Maximal Ideals in Ring of Continuous Functions 
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 $M \cap M_c$.)
 A: 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 :-)
A: Whoever told you that was mistaken; there is no way to prove this without using some special property of $[0,1]$ that is closely related to compactness.  To try to convince you of this, let me point out that the result is not true if you replace $[0,1]$ by $(0,1]$.  Indeed, if $R$ is the ring of continuous functions $(0,1]\to\mathbb{R}$, let $I\subset R$ be the set of functions which are identically $0$ on $(0,\epsilon)$ for some $\epsilon>0$.  Then $I$ is an ideal: if $f,g\in I$ with $f$ vanishing on $(0,\epsilon)$ and $g$ vanishing on $(0,\epsilon')$, then $f+g$ vanishes on $(0,\min(\epsilon, \epsilon'))$.  And if $f\in I$ and $g\in R$, then $gf$ vanishes everywhere that $f$ does, so $gf\in I$.
Since the constant function $1$ is not in $I$, $I$ is a proper ideal in $R$, so there is some maximal ideal $M$ containing it.  But for any $c\in (0,1]$, $M\neq M_c$, since we can take $\epsilon=c/2$ and find a continuous function $f$ which vanishes on $(0,\epsilon)$ but such that $f(c)\neq 0$, and then $f\in M$ but $f\not\in M_c$.
More generally, a similar counterexample can be given for any reasonably nice non-compact space.  To be precise, let $X$ be any completely regular space that is not compact and let $R$ be the ring of continuous functions $X\to\mathbb{R}$.  Then there exists a maximal ideal in $R$ which is not $M_c=\{f\in R:f(c)=0\}$ for any $c\in X$.  (Proof sketch: pick a point $x\in \beta X\setminus X$, let $I$ be the ideal of functions which vanish in a neighborhood of $x$, and let $M$ be any maximal ideal containing $I$.)
