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Let $R$ be a commutative semi-local ring (finitely many maximal ideals) such that $R/P$ is finite for every prime ideal $P$ of $R$ ; then is it true that $R$ is Artinian ring ? From the assumed condition , we get that $R$ has Krull dimension 0 ; so it is enough to ask : Is $R$ a Noetherian ring ? From the semi-local and $0$ Krull dimension condition , it also follows that $R$ has finite Spectrum . But I am unable to say whether all this really implies $R$ is Noetherian or not .

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Take $V=\oplus_{i=1}^\infty F_2$ and form the ring

$$ R= \left\{\begin{bmatrix}a&v\\0&a\end{bmatrix}\middle|\,a\in F_2, v\in V\right\} $$

It isn't noetherian because the image of $V$ contains infinite ascending chains of ideals. It's also local (with residue field $F_2$) and $0$-dimensional.

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As you say, the ring is zero dimensional because each $R/P$ is a finite domain, hence a field. Hence we get a map $R \to \prod R/m_i$, running over the maximal ideals. The target here is a product of finite fields, hence the image is Noetherian. The kernel is the nilradical of $R$, since the nilradical is the intersection of all primes, which is in this case reduces to the intersection of all maximal ideals.

So if $R$ is reduced, we are done.

If you can prove that the nilradical in your case is Noetherian, then you can conclude from the 2-3 theorem for Noetherian property in exact sequences.

I think you can get a non-reduced counter example like this:

$R = \mathbb{F}_p[x_1, x_2, \ldots, x_n, \ldots] / (x_1, x_2, \ldots)^2$.

This is a dimension zero local ring and the quotient by the maximal ideal is the finite field $F_p$.

(In words: $R$ is the quotient of the countably infinite dimensional polynomial ring over $F_p$ by the square of the maximal ideal corresponding to zero.)

However, there is an infinite ascending chain of ideals $(x_1) \subset (x_1,x_2) \subset (x_1,x_2,x_3) \ldots$.

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  • $\begingroup$ The example you worked out should be correct... I think it is even isomorphic with the thing I gave. But it might be a more comfortable description for some to see :) $\endgroup$ – rschwieb Aug 25 '17 at 16:45
  • $\begingroup$ @Area Man : What If we also assumed $J(R)=\{0\}$ ? Would the ring be Noetherian then ? $\endgroup$ – user Aug 25 '17 at 16:54
  • $\begingroup$ @users What is $J$? Being reduced means that the nilradical is trivial, which is the kernel of the map I describe in the first paragraph. $\endgroup$ – Lorenzo Aug 25 '17 at 16:58
  • $\begingroup$ @AreaMan : $J(R)$ means the Jacobson radical ... ah yes I see that in that case $R$ has to be Noetherian and even more .. $\endgroup$ – user Aug 25 '17 at 17:00
  • $\begingroup$ @rschwieb I agree they seem isomorphic, your ring is generated by 1 and a countable collection of elements that annihilate each other and square to zero. Interesting... it's much more comfortable for me to think in terms of polynomial rings than matrix rings (in matrix rings I always have to pause and think about how the multiplication works), at least when the relations in the ideal are simple enough so that there are no suprises. :) $\endgroup$ – Lorenzo Aug 25 '17 at 17:05

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