# 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.