Prove something about Boolean ring (1) Prove that for a Boolean ring $R$, the following are equivalent:
(a)$R$ is artinian; (b) $R$ is noetherian; (c) $R$ is finit; (d) $R$ is semisimple.
(2) Prove that if $_RM$ is an artinian or noetherian module over a Boolean ring $R$, then $M$ is semisimple.
 A: I guess you are probably thinking (a) implies (b) could be deduced from the Hopkins-Levitzki theorem, which is true. We can try to work around that, though, if you are sufficiently familiar with semisimple rings.
I'm hoping you know the basic characterizations of semisimple rings. Specifically, that they are finite products of full matrix rings over division rings, and that they are Artinian rings with Jacobson radical zero, and that all right ideals are direct summands.
Lemma #1: Boolean rings are commutative. You've probably already proven this, but if not, it is asked and answered many times on this site.
Lemma #2 : Prove Boolean rings have Jacobson radical zero. (Use the characterzation that $x\in J(R)$ iff $1-xr$ is a unit for all $r\in R$. Boolean rings are not friendly places for units.)
Lemma #3: Prove finitely generated right ideals of Boolean rings are principal. (This is also solved many times on the site, but I'll give you the usual hint: compare $(a,b)$ to $(a+b-ab)$ and apply induction.)
Hint #1: By Artin-Wedderburn, a commutative semisimple $R$ is a finite product of fields. Since it is boolean, these fields have to be boolean as well. How many boolean fields are there? (With this, you should conclude $R$ is finite.)
Hint #2: "Finite implies Artinian", of course.
Hint #3: "Artinian implies semisimple" should be easy given lemma 2 and the characterization of semisimple rings. You would now have established the equivalence of everything except "Noetherian."
Hint #4: Now show a Noetherian Boolean ring is semisimple. Let $T$ be a right ideal. Since $R$ is Noetherian, $T$ is finitely generated. Since $R$ is Boolean, $T$ is principal. But the element that generates it is idempotent... can you see why $T$ is a summand of $R$? Conclude $R$ is semisimple.
Hint #5: Since semisimple-->finite-->Noetherian, you now have "Noetherian" as one of the equivalent conditions.
