A (non-artificial) example of a ring without maximal ideals

Solution 1:

In this paper Mel Henriksen shows that a commutative ring $R$ has no maximal ideals iff (a) $J(R)=R$, where $J(R)$ is the Jacobson radical of $R$, and (b) $R^2+pR=R$ for every prime $p\in\Bbb Z$. He then gives three examples. One starts with a field $F$ of characteristic $0$ and forms the integral domain

$$S(F)=\left\{h(x)=\frac{f(x)}{g(x)}\in F(x):f(x),g(x)\in F[x]\text{ and }g(x)\ne 0\right\}\;;$$ its unique maximal ideal is $R(F)=xS(F)$, which has no maximal ideals.

This paper by Patrick J. Morandi also constructs some examples.

Solution 2:

After some research I came across this paper, which yields an affirmative answer to my second question. I quote Proposition 2.4 below.

Proposition 2.4 If $X$ is a completely regular Hausdorff space, then every maximal ideal of $C_0(X)$ is fixed. In fact every maximal ideal of $C_0(X)$ is of the form $M_x\cap C_0(X)$, where $M_x$ is a fixed maximal ideal in $C(X)$ and the point $x$ has a compact neighborhood.

The ideals $M_x$ are of the form $M_x=\{f\in C(X)\mid f(x)=0\}$. Now, clearly $\mathbb{R}$ satisfies the conditions of the theorem, so the maximal ideals of $C_0(\mathbb{R})$ are of the form $M_x\cap C_0(\mathbb{R})$ for any $x\in\mathbb{R}$. Furthermore, $C_c(\mathbb{R})$ is not contained in any of these maximal ideals, and so by the correspondence theorem for rings, $R:=C_0(\mathbb{R})/C_c(\mathbb{R})$ has no maximal ideals.