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Math Help - Why sufficient ?

  1. #1
    Aki
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    Why sufficient ?

    The following is an extract from "Introduction to Commutative Algebra" by Atiyah and MacDonald.


    Let A be the ring of all C^\infty functions on the real line, and let \mathfrak{a} be the ideal of all f which vanish at the origin. ...
    On the other hand f is annihilated by some element 1+\alpha( \alpha \in \mathfrak{a}) if and only if f vanishes identically in some neighborhood of 0.


    I see the condition is necessary since if

    f(x)=a_0+a_1 x + a_2 x^2 + \cdots

    and

    1+\alpha=1+b_1 x+b_2 x^2 + \cdots

    are the Taylor series expansions in the neighborhood of 0 of f(x) and 1+\alpha respectively,


    0=f(x)(1+\alpha)=(a_0+a_1 x + a_2 x^2 + \cdots)(1+b_1 x+b_2 x^2 + \cdots)

    =a_0+(a_0 b_1 + a_1)x+(a_0 b_2 +a_1 b_1 + a_2)x^2 + \cdots

    implies that a_0=a_1=a_2=\cdots=0.

    I don't understand why the condition is sufficient.
    Is there a C^\infty function which takes the value 1 at 0 and 0 elsewhere ?

    Any help would be appreciated.
    Thanks in advance.
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  2. #2
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    Re: Why sufficient ?

    Well if f\equiv 0 in some neighborhood of x=0 then the Taylor expansion of f at x=0 is just f\equiv 0, and this has an infinite radius of convergence. So f\equiv 0 on \mathbb{R} and hence is annihilated by 1+\alpha for any \alpha\in\mathfrak{a}
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  3. #3
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    Re: Why sufficient ?

    Quote Originally Posted by Aki View Post
    The following is an extract from "Introduction to Commutative Algebra" by Atiyah and MacDonald.


    Let A be the ring of all C^\infty functions on the real line, and let \mathfrak{a} be the ideal of all f which vanish at the origin. ...
    On the other hand f is annihilated by some element 1+\alpha( \alpha \in \mathfrak{a}) if and only if f vanishes identically in some neighborhood of 0.


    I see the condition is necessary since if

    f(x)=a_0+a_1 x + a_2 x^2 + \cdots

    and

    1+\alpha=1+b_1 x+b_2 x^2 + \cdots

    are the Taylor series expansions in the neighborhood of 0 of f(x) and 1+\alpha respectively,


    0=f(x)(1+\alpha)=(a_0+a_1 x + a_2 x^2 + \cdots)(1+b_1 x+b_2 x^2 + \cdots)

    =a_0+(a_0 b_1 + a_1)x+(a_0 b_2 +a_1 b_1 + a_2)x^2 + \cdots

    implies that a_0=a_1=a_2=\cdots=0.

    I don't understand why the condition is sufficient.
    Is there a C^\infty 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 f(1+a)\equiv 0 then, since a(0)=0 then, by continuity 1+a\neq 0 in a neighbourhood of 0, so f must be zero (at least) in this neighbourhood. Conversely if f is zero in say (-b,b) then pick any smooth a such that a(0)=1 and a(x)\neq 0 iff x\in (-b/2,b/2), clearly 1+(a-1) satisfies what you want (For examples of such a function a look up test functions).
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  4. #4
    Aki
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    Re: Why sufficient ?

    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.
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