Elements of a ring problem

1. Oct 6, 2014

PsychonautQQ

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. Oct 6, 2014

RUber

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.

3. Oct 6, 2014

PsychonautQQ

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
4. Oct 6, 2014

RUber

$ab+ba = 1,$ 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
5. Oct 6, 2014

RUber

I would not assume what you are trying to show at this point.

6. Oct 6, 2014

Fredrik

Staff Emeritus
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.

7. Oct 6, 2014

PsychonautQQ

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.

8. Oct 6, 2014

RUber

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 $a^2 = 1$.

9. Oct 6, 2014

Fredrik

Staff Emeritus
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.

10. Oct 6, 2014

PsychonautQQ

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.