Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

By definition a ring $R$ is Artinian if it is Artinian as $R$-module. I think the following is an example of an Artinian ring:

$\mathbb Q / \mathbb Z$

Ideals in it are of the form $(\frac1n)$ (since $(\frac{1}{n_1}, \dots , \frac{1}{n_k}) = (\frac{1}{lcm_i(n_i)})$).

Since we have $(\frac1n) \subset (\frac1m)$ if and only if $n$ divides $m$, every decreasing chain stabilises eventually since $n$ only has finitely many divisors.

in response to a comment: is this correct?

share|improve this question
3  
How is $\mathbb Q / \mathbb Z$ a ring? –  lhf Jul 21 '12 at 15:32
1  
Related to math.stackexchange.com/questions/142597/…. –  lhf Jul 21 '12 at 15:34
    
I'm confused :,( –  Rudy the Reindeer Jul 21 '12 at 16:45
    
Or is the question whether $\mathbb Q / \mathbb Z$ is an Artinian $\mathbb Z$-module? –  lhf Jul 21 '12 at 17:56
1  
What is the question? –  Qiaochu Yuan Jul 21 '12 at 18:20

1 Answer 1

up vote 3 down vote accepted

You'd have to tell me how $\mathbb Q/\mathbb Z$ is a ring. The notation suggests that $\mathbb Z$ is an ideal of $\mathbb Q$ (which it isn't) and that we're forming the factor ring.

One way to make commutative Artinian rings is to fix a field $k$ and look at finite dimensional $k$-algebras. Geometrically these correspond to finite sets of points in affine space. In the classical setting these will look like $\prod k$, but it's interesting to think about stuff like $\mathbb R[x]/(x^2 + 1) \simeq \mathbb C$ and $k[x]/(x^2)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.