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I'm struggling with the following exercise in my algebra course:

Let $A$ be a non-trivial ring not necessarily with multiplicative identity. Suppose that for every $a \in A \setminus \left\{ {0}\right\}$ there exists an unique $a' \in A$ such that $aa'a=a$. Prove that

1) $A$ has no zero divisors.

2) $A$ has a multiplicative identity $1$.

3) $A$ is a division ring.

I didn't make much progress on question 1) and 2) (i just played around with the given property and the definition of zero divisors and multiplicative identity but nothing worked). But, supposing 1) and 2), then the exercise 3) quickly follow as $A$ has no zero divisors and for every $a \in A \setminus \left\{ {0}\right\}$ exists an unique $a' \in A$ such that $aa'a=a=a.1$, so $a'a=1$.

Thank you.

u1571372
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  • See: http://math.stackexchange.com/questions/613022/prove-a-unital-ring-with-this-property-has-no-zero-divisors – G. H. Faust Dec 31 '13 at 01:59

1 Answers1

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Faust has linked you to an answer for 1), for 2) here's a hint: Pick $a$ to be nonzero, then $aa'a = a$ suggests that $aa'$ should be an identity for $A$. So if $b$ is nonzero you want to prove that $aa'b = b$. What nonzero element can you multiply $aa'b - b$ by on the left to get zero? If you can figure that out then as there are no zero divisors you'll have $aa'b - b = 0$.

Jim
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