First of all, this is a homework problem.
Let $C^{\infty}(\mathbb{R})$ denote the ring of smooth functions. Let $I_n$ denote the set of $f\in C^{\infty}(\mathbb{R})$ such that $$f^{(k)}(0)=0, \ 0 \leq k \leq n$$ where $f^{(k)}$ denotes the $k^{th}$ derivative. I need to show:
$1)$ $I_n$ is an ideal of $C^{\infty}(\mathbb{R})$. I showed $I_n$ is an additive subgroup. But I could not show the second axiom, namely: for $f\in I_n, \ g\in C^{\infty}(\mathbb{R})$, why is $fg \in I_n$. I have to show that $$(fg)^{(k)}(0)=0, \ 0 \leq k \leq n.$$ But I could not complete the proof.
$2)$ If $n > m,$ then $I_n \subsetneq I_m$. I think I have to find an example of $f \in C^{\infty}(\mathbb{R})$ such that $$f^{(k)}(0)=0, \ 0 \leq k \leq m \mbox{ but } f^{(m+1)}(0)\neq 0.$$ I couldn't think of an example.
$3)$ Show that $C^{\infty}(\mathbb{R})/I_0$ is isomorphic to $\mathbb{R}$. I defined a map $$\phi: C^{\infty}(\mathbb{R})/I_0 \to \mathbb{R}, \ f+I_0 \mapsto f(0).$$ I showed well-defined, homomorphic and bijective properties. Is this a correct isomorphism?
$4)$ Let $i:\mathbb{R}[x] \to C^{\infty}(\mathbb{R})$ be an inclusion map. Let $\pi: C^{\infty}(\mathbb{R}) \to C^{\infty}(\mathbb{R})/I_n$ be the quotient map. Show that composed map $\pi \circ i$ is surjective and determine its kernel. I could not prove surjectivity. Kernel is simply those $p(x)$ such that $p(x)\in I_n \Rightarrow p(x)=a_{n+1}x^{n+1}+...$ Is this correct?
Sorry for asking many questions here. Thanks in advance for your help!
These are all hints, but you should be able to turn them into a complete solution.
1) Use the generalized product rule, which you can prove by induction: $$(fg)^{(k)}(x) = \sum_{n=0}^k f^{(n)}(x)g^{(k-n)}(x).$$ 2) Think about polynomials.
3) Yes. (If you showed that it's a bijective homomorphism, then you shouldn't need to ask us if it's an isomorphism!)
4) What do elements in $C^\infty(\Bbb R)/I_n$ look like? We're killing off functions whose derivatives past $n$ are zero - so pick an element in each equivalence class whose $(n+1)$th derivatives and onwards are zero. Polynomials seem to fit the bill...