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Elements of a ring problem

  1. Oct 6, 2014 #1
    1. The problem statement, all variables and given/known data
    A) If ab+ba = 1 and a^3 = a in a ring, show that a^2 = 1

    2. Relevant equations
    none

    3. The attempt at a solution
    Little confused. If we know that a^3 = a, can't we just multiply each on the right or left side by a^-1 to get a^2 = 1? Or could we only do that if the ring is said to be commutative?

    edit: I realized that to show that a^2 = 1 in the way I mentioned above that a would have to be a division ring. Normal rings aren't guaranteed to have multiplicative inverses.
     
    Last edited: Oct 6, 2014
  2. jcsd
  3. Oct 6, 2014 #2

    RUber

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    I would start by multiplying the first equation by an a on the left, then do the same with an a on the right. Compare the resulting equations. Then there should only be one reasonable conclusion that satisfies ##a^3=a## as well.
     
  4. Oct 6, 2014 #3
    So I get b + aba = a and aba + b = a, (considering a^2=1) is there something obvious that I'm missing here? The two equations are the same thing are they not?
     
    Last edited: Oct 6, 2014
  5. Oct 6, 2014 #4

    RUber

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    ## ab+ba = 1,## so left multiplying gives ## a^2 b + aba = a## and right multiplying gives ## aba+ba^2=a##. Moving ##aba## to the other side shows that ##a^2b = a-aba = ba^2##.
    Similarly, you could right/left multiply ##a^2## to show that ## a^3b+a^2ba=a^2 = aba^2 + ba^3 ##. Substituting the fact that ## a^2b=ba^2 ## should allow you to cancel out some (edit) ##a^3##s.
     
    Last edited: Oct 6, 2014
  6. Oct 6, 2014 #5

    RUber

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    I would not assume what you are trying to show at this point.
     
  7. Oct 6, 2014 #6

    Fredrik

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    There aren't many many ways to use the two given equations. Let's start with ##ab+ba=1##. What can we do with this other than to insert the assumption ##a=a^3##? This gives us ##1=ab+ba=a^3b+ba^3##. How can we rewrite the right-hand side? Replacing ##a## with ##a^3## again doesn't look very promising, and replacing ##a^3## with ##a## just takes us back to where we started. So we're almost certainly going to have to use ##ab+ba=1## (possibly more than once) to rewrite the right-hand side.
     
  8. Oct 6, 2014 #7
    Update:

    I now have four (seven?) equations I'm moving around.

    1) a = (a^2)*b + a*b*a
    2) a = aba + b(a^2)
    3) 1 = (a^3)b + b(a^3)
    4) 1 = ab + ba

    Hence...
    5) ab + ba = (a^3)b + b(a^3)
    6) aba + b(a^2) = (a^2)*b + a*b*a

    I even multiplied 5 and 6 together to try to get something. Got messy.

    Hmmm.
     
  9. Oct 6, 2014 #8

    RUber

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    From 1 and 2 you have ##a^2b=ba^2##.
    Multiplying 4 by ##a^2## gives either ## a^2 = a^3b+a^2ba ## or ## a^2 = aba^2+ba^3##.
    Using the facts that ##a^3 = a## and ##a^2b=b^2a## you can reduce this down to something equivalent to ##a^2 = 1##.
     
  10. Oct 6, 2014 #9

    Fredrik

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    I was able to solve it using RUber's suggestion, but I also found a way to solve it without ever multiplying an equation by anything. So one piece of advice is to resist the temptation to multiply things together.
     
  11. Oct 6, 2014 #10
    Wow I got it. Thanks a ton for your help I appreciate it. Totally got it before Fredriks last comment :P. I'm studying this stuff independently cuz i'm a Math nerd and it's tough! It's fun though.
     
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