# Elements of a ring problem

## Homework Statement

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

none

## 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:

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?

Last edited:
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.

Last edited:
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
• 