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Let $\left(A,+, \cdot\right)$ be a ring with $1$ that satisfies the following condition:

For any nonzero $a\in A$, there exists a unique $b\in A$ such that $aba = a$.

Show that $b$ also satisfies $bab = b$, and that all nonzero elements of $A$ are invertible.

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closed as off-topic by user26857, Daniel W. Farlow, Kamil Jarosz, Pragabhava, John B Feb 12 '16 at 15:46

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The ring has no nonzero zero divisors: if $ac=0$ for nonzero $a,c$, then you can find a unique $b$ such that $a=aba=a(b+c)a$. But then $b=b+c$ implies $c=0$, a contradiction.

At this point, you can even prove that $A$, as long as it is not equal to zero, has an identity without assuming so in the hypotheses. Given any $a=aca\neq 0$, we see that $ac$ is an idempotent, and verify that $ac(acx-x)=0=(xac-x)ac$ implies $ac$ is an identity element for the ring. We write "$1$" for $ac$ hereafter.

Knowing this, you can conclude that $a(ba-1)=0=(ab-1)a$ implies $ba=ab=1$ when a is nonzero. Clearly then $b$ can only be $a^{-1}$ and it follows that $bab=b$ also holds.

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$a(bab)a=(aba)ba=aba=a$. So $bab$ does the thing that $b$ does, but $b$ is the unique element that does that, so $bab=b$.

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