Aa = a Implies R is Commutative

1. Jun 24, 2008

e(ho0n3

The problem statement, all variables and given/known data
Suppose that R is a ring and that aa = a for all a in R. Show that R is commutative.

The attempt at a solution
Let a and b be two elements from R. I'm having a hard time figuring this out without multiplicative inverses. All I've been able to derive is ab = aabb = abab so aabb - abab = a(ab - ba)b = 0.

2. Jun 24, 2008

tmc

You don't need multiplicative inverses. Start with (ab-ba), multiply by b in a few places, then reduce the 'bb' to 'bb=b', and you'll find ab-ba=0.

3. Jun 24, 2008

D H

Staff Emeritus
That means at least one of a, b, or ab-ba is zero. Look at the special cases where a or b is zero (is multiplication by zero commutative?) and then look at the general case where a and b are non-zero.

4. Jun 24, 2008

e(ho0n3

Thanks for the tip tmc. I got it know.

5. Jun 24, 2008

e(ho0n3

I can't assume that.

6. Jun 24, 2008

e(ho0n3

Now that I've bothered with the algebra, I've realized that I wrote the above prematurely. I have to start with an equation, not just with ab - ba, so I started with ab - ba = -(ba) + ab but this lead nowhere.

7. Jun 24, 2008

tmc

You're right, that doesn't work.

Try by contradiction: suppose ba ~= ab, and multiplying by b from either side will not change the inequality (assuming b non-zero; if b is zero, then ab=ba is trivial). You should pretty quickly find a contradiction along the lines of 0 ~= 0

8. Jun 24, 2008

D H

Staff Emeritus
Stupid zero divisors!
Doesn't the same problem with the stupid zero divisors arise here?

9. Jun 24, 2008

matt grime

What is (a+b)^2? What is (-1)^2 (we may as well assume it is unital)?

10. Jun 24, 2008

e(ho0n3

The problem with the contradiction is that the algebra with the not-equal sign is not well defined: How do you know a != b implies ab != bb? It is possible for ab = bb but a != b.

11. Jun 24, 2008

e(ho0n3

That's a good tip. If I consider a - b = (a - b)(a - b), I get the answer.

12. Jun 24, 2008

matt grime

Actually, it's quite easy to show that 1=-1 in this ring in many ways, so I'm worried that you're writing a-b in there. With what you wrote you get

a-b=(a-b)^2=a^2+b^2 -ab-ba = a+b -ab-ba

i.e. 0=2b-ab-ba

which is quite a long way from what you need to end up with (though it is still correct, and will lead to a proof).

13. Jun 24, 2008

e(ho0n3

I wouldn't say I'm a long way from what I need since I end up with ba = a(-b). If I can show that -b = b, I'm done. This unfortunately I don't know how to demonstrate. By the way, you can't assume R has a multiplicative identity.

14. Jun 25, 2008

HallsofIvy

Staff Emeritus
Then look at (b+b)2= b+ b.

15. Jun 25, 2008

matt grime

Or embed it in its unitalization - the same property of x^2=x will still hold. And you don't end up with that from considering (a=b)^2: where did the 2b go? That was my point about the minus signs, not the ab=-ba fact which comes up if you consider (a+b)^2 instead.

16. Jun 25, 2008

e(ho0n3

HallsofIvy: Good tip. Thanks.

matt grime: I don't understand your point. Nevertheless, I now have a satisfactory answer. Thanks.

17. Jun 25, 2008

matt grime

Which bit don't you understand? The important bit is that you should consider a+b and not a-b as it is simpler. (Although it is entirely equivalent and you prove the same thing: 2x=0 for all x). You claimed that by considering a-b you only needed to show that ab=-ba, but in doing so you must have assumed that 2b=0, i.e b=-b.

18. Jun 25, 2008

e(ho0n3

Right. Using a + b rather than a - b is simpler. I didn't follow your argument about unity and that -1 = 1. I guess you were trying to suggest that -a = a for all a in R.

19. Jun 25, 2008

matt grime

Any ring can be embedded in a unital ring. If R is a non-unital ring, then define R' as follows.

R' as a set is RxZ (the integers), i.e. ordered pairs (r,n) where r is in R and n an integer. Addition is defined as

(r,n)+(s,m)=(r+s,n+m)

and multiplication

(r,n)(s,m)=(rs+ns+mr,sm).

You can verify that (0,1) is the unit element, and that R is a subring of R' via the elements (r,0). If we now declare that x^2=x for all x in R' (i.e. form a quotient ring), we still have a unital ring with R as a subring with the same property. So wlog you may assume that R has a unit.

20. Jun 25, 2008

e(ho0n3

Interesting. Thank you for the clarification.