Maximal ideal in the ring of continuous functions from $\mathbb{R} \to \mathbb{R}$ Well, the problem I'm trying to solve is this: 

Let $A$ be the ring of all continuous functions from $\mathbb{R} \to \mathbb{R}$. Show that $M = \{f \in A: f(0)=0\}$ is a maximal ideal of $A$.

I tried to show that if $J$ is an ideal of A that properly contains M, so $J = A$: we know that there is a function $g \in J$ with $g(0) \neq 0$. So, we have that
$$J \supseteq \langle g, M\rangle=\{ag+f;a \in A, f\in M\}$$ 
I tried to show that 1 $\in$ $\langle g,M\rangle$. However I'd have to find some function $h(x)\in A$ such that $h(x)(ag+f) = 1$, but this function $h$ has to be different from 0 for all x $\in \mathbb{R}$, so I don't know how to find this function.
My second idea to solve this problem was prove that the quocient $A/M$ is a field.
We have that
$$A/M = \{h(x) + M; h(x) \in A\},$$ but again I have the problem to show that all elements of $A/M$ are units.
If someone could help, I would be really grateful (:
 A: If $J \subset A$ contains $M$ properly, then there must be some function $f \in J - M$, i.e., for which $f(0) = a \not= 0$.
Consider $f - a \in M \subset J$: we know $f-a$ is continuous, and evaluated at $0$ it returns $a-a = 0$.
Next, consider that $J$ is closed (as it is an ideal) under addition. Since $f \in J$ and $f-a \in J$, we find that the constant nonzero function $a = f - (f-a) \in J$. Since $a \neq 0$, we also have the continuous (constant) function $1/a \in A$, which means that $J$ contains $a(1/a) = 1$, since ideals are closed under multiplication with any element of the parent ring. But this means that $1 \in J$, so that $J = A$, whence $M$ is maximal as desired.
A: Your summary of what you need to show is incorrect: you aren't given $a$, $g$, and $f$ and need to find $h$ such that
$$ h (ag + f) = 1$$
Instead, the problem you actually need to solve is that you're given $g$ and need to find $h$, $f$, and $a$ such that
$$ h (ag + f) = 1 \qquad \qquad f(0) = 0$$
As you've already observed solving for $h$ is problematic, you can make life much easier for yourself by picking $h$ to be something simple and then trying to solve for $a$ and $f$.
But even this is making things harder than you need to. You asked to show that $1 \in \langle M,g \rangle$, and you know everything in $\langle M,g \rangle$ is of the form $ag + f$ with $f(0)=0$. So the problem you actually need to solve is, given $g$, to find $a,f$ such that
$$ ag+f = 1 \qquad \qquad f(0)=0$$
A: Consider $\varepsilon: A \to \mathbb R$ given by $\varepsilon (f)=f(0)$. This is a surjective homomorphism with kernel $M$. Since the image is a field, the kernel is a maximal ideal.
A: Let $R$ be the ring of all continuous functions from $\mathbb{R} \rightarrow \mathbb{R}$
Let $B=\{f \in R~|~f(0)=0\}$.
Now, we know that if $R$ is a commutative ring with unity and $I$ is an ideal, then, $R/I$ is a field if and only if $I$ is maximal.
If we prove that $R/B$ is a field, then $B$ is maximal. For which,we must prove that every non zero coset element $\{g(x) + B~|~g(x) \in R ,~~ g(x) \notin B \implies g(0) \neq 0\}$ has an inverse.
Note that $g(x)+B = g(0) + B$ as $g(x)-g(0) \in B$
and $(g(0)+B)~( (g(0))^{-1}+ B) = 1+B$
Hence, every element has an inverse..
