What are the zero divisors of $C[0,1]$? 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?
 A: 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.
A: 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.
A: 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.
A: 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 $(a,b)$.
