Properties of the ring of smooth function germs, question on proof.

Question

Let us denote by $C_n$ the ring of $C^{\infty}$ smooth function germs $f : (\mathbb R^n, 0) \to \mathbb R$ or the ring of analytic functions germs $f : (\mathbb C^n, 0) \to \mathbb C$. Denote by $\mathbb F$ either $\mathbb R$ or $\mathbb C$ in each case. Then the following holds:

Proposition:

(i) $C_n$ is a local ring with maxmial ideal $$ M_n := \{ f \in C_n : f(0) = 0 \}. $$

(ii) $M_n^k = ( x^{\alpha} = \prod_{i=1}^n x_i^{\alpha_i} : |\alpha| \ge k )$ and we have that $$ M_n^k = \left\{ f \in C_n : \frac{\partial^{|\alpha|} f}{\partial x^{\alpha}}(0) = 0, ~ \forall |\alpha| = 0,1,\ldots,k-1 \right\}. $$

(iii) For an ideal $I$ in $C_n$ we have: $$ \mbox{dim}_{\mathbb F} C_n/I < \infty \quad\mbox{iff}\quad \exists k < \infty \mbox{ such that }~ M_n^k \subseteq I. $$ The proof of (iii) relies on the two facts that

1) $\mbox{dim}_{\mathbb F} C_n / M_n^k < \infty$

and

2) the infinite chain $C_n \supseteq I + M_n \supseteq I + M_n^2 \supseteq \ldots$ has just a finite number of strict inclusions.

Why does both properties hold? I am quite new to this field, so this might be obvious, but I do not see it... so thanks for your help!

Answer 1

The quotient $C_n /M_n^k$ is finite-dimensional since every $f\in C_n$ is congruent to its Taylor polynomial of order $k-1$ modulo $M_n^k$. The space of polynomials (in $n$ indeterminates) of degree $< k$ is finite-dimensional.

If $C_n/I$ is finite-dimensional, only finitely many of the inclusions $I + M_n^k \supset I + M_n^{k+1}$ can be strict, since we always have

$$I \subset I + M_n^k \subset C_n,$$

and therefore

$$\dim_{\mathbb{F}} \bigl(C_n/(I+M_n^k)\bigr) = \dim_{\mathbb{F}}(C_n/I) - \dim_{\mathbb{F}}\bigl((I+M_n^k)/I\bigr) \leqslant \dim_{\mathbb{F}} (C_n/I).$$

The dimensions of $C_n/(I+M_n^k)$ are bounded independently of $k$, and each strict inclusion in the chain corresponds to a strict increas in the dimension of the quotient.

Answer 2

Mapping $f$ to the tuple of all its partial derivatives with $|\alpha|<k$ is a linear map from $C_n$ to a finite dimensional space with kernel $M_n^k$. Hence $\dim_F C_n/M_n^k<\infty$.

As we mod out less and less, we have $$\dim C/(I+M_n)\le \dim C/(I+M_n^2)\le\dim C_n/(I+M_n^2)\le\ldots\le \dim C_n/I$$ Hence if the last is finite, the sequence of the $\dim C/(I+M_n^k)$ must be eventually constant. Then $I+M_n^k=I+M_n^{k_0}$ for all $k\ge k_0$. Conclude $I+M_n^{k_0}\subseteq \bigcup (I+M_n^k)=I+\bigcup M_n^k=I$. The other dirction is clear: If $M_n^{k_0}\subseteq I$ for some $k_0$, then the sequence of $I+M_n^k$ becomes stationary at $k=k_0$.