Why sufficient ?

**Aki** · June 25th 2011, 03:20 AM

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.

**hatsoff** · June 25th 2011, 06:18 AM

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 $\alpha\in\mathfrak{a}$

**Jose27** · June 25th 2011, 09:29 AM

Originally Posted by Aki

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).

**Aki** · June 25th 2011, 09:53 PM

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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