# Elements of a ring problem

## Homework Statement

A) If ab+ba = 1 and a^3 = a in a ring, show that a^2 = 1

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## 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.

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RUber
Homework Helper
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.

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?

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RUber
Homework Helper
## 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.

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RUber
Homework Helper
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?
I would not assume what you are trying to show at this point.

PsychonautQQ
Fredrik
Staff Emeritus
Gold Member
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.

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.

RUber
Homework Helper
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##.

Fredrik
Staff Emeritus