Prime ideal of the ring of continuous functions on $[0,1]$ Let $A$ be the ring of continuous maps $f: [0,1] \rightarrow \mathbb{R}$. Prove that there exists a prime ideal $Q \in \operatorname{Spec} (A)$ which is not of the form $I_p = \{ f \in A \mid f(p)=0 \}$ with $p \in [0,1]$.
Hint: use localization.
 A: Consider the ideal $J\subset  A$ of functions $f\in A$ such that  the limit for $x$ tending to $\frac 12$ (but remaining $\neq \frac 12$ ) of $\frac {f(x)}{(x-\frac 12)^r}$ is $0$ for all $r=1,2,3,...$ .
Then $J$ is not prime, but is a reduced ideal (i.e. equal to its nilradical) .
So $J$ is the intersection of the prime ideals containing it.
Since the only $I_p$ containing $J$ is $I_{\frac 12}$ there and since $J\neq I_{\frac 12}$ there must exist some prime ideal $\mathfrak p\subset A$ with $\mathfrak p\neq I_p$ for all $p\in [0,1]$.
Remark: why $J$ is not prime
Consider in $A$ the functions $f(x)= \operatorname {max} (0,x-\frac 12)$ and $g(x)= \operatorname {max} (0,\frac 12 -x)$.
Then $fg=0\in J$ although  $f,g\notin J$. Hence the ideal $J$ is not prime.
A: I notice that the hint was to “use localization” and that the other answer does not seem to have utilized it. Just today I saw someone mention the probable desired trick, and it seems too ingenious not to share.   Here it is.
Take $S$ to be the subset of nonzero polynomial functions restricted to $[0,1]$. It is a mutiplicative set.  Therefore by basic commutative algebra, there is a prime ideal disjoint from $S$.  Clearly $P$ is not of the form described, since to do so would require it to contain $x-a$ for some $a\in[0,1]$.
