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

As a brief overview of the below, I am asking for:

1. An example of a ring with no maximal ideals that is not a zero ring.
2. A proof (or counterexample) that $R:=C_0(\mathbb{R})/C_c(\mathbb{R})$ is a ring with no maximal ideals.

A homework question in my algebra class earlier this year asked to exhibit a ring (necessarily without identity) without any maximal (proper) ideals. For solutions to this, it suffices to exhibit an abelian group $(G,+)$ without maximal (proper) subgroups. For, given such a group, define multiplication to be constantly zero. In this case, $G$ becomes a zero ring without maximal ideals (because ideals correspond to subgroups).

It is not particularly difficult to construct examples of the above. For example, consider $(\mathbb{Q},+)$. Another interesting example is $P=\{z\in\mathbb{C}\mid \exists n\in\mathbb{N}, z^{p^n}=1\}$ with standard complex number multiplication as "addition" (that is, as the abelian group operation).

However, any example constructed in this manner is a zero ring, and as such seems "artificial," which I admit, is not a rigorous term. I would like to find a somewhat less artificial example of a ring without maximal ideals. For a definition of "less artificial," let us start with "not a zero ring."

I have a candidate in mind, but I am having trouble explicitly proving that it has no maximal ideals. Let $C_0(\mathbb{R})$ denote the ring of continuous real-valued function on $\mathbb{R}$ vanishing at infinity. Let $C_c(\mathbb{R})$ denote the (two-sided) ideal of compactly supported functions. I believe that the ring $R:=C_0(\mathbb{R})/C_c(\mathbb{R})$ contains no maximal ideals, but I am having trouble showing it.

My intuition for this problem is as follows. Given a function $f\in C_0(\mathbb{R})$, $f(x)$ approaches zero at some "rate" as $x\to\pm\infty$ (possibly different based on $\pm$). Furthermore, for any given rate, we can find a function with a larger rate, in the sense that we can find a $g\in C_0(\mathbb{R})$ such that $f(x)=o(g(x))$ ($f$ is little-$o$ of $g$). Now, even if $f$ is non-vanishing, there is no $h\in C_0(\mathbb{R})$ such that $fh=g$, for any $h$ could not vanish at infinity. Thus the principal ideal generated by $f$ does not contain $g$. By iterating this process we could construct a strictly ascending chain of principal ideals.

Now, the idea is that the ring $R$ consists of these "rates" as described above. I know that this is not precise, or necessarily even correct. But it's my intuition. The previous paragraph shows that we can find an ascending chain of "rates," but a lot of work still needs to be done. If anyone can clean this up, it would be much appreciated.

• Thank you for your comment, which pointed out a flaw in my answer; I was thinking that maximal ideals in $C_0(\mathbb R)$ also have to be closed, but I don't think that is true. I have deleted it. – Jonas Meyer Jun 6 '12 at 13:09
• What does "vanishing at infinity" mean? Asymptotic to 0? That's what it seems to be... if so you can disregard this comment. – rschwieb Jun 6 '12 at 21:21
• @rschwieb: Yes, that's correct. For a locally compact space $X$, $C_0(X)$ is the space of continuous functions such that for every $\epsilon>0$, the set $\{x\in X\mid \vert f(x)\vert\ge\epsilon\}$ is compact. $C_0(X)$ can be viewed as a maximal ideal in $C(\hat X)$, where $\hat X$ is the one point compactification of $X$. – J. Loreaux Jun 6 '12 at 23:02

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.

• So this certainly answers my first question, but does it immediately answer the second? As I see it, I don't see any way of easily computing the Jacobson radical of $C_0(\mathbb{R})/C_c(\mathbb{R})$. I appreciate the links with all the examples :) – J. Loreaux Jun 6 '12 at 5:07

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.

protected by user26857Aug 31 '15 at 16:44

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