# Ring theory problem

1. Jan 25, 2008

### ehrenfest

1. The problem statement, all variables and given/known data
Let R be a ring that contains at least two elements. Suppose for each nonzero a in R, there exists a unique b in R such that aba=a.
Show that R has no divisors of 0.

2. Relevant equations

3. The attempt at a solution
Let a*c=0 where a,c are not equal to 0.
aba=a implies aba-a=0=nac where n is any integer which implies that a(ba-1-nc)=0
I am not seeing the contradiction.

2. Jan 25, 2008

### Dick

You are given that b is unique for each a. Show that if ac=0, aba=a then there is another value d not equal to b such that ada=a.

3. Jan 25, 2008

### ehrenfest

Next part of the question: Show that R has unity.

I need to show that there exists an element 1 of R such that 1a=a1=1 for all a in A.

We can use cancellation now that we showed R has no divisors of 0, so bab=b.

Let a_1, a_2 be nonzero. Then there exists b_1 and b_2 such that $a_1 b_1 a_1 = a_1$ and $a_2 b_2 a_2 = a_2$.

To complete the proof, I need to show that $a_1 b_1 = a_2 b_2 = b_1 a_1 = b_2 a_2$, right?

I can show that $b_1 a_1 = a_2 b_2$ from the fact that $a_1 b_1 a_1 a_2 = a_1 a_2 = a_1 a_2 b_2 a_2$ and similarly I can show that $a_1 b_1 = b_2 a_2$ but I am having trouble showing that the left and right identity are the same.

4. Jan 27, 2008

### Dick

Consider the set G of all elements of the ring such that x^2=x. The products you are talking about have that property. Now consider two elements such that x^2=x and y^2=y. So x^2*y=x*y^2. Now cancel your way down to x=y.

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