An idea $I$ in $C(X)$ is called $z$-ideal if $Z(f) \in Z(I) $ implies $f \in I$.
$Z(f) =\{x \in X : f(x) = 0 \}$
Can you help me to prove these problems?
Let $X$ be a finite discrete space in $C(X)$.
1: $f$ is a multiple of $g$ $\Longleftrightarrow$ $Z(f) \supset Z(g)$
2: Every ideal is a $z$-ideal.
3: Every ideal is principal, and , in fact, is generated by an idempotent.
Note that if $X = \{1,\ldots,m\}$, say, that $C(X)$ is just $\mathbb{R}^m$ in the ring structure of pointwise addition and multiplication. Any real function on $X$ is continuous and is uniquely described by the $m$-tuple of its values.
So if $Z(f) \supseteq Z(g)$, we define $h(i) = 0$ for $i\in Z(g)$, $h(i) = \frac{f(i)}{g(i)}$ for $ i \notin Z(g)$. Then $f = h \cdot g$ in $C(X)$. The other direction is always true: if $f = h\cdot g$ for some $h \in C(X)$, $g(x) = 0$ implies $f(x) = 0$ so that $Z(g) \subseteq Z(f)$.
That every $I$ is a $z$-ideal also follows: let $Z(f) = Z(g)$ for some $g \in I$, then by (i) implies $f$ is a multiple of $g$, so $f \in I$ as required.
To see that every ideal is principal: let $I \subseteq C(X)$ be a proper ideal, then consider $Z[I] = \{Z(f) : f \in I\}$. This is a family of subsets of the finite $X$ that is closed under finite intersections and enlargements. If $Z(f) = \emptyset$ then $f$ is a unit and so cannot be in $I$, so $\emptyset \notin Z[I]$. So $Z[I]$ is a filter on a finite set $X$, so $Z[I] \ni K= \cap Z[I] \neq \emptyset$. Then the function $g$ that is $0$ on $K$ and $1$ outside of $K$ is an idempotent that generates $I$. ($f \in I$ means $Z(g) =K \subseteq Z(f)$ so $f$ is a multiple of $g$. That $g \in I$ is because $I$ is a $z$-ideal.