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Denote the ring $\text{C}^{\infty}$ i.e. infinitely differentiable functions from $\mathbb{R}\mapsto\mathbb{R}$.

I have managed to prove that this ring is a Principal Ideal Domain.

I must also prove that $f(x)=e^{-x^{-2}}$ when $x\neq 0$ and $f(0)=0$ is in $\text{C}^{\infty}$.

It is clear that $0\in\text{C}^{\infty}$ and for $e^{-x^{-2}}$, I replaced the Taylor Series for $e^x$ and got terms which lie in $\text{C}^\infty$ when $x\neq 0$.

Now I must prove the ring is NOT Noetherian - So consider the set $I_n=\{f(x):f(x)=0, \forall x\geq n\}$. It follows on that the ascending chain of ideals in NOT stationary, and thus the ring is NOT Noetherian.

Finally, I must produce a homomorphism from $\text{C}^\infty\mapsto\mathbb{R}[x]$ using Taylor Series. So,

$$\phi:\text{C}^\infty\mapsto\mathbb{R}[x]$$ $$\phi(f)=\sum_{i=0}^{\infty}\frac{f^{(n)}(0)\ x^n}{n!}$$

For the case of $e^{-x^{-2}}$, we will end up with a Taylor Series of the form

$$e^{-x_0^{-2}}\sum g_i(x)$$

(due to the nature of the exponential function) where $x_0=0$, and thus the Taylor Series is $0$ everywhere (since $f(0)=0$, as defined before). Therefore $e^{-x^{-2}}\in\text{ker}(\phi)$.

Is everything here correct? Or have I made an error somewhere?

Naji
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    Your ring isn't even a domain (much less a principal ideal domain), as it has zero divisors. There's a function in it that's zero for all nonnegative $x$, and a function that's zero for all nonpositive $x$, and the product of those two functions is identically zero. – Gerry Myerson Mar 23 '15 at 09:54
  • Very good point actually – Naji Mar 23 '15 at 18:39

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The proof that the function $$ f(x)=\begin{cases} 0 & \text{if $x=0$}\\ \exp(-1/x^2) & \text{if $x\ne0$} \end{cases} $$ belongs to $C^{\infty}$ is completely wrong. It's true that $f$ is infinitely differentiable at every point $\ne0$, but the difficult part is exactly showing that it is also infinitely differentiable at $0$. You find a full proof in every textbook (and also on this site, I believe).

Also the fact that $C^{\infty}$ is not noetherian is not sufficiently justified. You have to prove that $I_n$ is a proper subset of $I_{n+1}$; not difficult, though.

You get a homomorphism $C^\infty\to\mathbb{R}[[x]]$ (the ring of formal power series) by considering the Taylor series expansion at $0$.

The image of $f$ (defined above) under this homomorphism is the zero series. Why? When you have completed the proof that $f\in C^{\infty}$ you'll see.

Phrases like “due to the nature of the exponential function” are not allowed in a proof. Besides, if $x_0=0$, $e^{-1/x_0^2}$ means nothing, so you can't use it.

Finally, a principal ideal domain is noetherian. So you can't have possibly proved it is a PID and that it is not noetherian.

egreg
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  • Sorry, what I meant was the ideal $I$ of functions vanishing at particular roots are principal - I didn't mean PID

    I will now look for a proof of an infinitely differentiable function

     – Naji Mar 22 '15 at 22:55
  • @Naji Your statement is false. The ideal of functions vanishing at $0$ is not principal. – egreg Mar 22 '15 at 22:59
  • But if a function f(x) vanishes at, say, $\alpha$, then f(x) multiplied by any function (in $C^\infty$) would also be $0$ at $\alpha$? – Naji Mar 22 '15 at 23:06
  • And with regards to the proof of $e^{-x^{-2}}$ being infinitely differentiable, I assume the proof would be similar to here

    http://math.stackexchange.com/questions/476195/infinitely-differentiable-functions-how-to-prove-that-e-frac1x2-1-has-d

     – Naji Mar 22 '15 at 23:07
  • @Naji See http://math.stackexchange.com/questions/119858/infinitely-differentiable-function – egreg Mar 22 '15 at 23:10