# Ring of continuous maps and prime ideals

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.

#### Solutions Collecting From Web of "Ring of continuous maps and prime ideals"

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.