For $f\in R$ let $Z(f)=\{x\in[0,1]:f(x)=0\}$, the zero-set of $f$. A family $\mathscr{F}$ of zero-sets of $[0,1]$ is a zero-set filter if
- $\varnothing\notin\mathscr{F}$,
- $\mathscr{F}$ is closed under finite intersections, and
- $Z\in\mathscr{F}$ whenever $Z$ is a zero-set and $Z\supseteq Z_0$ for some $Z_0\in\mathscr{F}$.
Suppose that $I$ is an ideal in $R$. If $F$ is a finite subset of $I$ such that $\bigcap_{f\in F}Z(f)=\varnothing$, let $g=\sum_{f\in F}f^2$; then $Z(g)=\bigcap_{f\in F}Z(f)=\varnothing$, so $1/g\in R$, and $1_R\in I=R$. Thus, for a proper ideal $I$ the family $\mathscr{F}=\{Z(f):f\in I\}$ is closed under finite intersections and does not contain $\varnothing$. Now suppose that $Z$ is a zero-set of $[0,1]$ such that $Z\supseteq Z(f)$ for some $f\in I$. Let $g\in R$ be such that $Z(g)=Z$; then $fg\in I$ and $Z(fg)=Z$, so $Z\in\mathscr{F}$. That is, $\mathscr{F}$ is a zero-set filter on $[0,1]$.
Conversely, suppose that $\mathscr{F}$ is a zero-set filter on $[0,1]$ and let $I=\{f\in R:Z(f)\in\mathscr{F}\}$; then $I$ is an ideal of $R$.
If $f,g\in I$, then $Z(f-g)\subseteq Z(f)\cap Z(g)\in\mathscr{F}$, so $f-g\in I$.
If $f\in I$ and $g\in R$, then $Z(fg)=Z(f)\cup Z(g)\supseteq Z(f)$, so $g\in I$, and $RI\subseteq I$. (In fact $RI=I$, since $R$ has a multiplicative identity.) Let $Z=\bigcap\mathscr{F}$; $[0,1]$ is compact, so $Z\ne\varnothing$, and $I\subseteq I_Z=\{f\in R:Z(f)\supseteq Z\}$. However, $I$ may be a proper subset of $I_Z$. For example, fix a zero-set $Z\subsetneqq[0,1]$, let $\mathscr{Z}=\{Z_n:n\in\omega\}$ be a family of zero-sets such that $Z_0\supseteq\operatorname{int}Z_0\supseteq Z_1\operatorname{int}Z_1\supseteq Z_1\supseteq\ldots\supseteq Z\;,$ and let $\mathscr{F}$ be the zero-set filter generated by $\mathscr{Z}$, and let $I$ be the associated ideal. Let $f$ be any member of $R$ such that $Z(f)=Z$; then $f\in I_Z\setminus I$.