My Math Forum  

Go Back   My Math Forum > College Math Forum > Abstract Algebra

Abstract Algebra Abstract Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
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
johnmath is offline  
 
February 27th, 2011, 01:46 PM   #2
Site Founder
 
julien's Avatar
 
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.
julien is offline  
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 , 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 (and therefore also -a²) commute with every element in the group. This is seen by the following manipulations, where a and b are arbitrary.



Now we are prepared to see why the manipulation above shows that R is a commutative ring. For a ? R, note that



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-
 is offline  
February 28th, 2011, 08:53 AM   #4
Global Moderator
 
CRGreathouse's Avatar
 
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.
CRGreathouse is offline  
February 28th, 2011, 09:05 AM   #5
Site Founder
 
julien's Avatar
 
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.
julien is offline  
February 28th, 2011, 10:47 AM   #6
Global Moderator
 
The Chaz's Avatar
 
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)
The Chaz is offline  
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-
 is offline  
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.
DrSteve is offline  
February 2nd, 2013, 09:18 AM   #9
Global Moderator
 
Joined: Dec 2006

Posts: 20,484
Thanks: 2041

See this.
skipjack is offline  
Reply

  My Math Forum > College Math Forum > Abstract Algebra

Tags
aaaa, prove, ring



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Prove a set is a commutative ring HairOnABiscuit Abstract Algebra 9 August 30th, 2012 09:13 PM
Ring and pseudo-ring cgouttebroze Abstract Algebra 5 August 14th, 2008 12:04 PM
prove prove prove. currently dont know where to post qweiop90 Algebra 1 July 31st, 2008 06:27 AM
Ring! payman_pm Abstract Algebra 3 September 23rd, 2007 11:09 AM
prove prove prove. currently dont know where to post qweiop90 New Users 1 December 31st, 1969 04:00 PM





Copyright © 2019 My Math Forum. All rights reserved.