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Suppose you have a ring $(C[0,1],+,\cdot,0,1)$ of continuous real valued functions on $[0,1]$, with addition defined as $(f+g)(x)=f(x)+g(x)$ and multiplication defined as $(fg)(x)=f(x)g(x)$. I'm curious what the zero divisors are.

My hunch is that the zero divisors are precisely the functions whose zero set contains an open interval. My thinking is that if $f$ is a function which is at least zero on an open interval $(a,b)$, then there exists some function which is nonzero on $(a,b)$, but zero everywhere else on $[0,1]\setminus(a,b)$. Conversely, if $f$ is not zero on any open interval, then every zero is isolated in a sense. But if $fg=0$ for some $g$, then $g$ is zero everywhere except these isolated points, but continuity would imply that it is also zero at the zeros of $f$, but then $g=0$, so $f$ is not a zero divisor.

I have a hard time stating this formally though, since I'm only studying algebra, and not analysis. Is this intuition correct, and if so, how could it be rigorously expressed?

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4 Answers

up vote 6 down vote accepted

You’ve the right general idea, but it’s a little more complicated than that: the zero-set of the function $f$ might be a Cantor set, which has no isolated points.

Suppose that $Z=\{x\in[0,1]:f(x)=0\}$ contains no non-empty open interval; $Z$ is closed, so this says that $Z$ is nowhere dense in $[0,1]$. Let $V=[0,1]\setminus Z$: $V$ is a dense open set in $[0,1]$. Now suppose that $fg=0$; clearly we must have $g(x)=0$ for every $x\in V$. But $V$ is dense in $[0,1]$, and $g$ is continuous, so $g(x)=0$ for every $x\in[0,1]$, and $f$ is not a zero-divisor.

Your argument in the other direction is correct: if $Z$ contains a non-empty open interval $(a,b)$, just let

$$g(x)=\begin{cases} 0,&\text{if }x\in[0,1]\setminus(a,b)\\\\ x-a,&\text{if }a<x\le\frac12(a+b)\\\\ b-x,&\text{if }\frac12(a+b)\le x<b\;. \end{cases}$$

Then $fg=0$, but $g$ is non-zero on $(a,b)$.

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Thanks Brian. I appreciate that explicit formula for $g(x)$ a lot. –  Linda Cortes Jun 9 '12 at 7:48
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Your conjecture is correct and your proof is nearly complete, too.

Assertion:
The zero divisors of $C[0,1]$ are the functions vanishing on some non-empty open interval $(a,b)\subset [0,1]$

Proof
Fix $f\in C[0,1]$

a) Suppose $f\mid(a,b)=0$ for some $0\leq a\lt b\leq 1$.
Then choose $c,d$ such that $a\lt c\lt d\lt b$ and a continuous function $g\in C[0,1]$ which is non-zero exactly on $(c,d)$ ( this is very easy to construct: take a piecewise affine function ). Then $fg=0$ even though $g\neq 0$, which means exactly that $f$ is a zero divisor.

b) Suppose that $f $ is a zero divisor, i.e. that $fg=0$ for some non zero $g\in C[0,1]$, and let us show that $f\mid(a,b)=0$ for some $(a,b)\subset [0,1]$ .
Since $g\neq 0$ there exists $x\in [0,1]$ with $g(x)\neq 0$ and by continuity of $g$ we can find some neighbourhood $(a,b)$ of $x$ with $g(y)\neq0$ for all $y\in (a,b)$.
But since $fg=0$ on $(a,b)$ this forces $f\mid(a,b)=0$, as announced.

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If $A\subset[0,1]$ is an arbitrary closed set, then $A$ - is the zero set of the continuous function $$ d_A(x)=\inf\{|y-x|:y\in A\} $$ It is indeed continuous, see this answer. And explanation that $A$ its zero set you can find here.

If $A$ and $[0,1]\setminus A$ is not dense in $[0,1]$, then $d_A$ is divisor of zero. Indeed consider one more continuous function $$ d_{[0,1]\setminus A}(x)=\inf\{|y-x|:y\in[0,1]\setminus A\} $$ This function is non-zero since $[0,1]\setminus A$ is not dense in $[0,1]$ and for all $x\in [0,1]$ $$ d_A(x)d_{[0,1]\setminus A}(x)=0 $$ You can generalize this construction to arbitrary metric spaces.

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If $f$ and $g$ are not identically zero and $f \cdot g = 0$ then $g^{-1}(\mathbb{R} \setminus \{0\})$ is open and non-empty and $f$ vanishes on this open subset. You already did the implication in the other direction. So $f$ is a zero divisor if and only if it is not identically zero and vanishes on some non-empty open set.

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