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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 $(\frac{1}{n_1}, \dots , \frac{1}{n_k}) = (\frac{1}{\text{lcm}_i(n_i)})$).

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

Is this correct?

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How is $\mathbb Q / \mathbb Z$ a ring? – lhf Jul 21 '12 at 15:32
Related to…. – 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
What is the question? – Qiaochu Yuan Jul 21 '12 at 18:20
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 $k[x]/(x^2)$.

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