We know maximal ideals in $C[0,1]$, But what about prime ideals which are not maximal.
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The below answer is by Martin Brandenburg.
If $R$ is a reduced commutative ring, then the following statements are equivalent:
- $\dim(R)=0$
- Every prime ideal of $R$ is maximal.
- For every $a \in R$ we have $(a^2)=(a)$.
- For every $a \in R$ there is some unit $u \in R$ such that $ua$ is idempotent.
In that case, $R$ is called von Neumann regular. The proof of the equivalences is not so hard. 1. $\Leftrightarrow$ 2. is trivial, 2. $\Rightarrow$ 3. may be reduced to the case of a reduced $0$-dimensional local ring, which has to be a field, for which the claim is obvious, $3. \Rightarrow 2.$ If $\mathfrak{p}$ is a prime ideal, in $R/\mathfrak{p}$ we have $a \equiv a^2 b$ for some $b$, hence $a \equiv 0$ or $1 \equiv ab$, which shows that $R/\mathfrak{p}$ is a field. I leave the equivalence to $4.$ as an exercise.
Applying this to $R=C(K)$ for a perfectly normal space $K$, we see that $\dim(R)=0$ iff $K$ is finite discrete (use that every closed subset of $K$ is the zero set of some $f \in C(K)$, which has to be open-closed by 4.).
In particular, $C[0,1]$ has (lots of) prime ideals which are not maximal. But I don't think that you can write them down explicitly. One can show that every norm-closed prime ideal is maximal (for example using Gelfand duality).
My question is, Is it really "not possible" to write a non maximal ideal prime ideal explicitly. Pardon me I have misunderstood the answer.
I would love to see a non maximal prime ideal written explicitly!!
