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By definition, an Artinian ring is a ring that satisfies the descending chain condition on ideals. In practice, how to check if a ring is Artinian? For example, let $R$ be the quotient of the commutative ring $\mathbb{C}[x_1,x_2,x_3]$ the ideal $I$ generated by $x_1^2-1, x_2^2-1, x_3^2-1, (x_1-x_2)(x_1-x_3)$. How to check if $\mathbb{C}[x_1,x_2,x_3]/I$ is Artinian? Thank you very much.

LJR
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    In this case it is a finite dimensional $\mathbb{C}$-algebra. – Gunnar Sveinsson Sep 29 '19 at 15:53
  • You can also look at the Krull dimension of $R$. In this case notice that the height of the ideal $I$ is $3$ since every prime ideal containing it must contain three out of $x_i\pm1$, $i=1,2,3$, and these are maximal ideals. – user26857 Oct 01 '19 at 07:11

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Most examples I'm aware of in practice fall under the following two special cases.

Exercise 1: Every finite-dimensional algebra over a field $k$ is artinian.

Exercise 2: Every finite ring is artinian.

In this case, as Gunnar says in the comments, $R$ is a finite-dimensional algebra over $\mathbb{C}$. You can tell this because, after quotienting by $x_1^2 - 1, x_2^2 - 1, x_3^2 - 1$, the result is spanned by monomials where each $x_i$ occurs with an exponent of at most $1$, of which there are finitely many (in fact exactly $8$): $1, x_1, x_2, x_3, x_1 x_2, x_2 x_3, x_1 x_3, x_1 x_2 x_3$.

Community
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Qiaochu Yuan
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    And there are partial converses: A finitely generated artinian commutative algebra over a field is finite dimensional, and a finitely generated artinian commutative ring is finite. (Actually, I don't know whether the commutative hypotheses are needed...) – Eric Wofsey Sep 30 '19 at 22:18