It is well-known that every semisimple ring (with identity) is Artinian. But it seems that semisimple rngs (without identity) may not be Artinian.
Definition of semisimple ring: A ring/rng is semisimple if it is a semisimple module (which is the direct sum of simple modules) over itself.
Here is an elementary counterexample:
Let $R = \{S|S \subsetneqq \mathbb{Z}, |S| < \infty\}$ be a set whose elements are finite sets of integers, i.e., $R \subsetneqq 2^\mathbb{Z}$. Addition and Multiplication of $R$ are defined as symmetric difference and intersection, respectively. Then, $R$ forms a communicative boolean rng without an identity. The zero element of $R$ is the empty set $\emptyset$. $R$ is not a finitely generated module over itself.
$R$ is semisimple as $R = \oplus_{n \in \mathbb{Z}} \{\emptyset, \{n\}\}$, i.e., $R$ is the direct sum of simple modules that consist of two elements, one is the empty set and another one is a singleton. To show that $R$ is not Artinian, consider $T \subset \mathbb{Z}$ and define $I(T) := \{S|S \subsetneqq T \subset \mathbb{Z}, |S| < \infty\}$. Then $I(T)$ is an ideal of $R$. If $S \subsetneqq T$, then $I(S) \subsetneqq I(T)$. Take $T_{k \geq 0} = 2^k\mathbb{Z}$ ($T_k$ is the set of integers divided by $2^k)$, then $\{I(T_0), I(T_1), I(T_2), …\}$ forms an unstable descending chain of ideals of $R$, so $R$ is not Artinian.
My questions are:
(1) Is my conclusion "Semisimple rng may not be Artinian" correct?
(2) Is my counterexample correct?
Best Answer
Seems all right. To clarify a bit, this is a more general framework:
Given simple rings $R_1, R_2, \cdots, $ (in your case $R_i\simeq \mathbb Z/2\mathbb Z$), then the rng $\oplus_{i=0}^\infty R_i$ is semi-simple but not artinian, as $I_n = \oplus_{i=n}^\infty R_i$ forms an infinite proper descending chain of ideals.