For a commutative Artinian unital ring, it is well known that every ideal contains at least a minimal ideal, a non-zero ideal that dose not contain a proper non-zero ideal. In general, not Artinian case, are rings with above property important or are they a famous class of rings, or are there any characterization or description for them?

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If every nonzero right ideal of a ring $R$ contains a minimal right ideal, you usually express this by saying that $R$ has an essential right socle.

This expression is frequently used in the literature: try for example this query in google books.

I am not aware of alternative characterizations of this, although it is used in conjunction with other conditions to make characterizations. I think I saw it frequently used in Faith's work on pseudo-Frobenius and finitely-pseudo-Frobenius rings.

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If $A$ is a noetherian commutative domain and is not a field, then $A$ has no minimal non-zero ideal at all.
Let $\mathfrak a$ be a minimal ideal. We have $\mathfrak a^2=\mathfrak a$ and thus by NAKAYAMA there exists $e\in \mathfrak a$ with $a=ea$ for all $a\in \mathfrak a$. But then $e=e^2$, hence $e=0$ or $e=1$ so that $\mathfrak a=0$ or $\mathfrak a=A$.

Geometric interpretation
Given an integral noetherian affine scheme $X$, no closed subscheme $\emptyset \subsetneq Y\subsetneq X$ is maximal .

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  • $\begingroup$ Or positively formulated: Every noetherian commutative domain which has a minimal ideal is a field. $\endgroup$ – HeinrichD Dec 17 '16 at 11:10
  • $\begingroup$ Seems a bit overkill! If $aR=(aR)^2$, then $a=aras$ for some $r,s\in R$, and cancelling you get $1=ras$ and $a$ is a unit. So a (possibly noncommutative, possibly nonNoetherian) domain with a minimal ideal is a division ring. I understand the additional hypotheses might be necessary for the geometric interpretation, but Nakayama is certainly not necessary for the proof :) $\endgroup$ – rschwieb Dec 17 '16 at 14:25
  • $\begingroup$ I don't think any justification is necessary, and I found it to be really interesting! I just didn't want anyone to get a false impression about the conditions or difficulty of proving this phenomenon. $\endgroup$ – rschwieb Dec 17 '16 at 18:46
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    $\begingroup$ Dear @schwieb: I agree that your calculation is more elementary and thus more satisfying. I'm just so used to Nakayama that I hadn't thought about doing an explicit calculation... $\endgroup$ – Georges Elencwajg Dec 17 '16 at 18:51

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