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From http://en.wikipedia.org/wiki/Ascending_chain_condition:

"In mathematics, the ascending chain condition (ACC) and descending chain condition (DCC) are finiteness properties satisfied by some algebraic structures, most importantly, ideals in certain commutative rings.[1][2][3]

Notes

1 ^ Hazewinkel, Gubareni & Kirichenko (2004), p.6, Prop. 1.1.4.
2 ^ Fraleigh & Katz (1967), p. 366, Lemma 7.1
3 ^ Jacobson (2009), p. 142 and 147

References

    Atiyah, M. F., and I. G. MacDonald, Introduction to Commutative
    Algebra, Perseus Books, 1969, ISBN 0-201-00361-9
        Michiel Hazewinkel, Nadiya Gubareni, V. V. Kirichenko. Algebras, rings and modules. Kluwer Academic Publishers, 2004. 
    ISBN 1-4020-2690-0

        John B. Fraleigh, Victor J. Katz. A first course in abstract algebra. Addison-Wesley Publishing Company. 5 ed., 1967. 
    ISBN 0-201-53467-3

        Nathan Jacobson. Basic Algebra I. Dover, 2009. ISBN 978-0-486-47189-1"

I have the 7th edition of Fraleigh and Katz but can't find the equivalent reference. I'm nearly sure that in another Wikipedia page it spells out what kind of ideals and commutative (semi?)rings but I can't find it any more (tab deleted somehow and can't get it back). It also spells out why this is true. Can anyone please point me to any references?

Thanks very much. (Sorry about the references to Wikipedia pages.)

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  • $\begingroup$ Personally I think the Atiyah and MacDonald chapters on "Chain Conditions","Noetherian Rings" and "Artin Rings" are the cleanest exposition of the basic theory. $\endgroup$ Aug 20, 2013 at 17:31
  • $\begingroup$ What I meant was an explanation of why these structures have the chain conditions short of a formal proof $\endgroup$
    – mike
    Aug 20, 2013 at 17:41
  • $\begingroup$ @mike For both rings and semirings, there are ones that do and ones that do not have each of ACC and DCC on their ideals. There is no proof that "those structures have the chain conditions." $\endgroup$
    – rschwieb
    Aug 20, 2013 at 17:46
  • $\begingroup$ @rshwieb I suppose that follows directly from the related posets then? $\endgroup$
    – mike
    Aug 20, 2013 at 17:53
  • $\begingroup$ @mike Yeah, you would have to prove whether or not a ring has the ACC or DCC on ideals. For example, a field is Artinian and Noetherian, the polynomial ring $\Bbb Z[x]$ is Noetherian but not Artinian, and a product of infinitely many nonzero rings is not Noetherian or Artinian. $\endgroup$
    – rschwieb
    Aug 20, 2013 at 18:09

2 Answers 2

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The ACC and DCC are conditions on a poset, not on the elements of the poset. For most algebraic objects, the important subobjects form a poset under inclusion. There are examples of rings, modules, groups and semirings which do have and which do not have the ACC and/or DCC.

Commutative rings whose ideals satisfy the ACC are called Noetherian rings. Commutative rings whose ideals satisfy the DCC are called Artinian rings.

I'm pretty sure the identical definitions are carried over for ideals of semirings.

For noncommutative rings, there is a notion of right/left Artinian/Noetherian based on the lattice of right ideals and the lattice of left ideals.

Groups with the ACC on subgroups are called Noetherian, and groups with the DCC on subgroups are called Artinian. The same goes for modules and their submodules.

More generally lattices, which are also posets, can be labeled Artinian/Noetherian according to whether or not they have the DCC/ACC on their members.

Bonus: There is one freaky thing you might like, however. It turns out that for rings, if the ring satisfies the DCC on right ideals, it also satisfies the ACC on right ideals. This is the Hopkins-Levitzki theorem, if you ever have the time to study it :) It's false for groups and modules. I'm not sure if it's false for semirings or not!

Non algebraic example: A topology is called Noetherian if it satisfies the ACC on open sets. I've never actually seen Artinian topological spaces studied, but a quick google search yields hits, so maybe that exists too.

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    $\begingroup$ Thanks that makes a lot of sense in terms of what I've read already. $\endgroup$
    – mike
    Aug 20, 2013 at 17:42
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    $\begingroup$ Thanks for the edit. That clarifies some confusion I had. $\endgroup$
    – mike
    Aug 20, 2013 at 17:47
  • $\begingroup$ @mike Glad to help :) Wasn't sure if I was heading the right way or not. $\endgroup$
    – rschwieb
    Aug 20, 2013 at 18:00
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For example, you can check these notes.

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  • $\begingroup$ Thanks very much for the lightning response. It will take me a while to digest. Can you tell me the reference for the notes please? $\endgroup$
    – mike
    Aug 20, 2013 at 17:37
  • $\begingroup$ This is chapter 13 in the course math.northwestern.edu/~len/d70/index.html . $\endgroup$ Aug 20, 2013 at 17:40
  • $\begingroup$ Thanks for the reference. This is all great! $\endgroup$
    – mike
    Aug 20, 2013 at 17:46

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