In the ring of continuous functions $\mathbb{R} \to \mathbb{R}$, the set of all $f$ with $f(0) = 0$ is a maximal ideal. 
Problem
Let $R$ be a ring of continuous functions from $\mathbb{R}$ to $\mathbb{R}$. Show that $A = \{f \in R \mid f(0)=0\}$ is a maximal ideal of $R$.

If $f,g \in A$, then $f-g(0) = f(0)-g(0) = 0$ . So $f-g$ is in $A$. Also $f(0) h(0) = 0$ for all $h \in R$. So $A$ is an ideal.
Attempt to show $A$ is maximal
If  $R/A$ is a field then $A$ is maximal.
Let $f \notin A$. Then $f(0) \neq 0$. Let $g(x)=0$. Then $f-g \in A$. So $f+A = g+A$. $1/g$ is the inverse of $g$. So $f$ has unity (not sure about this).
Is this correct?
 A: Hint: Use the first isomorphism theorem and a well chosen map $R \to \mathbb{R}$.
A: Notice that $C(\mathbb{R})$ is a vector space and that $A \dotplus \mathbb{R}1  = C(\mathbb{R})$.
Let $I$ be an ideal in $C(\mathbb{R})$ such that $A \subsetneq I$. Notice that $I$ is also a subspace of $C(\mathbb{R})$, because $\lambda f = (\lambda 1) f$.
Pick $f \in I \setminus A$. There exist unique $g \in A$ and $\lambda \in \mathbb{R}$ such that $f = g + \lambda 1$. Since $f \notin A$ we have $\lambda \ne 0$. Therefore
$$1 = \frac1{\lambda}(f-g) \in I$$
so $I = C(\mathbb{R})$.
A: Your proof that $A$ is an ideal works, although strictly speaking, you also need to check that $0 \in A$ (which shouldn’t be a problem).
Your proof that $A$ is maximal does not work in its current form.
You’re right that it sufficies to show that $R/\!A$ is a field, and that for this you need to show that for $f \in R$ with $f \notin A$ the element $f + A \in R/\!A$ has an inverse.
The problem is that your choice of $g$ and how you work with it makes no sense:


*

*You choose $g(x) = 0$, i.e. $g = 0$ is the (constant) zero function.
But then
$$
  (f - g)(0) = f(0) - g(0) = f(0) \neq 0
$$
and therefore $f - g \notin A$ (and thus also $f + A \neq g + A$).

*The function $1/g$ does not exists.

Your approach can be fixed by making the right choice for $g$:

 For $c := f(0)$ you choose $g$ as the constant function $g(x) = c$.
 Then
 $$(f - g)(0) = f(0) - g(0) = c - c = 0,$$
 so that $f - g \in A$ and thus $f + A = g + A$.
 The function $g$ has in $R$ a multiplicative inverse $1/g$ given by $(1/g)(x) = 1/c$ because $c \neq 0$.
 It follows that $1/g + A$ is a multipicative inverse to $g + A = f + A$ in $R/\!A$.

A different approach to the problem would be to use a suitable isomorphism theorem:

 The set $A$ is the kernel of the ring homomorphism $\varphi \colon R \to \mathbb{R}$ given by $\varphi(f) = f(0)$.
 This already shows that $A$ is an ideal.
 The ring homomorphism $\varphi$ is surjective, so it follows from the first isomorphism theorem that
 $$ R/\!A = R/{\ker(\varphi)} \cong \operatorname{im}(\varphi) = \mathbb{R}$$
 is a field.
 This shows that $A$ is maximal.

