Let $X$ be a locally compact Hausdorff space which is not compact, and let $C(X)$ be the ring of continuous functions $X\rightarrow\mathbb{R}$. Show that there exists a maximal ideal of $C(X)$ which is not of the form $\mathfrak{m}_x=\{f\in C(X):f(x)=0 \}$.
My thoughts on the problem:
The proof for $X$ compact showing that all ideals must be of the form $\mathfrak{m}_x$ for some $x\in X$ uses contradiction. So I don't think its proof could help me much in computing a maximal ideal like that.
We have that open intervals in $\mathbb{R}$ are locally compact, but not compact. So this would be a prime candidate to get examples from. I was thinking that perhaps an ideal of unbounded continuous functions would work since these are the main examples of continuous functions I can think of that can't happen over compact spaces, but this does not form an ideal since $0\cdot f=0$ is no longer unbounded.
Best Answer
The answer linked to in the comments (Is there any non-compact space π such that every maximal ideal of $C(X,\mathbb{R})$ is of the form $\{f\in C(X,\mathbb{R}) : f(a) = 0\}$?) shows that the maximal ideals in $C(X)$ which are not of the form $\mathfrak{m}_x$ are in bijection with the free $Z$-ultrafilters.
So it suffices to find a free $Z$-ultrafilter on $X$. To do this, we wish to find a family of zero-sets $\mathcal{Z} = \{Z_x\mid x\in X\}$ such that $\mathcal{Z}$ has the finite intersection property (for all $x_1,\dots,x_n\in X$, $\bigcap_{i=1}^n Z_{x_i}\neq \emptyset$), and such that for all $x\in X$, $x\notin Z_x$. Indeed, suppose we have such a family $\mathcal{Z}$. Then $\mathcal{Z}$ generates a proper $Z$-filter (consisting of all zero-sets which contain a finite intersection of zero-sets from $\mathcal{Z}$: the finite intersection property ensures that $\emptyset$ is not in this $Z$-filter), and this $Z$-filter extends to a $Z$-ultrafilter $\mathcal{U}$ by Zorn's lemma. Finally, $\mathcal{U}$ is free, since $\bigcap_{Z\in \mathcal{U}} Z\subseteq \bigcap_{x\in X} Z_x = \emptyset$.
Fix some open cover $\{O_i\mid i\in I\}$ which has no finite subcover. Let $x\in X$. Then there is some $i\in I$ such that $x\in O_i$. We will write $O_x$ for $O_i$. Let $C_x = X\setminus O_x$. Since $X$ is locally compact Hausdorff, it is completely regular, so there is a continuous function $f_x\in C(X)$ such that $f_x(x) = 1$ and $f_x|_{C_x} = 0$. Let $Z_x = Z(f_x)$. We have $x\notin Z_x$, as desired.
It remains to show that $\mathcal{Z} = \{Z_x\mid x\in X\}$ has the finite intersection property. Suppose for contradiction that there are $x_1,\dots,x_n\in X$ such that $\bigcap_{j=1}^n Z_{x_j}= \emptyset$. Since $Z_{x_j} = Z(f_{x_j}) \supseteq C_{x_j}$, we have $\bigcap_{j=1}^n C_{x_j} = \emptyset$. But then taking complements, $\bigcup_{j=1}^n O_{x_j} = X$, contradicting the assumption that $\{O_i\mid i\in I\}$ has no finite subcover.