Prime ideals in $C[0,1]$

Solution 1:

If $R$ is a reduced commutative ring, then the following statements are equivalent:

1. $\dim(R)=0$
2. Every prime ideal of $R$ is maximal.
3. For every $a \in R$ we have $(a^2)=(a)$.
4. 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).

Solution 2:

Here's just one example.

Let $(x_n)$ be a sequence that converges to a point $x$, and let $U$ be an ultrafilter on $\mathbb{N}$. Intuitively, an ultrafilter is just a funny way of dividing the subsets of $\mathbb{N}$ into "small" and "large". So let $I$ be the ideal of functions $f$, such that $\{n | f(x_n) = 0\}$ is in $U$, i.e. that have, according to $U$, "many zeroes".

You can check, from the definition of ultrafilter, that this is indeed a prime ideal (basically because the intersection of two large sets is large and the union of two small sets is small). Now if the ultrafilter is not principal (i.e. not generated by one of those points) then the ideal won't be maximal.

Solution 3:

another way to construct a prime wich is not maximal is the following:

Consider the multiplicative set of continuos function having just a finite number of zeros in [0,1]. Since this is a multiplicative set not containing 0 there's a prime ideal not intersecting him. But then since it does not intersect him, it cannot be the set of functions that kills one point(indeed the function $x-pt$ has finite number of 0(just 1) and is in that ideal): but those are all the maximals(by the Banach "nullstellensatz")

Solution 4:

Fix a point $c\in (0, 1)$. Define $I := \{ f \in C[0,1] : Z(f)$ contains an open nbd of $c \}$ where $Z(f)$ denotes the zero set of $f$. It can be easily checked that $I$ is a radical ideal and $I \subsetneq M_c := \{ f \in C[0,1] : f(c) = 0 \}$. Also note that $I \nsubseteq M_b$ for any $b \in [0, 1], b\neq c$. Now use the following fact: In a commutative ring with unity, a radical ideal is an intersection of prime ideals. So, $I$ is an intersection of prime ideals, but clearly it's not an intersection of maximal ideals only. Thus there must exists at least one non-maximal prime.