Maximal ideal in the ring of continuous functions from $\mathbb{R} \to \mathbb{R}$

Question

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 (:

Answer 1

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.