Special kind of commutative semi-local ring

Question

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 .

Answer 1

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.

Answer 2

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 $(x_1) \subset (x_1,x_2) \subset (x_1,x_2,x_3) \ldots$.