Quotient of infinite product of fields which is semi-artinian

Question

Let $I$ be an infinite set (e.g. $I=\mathbb{N}$) and let $k$ be a field. The ring $R=\prod_I k$ is a notorious example of a ring which is absolutely flat (alternatively said von Neumann regular) and not semi-artinian. Its Zariski spectrum corresponds to the set of ultrafilters of $I$, endowed with the Stone topology. These are all known results explained in various questions here on Math StackExchange.

My question is the following:

We consider $J=\sum_I k$, the ideal consisting of the direct sum of the fields. Is the quotient $R/J$ semi-artinian?

I would think that this is false, but I am not sure how to prove it since I am not familiar at all with semi-artinian rings.

An associated question is:

Does there exist an ideal $L\subset R$ such that the quotient $R/L$ is nontrivially semi-artinian?

Here nontrivially means that the Zariski spectrum of $R/L$ should not be finite. This avoids cases as taking $L$ to be a maximal ideal so that $R/L\cong k$. I believe that if I can answer the first question the solution to second should follow relatively easily.

Answer 1

No, $R/J$ is not semi-artinian. In fact, $R/J$ has no minimal non-zero ideals, so considered as a module over itself, it has no non-zero simple submodule.

Suppose for contradiction that $J'$ is a minimal non-zero ideal in $R/J$. Then $J'$ pulls back to an ideal $J''\subseteq R$ which is minimal among those properly containing $J$. Since $J\subsetneq J''$, there is some $x = (x_i)_{i\in I}\in J''\setminus J$. Let $X\subseteq I$ be the support of $x$, $X = \{i\in I\mid x_i \neq 0\}$. Since $x\notin J$, $X$ is infinite. Now let $Y\subseteq X$ be an infinite and coinfinite subset of $X$, and let $y = (y_i)_{i\in I}$, where $y_i = 1$ if $i\in Y$ and $y_i = 0$ otherwise. Since the support of $y$ is $Y$, which is infinite, $y\notin J$. And for any $z\in J$ and $r\in R$, the support of $z+ry$ contains at most finitely many elements not in $Y$, so $x\notin J+\langle y\rangle$. Thus $J\subsetneq J+\langle y\rangle \subsetneq J''$, contradicting minimality of $J''$ among ideals properly containing $J$.

Answer 2

I think I found a solution for my second question. I am posting it here for those interested. The conclusion is that the answer is no. For any ideal $L\subset R$ the quotient $R/L$ will be either finite or it will not be semi-artinian.

For now, I will consider $R$ to be a generic absolutely flat ring. I will need two non-trivial notions. The first one is the Loewy series of a module (see https://en.wikipedia.org/wiki/Loewy_ring). I will denote the socle of a $R$-module $N$ by $s(N)$, and for the Loewy series I will use $s_{\alpha}(N)$ to refer to its $\alpha$-th step (here $\alpha$ is an ordinal number).

The second notion is the one of Cantor-Bendixon rank (https://en.wikipedia.org/wiki/Derived_set_(mathematics)) of a topological space. I will denote the Cantor-Bendixon derivative of a space $X$ by $\delta X$, then for a general ordinal $\alpha$ the space $\delta^{\alpha}X$ will be the $\alpha$-th iterated derivative of $X$.

First of all, it holds that a ring $S$ is semi-artinian if and only if the Loewy series of $S$ considered as module over itself converges.

It can be proved that for an absolutely flat ring $R$ there is a bijection between its non-zero minimal ideals and the isolated points of $\text{Spec}(R)$. If we start from a simple ideal $I$, the condition of $R$ being absolutely flat implies we can write $I=(a)$ with $a\in R$ idempotent. Then we have $R=(a)\oplus (1-a)$ and we can show $J=(1-a)$ is a maximal ideal. It easily follows $\{ J\}=U(a)$, hence this is an isolated point. Vice versa, if we start from $J$ an isolated point in the Zariski spectrum, since $U(r)$ for $r\in R$ provides a basis for the topology we must have $\{J\}=U(a)$ for some element $a$. Again, using $R$ is absolutely flat, we deduce $a$ can be taken to be idempotent. From here, we can prove that $J=(1-a)$, $R=(a)\oplus J$ and maximality of $J$ implies $(a)$ is a simple ideal.

From this bijection, we can deduce that $\text{Spec}(R/s(R))\cong \{ J\in \text{Spec}(R) : s(R)\subseteq J \}$ coincides with $\delta \text{Spec}(R)$. That is, the requirement of including the socle implies that the ideal $J$ must contain all the simple ideals of $R$, hence we are eliminating the isolated primes.

This homeomorphism can be generalized by transfinite induction to $\text{Spec}(R/s_{\alpha}(R))\cong \delta^{\alpha}\text{Spec}(R)$ for any ordinal $\alpha$.

Thus the absolutely flat ring $R$ being semi-artinian is equivalent to the existence of an ordinal $\alpha$ such that $\delta^{\alpha}\text{Spec}(R)=0$. That is, its Zariski spectrum must be scattered.

Now, let us go back to $R=\prod_{\mathbb{N}}k$. If $L\subset R$ is an ideal, we have $\text{Spec}(R/L)$ consists in the closed subset of $\text{Spec}(R)$ given by the ideals containing $L$. As mentioned above, we have $\text{Spec}(R)\cong \beta \mathbb{N}$ is the space of ultrafilters on the natural numbers with the Stone topology. We have that the isolated points of $\beta \mathbb{N}$ coincide with the principal ultrafilters, hence $\delta \beta \mathbb{N}$ consists in the non-principal ultrafilters. This subspace is actually perfect, i.e. $\delta^2 \beta \mathbb{N}= \delta \beta \mathbb{N}$.

It is known fact that the closed subsets of $\beta \mathbb{N}$ are either finite, or they contain a topological copy of $\beta \mathbb{N}$. A space containing a perfect subspace cannot be scattered. Thus, in the second case $\text{Spec}(R/L)$ cannot be scattered, hence $R/L$ cannot be semi-artinian.