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(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.

mazicai
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  • A Boolean ring is a ring $R$ with $a^2=a$ for each a belongs to $R$. – mazicai Jan 14 '14 at 14:26
  • I know that the proof from (a) to (b) is obvious by some theorem, but that is what I will learn latter, and it doesn't use the condition that $R$ is a Boolean ring. – mazicai Jan 14 '14 at 14:30
  • I am a first learner,thanks for any detail proof or hint – mazicai Jan 14 '14 at 14:32

1 Answers1

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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.

rschwieb
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  • Thanks for your detail hints,in fact,the question above is the last question in GTM13, page 133, and the Jacobson radical is the content in next chapter, anyway thanks again for your answer.And I will learn it by myself. – mazicai Jan 14 '14 at 15:49
  • @马紫菜 Whenever you pull a problem from a specific source, it's a good idea to include the source in your post. I might have answered even more accurately if I had known the source. – rschwieb Jan 14 '14 at 16:11
  • Ok,I will include the sourse in my post next time. – mazicai Jan 15 '14 at 00:06
  • I have just understood the first small question, but I'm still confused with the second one.I know that $Rad M=0$ and $M$ is artinian implies that $M$ is semisimple,whether can I get $Rad M=0$ from the condition or not? If not,how can I figue it out? Thanks! – mazicai Jan 17 '14 at 01:07
  • @马紫菜 I gave a hint for that on lemma 2... Did you try it? – rschwieb Jan 17 '14 at 03:34
  • Yes, I tried it.And I figued it out.Thank you. – mazicai Jan 17 '14 at 16:07