Rings with no Maximal Ideals

In this note we give examples of a ring that has no maximal ideals. Recall that, by a Zorn’s lemma argument, a ring with identity has a maximal ideal. Therefore, we need to produce examples of rings without identity. To help motivate our examples, let S be a ring without identity. We may embed S in a ring R with identity so that S is an ideal of R. Notably, set R = Z⊕S, as groups, and where multiplication is given by (n, s)·(m, t) = (nm, nt+ms+st). It is easy to show that R is indeed a ring, that (1, 0) is the identity of this ring, and that {(0, s) : s ∈ S} is an ideal of R that is isomorphic, as a ring, to S. Thus, any ring without identity may be viewed as an ideal in a ring with identity. We will then search for ideals in rings with identity as candidates for rings with no maximal ideals. The most common example in textbooks of a ring with no maximal ideals is to start with the group Z(p∞), which is the subgroup of Q/Z of elements of order a power of the prime p. It is known that the subgroups of Z(p∞) form an infinite increasing chain, and so there is no maximal subgroup. By defining multiplication in Z(p∞) by x · y = 0, this group becomes a ring with no maximal ideals. However, there are simpler examples of groups with no maximal subgroups. In fact, we prove in the proposition below that any divisible group has no maximal subgroups. To prove the proposition, we prove the following lemma, which is a common problem in an introductory group theory course. A group is said to be simple if it has no nontrivial normal subgroups. If the group is Abelian, then this is equivalent to the group having no nontrivial subgroups. From the fundamental homomorphism theorems, if A is a subgroup of an Abelian group G, then the subgroups of G/A are in 1-1 correspondence with the subgroups of G that contain A. Thus, if A is a maximal subgroup of G, then G/A is simple.