TITLE: Maximal Ideals in Ring of Continuous Functions
QUESTION [13 upvotes]: Dummit and Foote, 7.4.33(a): Let $R$ be the ring of all continuous functions $[0,1] \to \mathbb{R}$ and let $M_c$ be the kernel of evaluation at $c \in [0,1]$, i.e. all $f$ such that $f(c) = 0$. Show that if $M$ is a maximal ideal in $R$ then $M = M_c$ for some $c$. 

I am aware of proofs that use the compactness of $[0,1]$; however I have been told that a simpler proof exists using only the isomorphism theorems for rings along with basic facts about ideals. I have tried on my own to prove it this way but still have no success; is it possible?
(My attempts go something like: since $M$ is maximal, the quotient ring $R/M$ is a field. Then it would be nice to use the fact that $R/M$ has no zero divisors, but I can't quite make it work. Alternatively, consider $M \cap M_c$.)

REPLY [3 votes]: Maximal ideals in the ring of continuous functions were studied by Edwin Hewitt in
Hewitt, Edwin. Rings of real-valued continuous functions. I. Trans. Amer. Math. Soc.  64,  (1948). 45–99.
He called such ideals hyper-real.  They are not always principal :-)