Example of a ring without unity that does not have a maximal ideal?

Question

This is a followup question.

I know that if we assume Zorn's lemma, any ring with unity has a maximal ideal. I also know that this may not be true for a ring without unity.

Does anyone have an example of a ring without a maximal ideal, preferably with a "nice, simple" construction that doesn't involve the Axiom of Choice?

AnyMouse: well of course any ring that is a field has only one proper ideal, which is maximal: { 0 }. Also, recall that a maximal ideal of a ring cannot be the ring itself, by definition; a maximal ideal must be a proper ideal. Right? :)

Answer 1

I've had to ponder this question for some homework before. Here is a link to an excerpt from that assignment http://www.math.uiuc.edu/~tlesaul2/Yahoo.pdf
It gets a little wordy. The idea is to define the ring trivially by making all multiplications 0. Note that there is no multiplicative identity. Now an ideal is just an additive subgroup so you can start with any group with no maximal subgroups. You can do this with a lot of common infinite groups but for some of them its harder to prove that they don't have a maximal subgroup. It all sounds a little cheap but I asked my prof about it and this is exactly what he did as well.

Answer 2

I'm bothered by your question -- because I know there was a time when I could answer it. But it has been more than 30 years since I studied Ring Theory.

So -- I will throw something out ther that may be a solution -- but might also be way off.

Consider the set of numbers made up of ordered pairs (a,b) -- where a is in the set of even integers and b is in the set of integers that are a multiple of three. Define addition as pairwise addition, and multiplication as pairwise multiplication.

I'm pretty sure this defines a ring, and know it has no multiplicative identity. I think it also has no maximal ideals -- but I will leave it to you to test it out.

Answer 3

\R itself doesn't have a nondegenerate maximal ideal. ( The only ideals are {0} and \R itself )

Assume that I is a nontrivial maximal ideal in \R. Let x be a nonzero element of I. Then, as I is an Ideal, (1/x) * x = 1 is in I.
Thus I = \R.

The "maximal ideal" you obtain when you utilize Zorn's Lemma could be the entire ring itself...

Answer 4

Um the ring my ex gave me really didn't signify unity hehe