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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 $P⊂M_α$ ?

Eric Wofsey
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Every maximal ideal $M_\alpha$ in $C[0,1]$ contains a non-maximal prime ideal. Indeed, in your construction, you can instead define $S$ as the set of functions of the form $fg$ where $f$ is a nonzero polynomial and $g(\alpha)\neq 0$, and the rest of the argument still works. To justify that $T$ is nonempty, note that $0\not\in S$ since if $g(\alpha)\neq 0$, $g$ is nonzero on an entire open neighborhood of $\alpha$, and so $fg$ vanishes at only finitely many points of that neighborhood if $f$ is a nonzero polynomial. The key step at the end $g,h\notin P\implies gh\not\in P$, which you did not prove, works for any multiplicatively closed set $S$: if $g,h\notin P$, then by maximality of $P$, $(g)+P$ and $(h)+P$ both intersect $S$, but then by multiplying we find that $((g)+P)((h)+P)\subseteq (gh)+P$ also intersects $S$ and so $gh\notin P$.

More generally, for any topological space $X$, if $\alpha\in X$ and there is a function $f\in C(X)$ such that $f(\alpha)=0$ but $f$ does not vanish identically in any neighborhood of $X$, then a similar argument shows there is a prime ideal properly contained in the maximal ideal $M_\alpha$ of functions vanishing at $\alpha$: just apply the argument with $S$ the set of functions of the form $f^ng$ where $g(\alpha)\neq 0$, to obtain a prime ideal $P$ contained in $M_\alpha$ which does not contain $f$. The assumption that $f$ does not vanish identically in any neighborhood of $\alpha$ is needed to guarantee that $T$ is nonempty. In fact, by a more complicated construction, you can find an infinite chain of such prime ideals $P\subseteq M_\alpha$; see my answer at Finding a space $X$ such that $\dim C(X)=n$..

Eric Wofsey
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