# Which are the possible maximal ideals in $C[0,1]$ containing a prime which is not maximal?

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 $$P⊂M_α$$ ?

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 $\dim C(X)=n$..