TITLE: Maximal ideals in the ring of real functions on $[0,1]$
QUESTION [21 upvotes]: Let $S$ be the ring of all continuous functions from $[0,1]$ to $\mathbb R$. How to prove that all maximal ideals of $S$ have the form $M_{x_0}=\{f\in S \mid f(x_0)=0\}$? 
Thanks in advance.

REPLY [4 votes]: Here is another, slightly silly, way to see this. One can use the Stone-Weierstrass Theorem.
Let $R$ denote the ring of continuous functions on $[0,1]$, and let $M$ be a proper maximal ideal. For the sake of contradiction, suppose there is no $x\in [0,1]$ such that $f(x) = 0$ for all $f\in M$.
Observe that $M$ is an algebra of real continuous functions on $[0,1]$. By assumption, it vanishes at no point of $[0,1]$. Suppose there are $x_1$, $x_2 \in [0,1]$ such that $f(x_1) = f(x_2) \neq 0$ for all $f\in M$. Then the property of ideals tells us $x\cdot f(x) \in M$, so $$x_1 f(x_1) = x_2 f(x_2) \implies x_1 = x_2$$ Thus, $M$ separates points on $[0,1]$. The Stone-Weierstrass Theorem implies $M$ is dense in $R$. In particular, there is some $g\in M$ such that $g>0$ on $[0,1]$. Then $1/g \in R$, so $g \cdot (1/g) = 1 \in M$. Hence $M = R$, contradiction.