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 February 25th, 2011, 08:01 AM #1 Member   Joined: Jun 2010 Posts: 64 Thanks: 0 Let R be a ring, and for all a in R, a*a*a=a prove that R is Let R be a ring, and for all a in R, a*a*a=a prove that R is commutative ring
 February 27th, 2011, 01:46 PM #2 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 Re: Let R be a ring, and for all a in R, a*a*a=a prove that Interesting problem ... have you got any answer for this ? It seems that exploiting the multiplicative law isn't enough to conclude. Playing with the additive law (group structure), we find that 6a = 0 for every a in R. It doesn't allow us to conclude, though.
 February 28th, 2011, 08:12 AM #3 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: Let R be a ring, and for all a in R, a*a*a=a prove that Hello johnmath. I agree with julien, this is a really interesting (and tough) problem. I have been working on it off and on in my breaks since some time yesterday, and I think I finally have it cracked. There are several observations to make. The first thing to notice is that even powers of a are always equal to a², while odd powers equal a. This is quite clear from the given restriction on the ring. Note also that in our ring, if a² = 0, then a = 0, since 0 = a?a² = a³ = a. The next thing to notice is that a² (and therefore also -a²) commute with every element in the group. This is seen by the following manipulations, where a and b are arbitrary. \begin{align*} (a^2b-a^2ba^2)^2 \ &=\ a^2ba^2b-a^2ba^2ba^2-a^2ba^2b+a^2ba^2ba^2 \ &=\ 0 &\\ (ba^2-a^2ba^2)^2 \ &=\ ba^2ba^2-ba^2ba^2-a^2ba^2ba^2+a^2ba^2ba^2 \ &=\ 0 &\qquad \Rightarrow\\ a^2b \ &=\ ba^2.&& \end{align*} Now we are prepared to see why the manipulation above shows that R is a commutative ring. For a ? R, note that \begin{align*} (a^2+a) \ &=\ (a^2+a)^3 \ =\ (a^2+a)^2+(a^2+a)^2 \qquad \Rightarrow\\ a \ &=\ (a^2+a)^2 + (a^2+a)^2 - a^2 \end{align*} which implies that a commutes with every element of R, since a is a sum of squares, which we know from the previous manipulations each of which commutes with every element of R. I hope this helps out, it was at least a very good exercise! -Ormkärr-
 February 28th, 2011, 08:53 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Let R be a ring, and for all a in R, a*a*a=a prove that ^ Ormkärr is my new hero. I also thought this problem was interesting but didn't have time to attempt a solution. It's not at all clear that I would have succeeded.
 February 28th, 2011, 09:05 AM #5 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 Re: Let R be a ring, and for all a in R, a*a*a=a prove that I searched for this problem in my mathematical library ... it appeared in Bourbaki's Algebra I (kind of a bible and problems there are known to be difficult). Suggested approach by Bourbaki is more algebraic than computational. Great solution, Ormkärr ! I definitely wouldn't have cracked that one in so little time.
 February 28th, 2011, 10:47 AM #6 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Let R be a ring, and for all a in R, a*a*a=a prove that Excellent! (I'm also commenting for my quick future reference to this thread)
 February 28th, 2011, 01:36 PM #7 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: Let R be a ring, and for all a in R, a*a*a=a prove that Thanks for the friendly comments guys, but let me assure you, I spent a good few pages scribbling without much success before I came up with this solution. I dug around a little bit on the Internet just now and found a thread in another math forum (treason, I know!) which seems to validate my approach, although I don't understand what they are using the second step for. There might be, as julien says, many ways to go about this. The comments on the linked page also seem to imply that conditions of this general type can be given to guarantee commutativity. Pretty interesting, if you ask me. -Ormkärr-
 February 28th, 2011, 07:58 PM #8 Senior Member   Joined: Nov 2010 From: Staten Island, NY Posts: 152 Thanks: 0 Re: Let R be a ring, and for all a in R, a*a*a=a prove that Very nice problem, and nice solution Ormkarr. I've had very little experience with rings. It's interesting how such a simple-looking problem could be so difficult.
 February 2nd, 2013, 09:18 AM #9 Global Moderator   Joined: Dec 2006 Posts: 20,484 Thanks: 2041 See this.

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