2

Is there some sorts of Krull's theorem (that every ring has maximal ideal) for rings that do not have multiplicative identity (unit)? So I know that non-unital rings do not satisfy Krull's theorem, but for some types of non-unital rings, theorem does get satisfied. So what is it?

Edit: Wikipedia seems to mention the case with regular ideal, but does not explain it.

Kampas
  • 39
  • You really have to say what sort of result you're expecting. Some rngs will have maximal ideals, some will not. – rschwieb Aug 13 '14 at 13:01

2 Answers2

2

As you noted, there is no version of Krull's theorem that asserts the existence of maximal ideals in all rngs. The best you can hope for is truth in certain subclasses of rngs. One obvious example is if you ask for the maximal condition on right/left or two-sided ideals.

Here's the result that Wikipedia is referring to in Jacobson's Structure of rings page 6.

enter image description here

Of course, we've got no guarantee a proper modular right ideal exists in a given rng. Not even $\{0\}$ is guaranteed to be modular. In fact, $\{0\}$ being a modular right and left ideal is exactly saying that the ring has an identity.

rschwieb
  • 160,592
0

I found a theorem given in David Burton's book: "A first course in rings and ideals" where he proves a "non-unitary" version of Krull theorem. It says that every nonzero finitely generated ring $R$ has a maximal ideal. The proof is essentially the same than for the usual version of Krull theorem. So this result works for non-unitary rings, but actually this is just a particular case of Krull theorem for modules: Maximal submodule in a finitely generated module over a ring

Community
  • 1
Xam
  • 6,338