Let $A=C[0,1]$. We know that the maximal ideals of $A$ are of the form $$M_α=\{f∈A \mid f(α)=0\}, \ α∈[0,1].$$ Now we show that there is a prime ideal $P$ which is not maximal in $A$.

Consider $S$ the set of all monic polynomials and let $T$ be the set of ideals of $A$ which do not meet $S$.

$T$ is nonempty as the zero ideal is in $T$.

By Zorn’s lemma we can find an ideal which is maximal element in $T$, say this ideal is $P$.

Now $P$ must be prime (since $g∉P,~h∉P \implies gh∉P~$).

Also $P$ is not maximal.

If it is so, it is one of $M_α$, then $f(x)=x-α$ belongs to the intersection of $S$ and $T$, which is a contradiction.

So $P$ is a prime ideal but not maximal ideal in $A$.

We know that every prime ideal is contained in some maximal ideal. $M_α$ are the only maximal ideals in $A$.

Question.What are the possible $M_α$ such that $P⊂M_α$ ?