Let $I_c$ denote the functions that vanish at some fixed $c\in\mathbb{R}$. Is $I_c$ a maximal ideal?

Question

Let $A$ be the ring of continuous functions from $\mathbb{R}$ to $\mathbb{R}$ and let $I_c$ denote the functions that vanish at some fixed $c\in\mathbb{R}$. Is $I_c$ a maximal ideal? Give an example of a proper non-zero ideal of $A$ that is not of the form $I_c$ for some $c\in\mathbb{R}$.

My attempt: I think $I_c$ is not maximal, because we can consider the ideal $I_{c,d}$ of functions that vanish at fixed $c,d\in\mathbb{R}$. $I_c$ is clearly contained in $I_{c,d}$, but $I_{c,d}\neq A$ because $f(x)=1$ is not in $I_{c,d}$. Also, I think this ideal $I_{c,d}$ can answer the second part of the question, but it seems like there ought to exist a better example.

Is this right? Or at least a start in the right direction?

Any help appreciated!

Answer 1

Consider the ring homomorphism $A \to \mathbb{R}, f \mapsto f(c)$. This homomorphism is surjective and its kernel is $I_c$. By the first isomorphism theorem, we have $A / I_c \cong \mathbb{R}$, as the quotient by $I_c$ is a field, $I_c$ must be maximal.

Answer 2

This can also be proved by slightly simpler means: i.e., without reference to homomorphisms, kernels, or any of the isomorphism theorems:

Suppose that you have an ideal $J \subset A$ that contains the given ideal, $I_c$, properly. So, there must be some continuous function $f \in J$ for which $f(c) = d \neq 0$. Next, consider the continuous function $f-d$, which is in the original ideal: it is continuous since it is simply the difference between a continuous function and a constant, and when evaluated at $c$ it gives $d - d = 0$ as desired.

Since $J$ is an ideal, it is closed under addition. We know that $f \in J$ and $f - d \in I_c \subset J$; so, in particular, $f - (f-d) = d$, i.e., the constant function $d \neq 0$, is an element of $J$. The ideal $J$ is also closed under multiplication with elements in $A$; for example, the continuous (constant) function $1/d \in A$, which means that $d(1/d) = 1 \in J$. But the only ideal of $A$ that contains $1$ is $A$, itself; so, $J=A$, and $I_c$ is maximal as desired.