# Talk:Artinian ring

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## Counterexample

The article presently claims that R is Artinian iff R/rad(R) is a direct product of finitely many fields. This is false; to construct a counterexample, let k be a field, and let k[x1, x2, x3, ...] be a polynomial ring over k in infinitely many indeterminates. Let J be the ideal (x1) + (x1, x2)2 + (x1, x2, x3)3 + ..., and let R = k[x1, x2, x3, ...]/J.

R is clearly not Noetherian: (x1), (x1, x2), (x1, x2, x3), ... is an infinite ascending chain of ideals. Nor is R Artinian: (x1), (x1x2), (x1x2x3), ... is an infinite descending chain of ideals. But R/rad(R) is k: Every monomial in x1, x2, ... is nilpotent by the definition of J, hence every polynomial is nilpotent, and so rad(R) = (x1, x2, x3, ...).

I suspect that the correct statement is that R is Artinian iff R is Noetherian and R/rad(R) is a product of finitely many fields, but I don't immediately see how to prove this. For the moment I've removed all mention of this result from the article. Ozob (talk) 05:25, 19 December 2008 (UTC)

You know, you're right and what's funny is that I wrote this, considered looking it up, and decided I remembered it correctly. And now you come up with this after less than a day. Here's a sketch of a proof (with A an artinian ring):
1. Prove that every prime ideal in A is maximal
2. Prove that A has only finitely many prime ideals
3. Prove that rad(A) is nilpotent (obviously it contains only nilpotent elements; show that rad(A)k = 0 for some k)
4. Prove the following lemma: if M1, ..., Mn are maximal ideals (in some ring R) whose product is zero, then R is Noetherian if and only if it is Artinian (look at the successive quotients in the chain of partial products of the Mi).
5. Thus A Artinian implies A Noetherian and A / rad(A) is a product of finitely many fields. Conversely. the lemma shows that such a ring is automatically Artinian.
This proof comes out of Atiyah-MacDonald, which I'll add back to the article as a ref. Thanks for getting on my case! Ryan Reich (talk) 05:52, 19 December 2008 (UTC)
I changed rad to nil to avoid confusion with the jacobson radical. R/J(R) artinian semisimple is called semilocal and is a very common hypothesis. For instance, a semiartinian ring is semilocal with J(R) T-nilpotent, and semiperfect if it is semilocal and idempotents lift mod J(R), and artinian iff it is semilocal and J(R) is nilpotent. Presumably the last one is mildly interesting in the commutative case. JackSchmidt (talk) 06:29, 20 December 2008 (UTC)

## ${\displaystyle \operatorname {Spec} A}$ is finite and discrete

In the section "Commutative Artinian rings", we can read :

• ${\displaystyle \operatorname {Spec} A}$ is finite and discrete.
• ${\displaystyle \operatorname {Spec} A}$ is discrete.

Is it important here to repeat two times that ${\displaystyle \operatorname {Spec} A}$ is discrete ?

--160.178.210.85 (talk) 18:16, 29 November 2016 (UTC)

@User talk:160.178.210.85 Well, I can easily see the equivalence to the first line, but the equivalence to the second line seems like a bit more work. To me, that makes both sets of conditions interesting and valuable. Are you saying that you'd delete the first bullet point you listed above? Rschwieb (talk) 19:12, 29 November 2016 (UTC)
@User:Rschwieb I'm sorry. I misread this part of the article. I thought that this section was just stating something like "the following bullet points are true for Artinian rings", in which case saying that "${\displaystyle \operatorname {Spec} A}$ is finite and discrete" is enough and no need to repeat that "${\displaystyle \operatorname {Spec} A}$ is discrete". --160.178.210.85 (talk) 23:15, 29 November 2016 (UTC)