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Jul 18, 2019 at 10:14 vote accept Dmitry Kerner
Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Apr 11, 2017 at 5:54 history edited Todd Leason CC BY-SA 3.0
Added answer to the second question.
Apr 4, 2017 at 9:03 comment added Dmitry Kerner Sorry for the confusion, $C^{\infty}(\Bbb{R}^1,0)$ is the ring of germs of infinitely differentiable functions (defined on some neighborhoods of $0\in \Bbb{R}^1$)
Apr 4, 2017 at 8:15 comment added Todd Leason So $C^\infty(\mathbb{R},0)$ is the (one-element) set of functions $\mathbb{R}\to \{0\}$ ?
Apr 4, 2017 at 8:06 comment added Dmitry Kerner Well, I meant a local ring with the non-trivial maximal ideal. (0 denotes the zero ideal)
Apr 4, 2017 at 7:58 comment added Todd Leason In my example $R$ is a field and hence local. BTW: What's the meaning of the $0$ in $C^\infty(\mathbb{R},0)$ ?
Apr 3, 2017 at 19:34 comment added Dmitry Kerner I meant mostly the local rings, and your example is not local, right? If the local ring has no flat functions, i.e. $\mathfrak{m}^\infty=\{0\}$, then the power series can be controlled by $\mathfrak{m}$-adic topology. But for the ring $C^\infty(\Bbb{R}^1,0)$ the situation seems to be tricky.
Apr 3, 2017 at 0:05 history answered Todd Leason CC BY-SA 3.0