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?

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.

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 $(a,b)$.

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.

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.