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I'm trying to understand links between Artinian property of a ring $R$ and its spectrum. I suppose that $R$ is noetherian.

This answer is perfect, particulary: TFAE
5) $R$ is Artinian
6) $\operatorname{Spec}(R)$ is discrete
7) $\operatorname{Spec}(R)$ is finite and discrete

It is clear for me that (5) is equivalent with $\operatorname{Spec}(R)$ is $T_1$ i.e. points are closed so I get: (6) $\Rightarrow$ (5)

It is clear that (7) $\Rightarrow$ (6).

Any idea to conclude?

user26857
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Gabriel Soranzo
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    If $R$ is Artinian, all prime ideals are maximal (since an Artinian domain is a field), and there are finitely many of them (take a minimal element of the set of finite intersections of maximal ideals, then prove there are no other maximal ideals). It is then easy to see that $\mathrm{Spec}(R)$ is discrete. – Alvaro Martinez Nov 21 '19 at 21:36
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    Thanks! $\operatorname{Spec}(R)$ is discrete because it's a finite space builded with closed points – Gabriel Soranzo Nov 21 '19 at 21:54

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Let $R$ be artinian, with Jacobson radical $J$. Then $J$ is nilpotent and $R/J$ is product of fields. We can lift these idempotents to $R$, and hence $R$ is a finite product of local rings.

Assume therefore that $R$ is local artinian, with maximal ideal $J$. Since $J$ is nilpotent, it is contained in every prime ideal, so is the unique prime. Thus $\mathrm{Spec}\,R$ is a single point.

Andrew Hubery
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