$U_{c} = \{f \in C[a, b] \mid f (c) = 0 \}$, where $c$ is a fixed number from $[a, b]$, is a maximal ideal of the ring of continuous functions $C[a, b]$. Indeed, $f \longmapsto f(c)$ is a homomorphism from $C[a, b]$ onto $\mathbb{R}$ with kernel $U_{c}$, so that $C[a, b]/U_{c} \cong \mathbb{R}$ is a field.
It turns out that the $U_{c}$'s are actually the only maximal ideals of $C[a, b]$.
Can you give me guidance on why the following statement is correct?
The $U_{c}$'s are actually the only maximal ideals of $C[a, b]$.
Use that $[0,1]$ is compact. If $I$ is an ideal not contained in any $U_c$, there is $f_c\in I$ with $f_c(c)\ne0$. Then by a compactness argument, there are $c_1,\ldots,c_n$ such that for all $x\in [a,b]$ there is some $j$ with $f_{c_j}(x)\ne0$. Now consider $F=\sum_{j=1}^n f_{c_j}^2$. Prove that $F\in I$, but $F$ is a unit in $C[a,b]$.