Maximal Ideals in Ring of Bounded Functions I have a creeping suspicion that this is a stupid question, but here goes:
Consider the answer given by Eric Wofsey here. In particular, it is stated that: "The difference between $C_b(X)$ and $C(X)$ is that for $C_b(X)$, the residue fields for all of these maximal ideals are just $\mathbb{C}$..." where, for our purposes, we can just take $X= \mathbb{R}$. 
My question is this: How exactly can a maximal ideal $I \subseteq C_b(\mathbb{R})$ have residue field $\mathbb{C}$? Suppose this is the case, so that $C_b(\mathbb{R})/I \cong \mathbb{C}$. Then $C_b(\mathbb{R})/I$ is algebraically closed, so any non-constant polynomial $p \in (C_b(\mathbb{R})/I)[x]$ has a root in $C_b(\mathbb{R})/I$. Consider then the polynomial given by $p(x)= [1]+ [1]x^2$, where $[1]$ is to be interpreted as the "equivalence class of the constant function $f(y)=1$". This must have a root in $C_b(\mathbb{R})/I$ unless it is constant. Suppose there is such a root $[f]$. Then $[1]+[1][f]^2=[0]$, so $[1+f^2]=[0]$, hence $1+f^2 \in I$. But $1+f^2(x) \geq 1>0$ for all $x \in \mathbb{R}$, since $f$ is real-valued. Also, $\frac{1}{f^2+1}=(f^2+1)^{-1}$ is bounded, so $1+f^2$ is a unit in $C_b(\mathbb{R})$. Then $I$ contains a unit and thus can't be maximal.
So $p$ is constant then. But that can't be the case either, because then we could take two constant functions $a, b$ with $a \neq b$ and have $[1+a^2]=[1+b^2]$, so $a^2-b^2 \neq 0$ and $a^2-b^2 \in I$, so again $I$ contains a unit, contradicting maximality.
There must be something wrong with my reasoning above and I'd appreciate it if someone could point out my error.

The error that I'm seeing is that $f(x) \in C_b(\mathbb{R})$ should not be purely real-valued. $C_b(\mathbb{R})$ as defined must include complex-valued functions.
More specifically, either you choose $C_b(\mathbb{R})$ to be defined as continuous real-valued bounded functions, in which case residues of maximal ideals is $\mathbb{R}$ rather than $\mathbb{C}$, and hence $1 + x^2$ has no root, or you choose it to be complex and then you get the above issue.
More generally, let $C_b(X;F)$ indicate the ring of continuous bounded functions on a Hausdorff top. space $X$ with values taken in $F$, which is some locally compact topological field (maybe with additional conditions I cannot recall; really this level of abstraction is not my forte). Then the residues of maximal ideals can be identified with $F$.
