point of $X_n$ (so it is in the closure of $X_n\setminus\{x\}$) then it is easy to see that $x\in X_{n+1}$. I would guess that the closed ideals biject with chains of subsets with these properties, but I have not tried to prove that.</p> http://mathoverflow.net/questions/77273/ideals-of-the-ring-of-smooth-functions/77317#77317 Answer by Reimundo Heluani for Ideals of the ring of smooth functions Reimundo Heluani 2011-10-06T01:06:34Z 2011-10-06T02:50:39Z <p>In 1948 Whitney proved his ideal (spectral) theorem [1] describing the closed ideals</p> <p>Let $M$ be an $n$-dimensional manifold. for each point $p \in M$ and each natural $k$ we define $N(k)$ to be the number of (up to $n$) tuples $m$ such that $|m| \leq k$. Define the map $J_p^k: C^\infty(M) \rightarrow \mathbb{R}^{N(k)}$ by assigning to $f$ the $m$-jets of $f$ at $p$ up to $|m|=k$. </p> <p>If $I$ is an ideal of $C^\infty(M)$ then its closure is the ideal of functions $f$ such that for each $p$ in $M$ and $k \geq 0$ then $J^k_p f \in J^k_p(I)$. </p> <p>So in some sense the closed ideals are like $I_\infty$ in Neil's answer. </p> <p>[1] H.Whitney. On ideals of differentiable functions. American Journal of Mathematics. Vol. 70, No. 3, pp. 635-658 (1948)</p>