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Let $R=C([0,1])$ be the ring of all real valued continuous functions on [0,1].

A prime ideal $p\in Spec(R)$ is called associated prime if it arises as annihilator of some element of $R$, that is $p=ann(x)$ for some $x\in R$.

My question is, how do we find the associated prime of $R=C([0,1])$.

Any hints..

Edits: Let us denote the set of all associated prime of $R$ by $AP(R)$

It is easy to observe that $p\in AP(R) $ if and only if there is a injective map $$i:\frac{R}{p}\longrightarrow R$$

We know $R/p$ is an integral domain, can we deduce something from here.

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    If I am not mistaken, $R = C[0,1]$ has no associated primes at all. Any $0 \neq f \in R$ is nonzero on some open interval $I$. Take some functions $0 \neq g,h \in R$ which are zero outside $I$ such that $gh = 0$. Then obviously $gh \in \mathrm{ann}(f)$ but neither $g$ nor $h$ lies in $\mathrm{ann}(f)$. – 2017-02-01
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    @Dune Oh! You are right. So it means AP(R) is empty, That gives us another way to show that R is non Noetherian. – 2017-02-01
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    Exactly! More precisely we can say that the set $\{ \mathrm{ann}(f) : 0 \neq f \in R \}$ has no maximal element. – 2017-02-01

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