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I was reading Jean Dieudonné's nice counterexample about permutations of regular sequences (see here), then the following question came to my mind:

What is the difference between the ring of germs of $C^\infty$ functions at $0\in\mathbb R$ (with values in $\mathbb R$), and the ring of germs of infinitely differentiable functions at $0$?

Note that the former ring is strictly contained in the latter: in fact, the elements of the former ring are (equivalence classes of) functions $f\in C^\infty(V)$, where $V$ is an open interval containing $0$. On the other hand, using the functions $x^n\sin(1/x)$, which belong to $C^{n-1}(\mathbb R)$ but don't have $n$-th derivative at $0$, and patching adequately (I know I am capable to do this, but I am lazy now), we can construct a function $g:\mathbb R\to\mathbb R$ such that $g^{(n)}(0)$ exists for all $n$, yet $g$ is not smooth at any neighborhood of $0$.

BONUS TRACK

The same question, this time comparing the ring of germs of holomorphic functions near $0\in\mathbb C$, that is $\mathbb C\{X\}$, with the ring of germs of complex functions with derivatives of all orders at $0\in\mathbb C$ ("holomorphic at $0$"?).

Matemáticos Chibchas
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    What sort of difference are you interested in? Algebraic properties? The difference in the definition is straightforward, for the germs of $C^\infty$ functions, you have one $\varepsilon > 0$ such that the representative is $C^\infty$ on $(-\varepsilon,\varepsilon)$, while for the germs of functions infinitely differentiable at $0$ you have a (non-increasing) sequence $\varepsilon_n > 0$ such that for a representative $f$, $f^{(n)}$ is differentiable on $(-\varepsilon_n,\varepsilon_n)$. Concerning the bonus track, is it "complex differentiable at $0$" once, or infinitely often? – Daniel Fischer Jul 03 '14 at 15:13
  • @DanielFischer Yes, I am interested in the algebraic properties. Regarding the complex case, you catch me, I naively thought that complex differentiability at $0$ implies differentiability of all order. I am going to edit the question. – Matemáticos Chibchas Jul 03 '14 at 17:03
  • In that case, for the bonus track the answer is "No difference". If $f$ is twice complex differentiable at $0$, that means there is a neighbourhood $U$ of $0$ such that $f$ is (complex) differentiable on $U$. But then $f$ is holomorphic on $U$. – Daniel Fischer Jul 03 '14 at 17:14
  • @DanielFischer Thanks a lot for your partial answer. – Matemáticos Chibchas Jul 03 '14 at 17:16

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