Well if in some neighborhood of then the Taylor expansion of at is just , and this has an infinite radius of convergence. So on and hence is annihilated by for any
The following is an extract from "Introduction to Commutative Algebra" by Atiyah and MacDonald.
Let be the ring of all functions on the real line, and let be the ideal of all which vanish at the origin. ...
On the other hand is annihilated by some element ( ) if and only if vanishes identically in some neighborhood of 0.
I see the condition is necessary since if
and
are the Taylor series expansions in the neighborhood of 0 of and respectively,
implies that .
I don't understand why the condition is sufficient.
Is there a function which takes the value 1 at 0 and 0 elsewhere ?
Any help would be appreciated.
Thanks in advance.
This works as a guide, but it is by no means a rigorous argument: Not every smooth function is analytic.
For the problem, if then, since then, by continuity in a neighbourhood of , so must be zero (at least) in this neighbourhood. Conversely if is zero in say then pick any smooth such that and iff , clearly satisfies what you want (For examples of such a function look up test functions).
Hello, hatsoff, Jose27.
Thank you for you help.
Given your suggestions, I read some articles on Wikipedia such as "Distribution","Bump Function","Non-Analytic Smooth Function". And I found
if K is an arbitrary compact set in n dimensions and U is an open set containing K, there exists a bump function φ which is 1 on K and 0 outside of U.
This proves the sufficiency of the condition.
Thank you again.