Thread: Why sufficient ?

1. Why sufficient ?

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

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

I see the condition is necessary since if

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

and

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

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

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

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

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

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

Any help would be appreciated.

2. Re: Why sufficient ?

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

3. Re: Why sufficient ?

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

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

I see the condition is necessary since if

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

and

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

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

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

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

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

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

Any help would be appreciated.
This works as a guide, but it is by no means a rigorous argument: Not every smooth function is analytic.

For the problem, if $\displaystyle f(1+a)\equiv 0$ then, since $\displaystyle a(0)=0$ then, by continuity $\displaystyle 1+a\neq 0$ in a neighbourhood of $\displaystyle 0$, so $\displaystyle f$ must be zero (at least) in this neighbourhood. Conversely if $\displaystyle f$ is zero in say $\displaystyle (-b,b)$ then pick any smooth $\displaystyle a$ such that $\displaystyle a(0)=1$ and $\displaystyle a(x)\neq 0$ iff $\displaystyle x\in (-b/2,b/2)$, clearly $\displaystyle 1+(a-1)$ satisfies what you want (For examples of such a function $\displaystyle a$ look up test functions).

4. 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.