Prove that $R$ has no zero divisors and $R$ has an identity.

Question

Let $R$ be a ring with more than one element such that for each nonzero $a \in R$ there is a unique $b \in R$ such that $aba=a$. Prove that $R$ has no zero divisors and $R$ has an identity.

Answer 1

Servaes's answer has shown the non-existence of zero divisors. Now let $a \not =0, a \in R$, so $\exists b \in R$ with $aba=a$. Let $ab=t, ba=u$, where $t, u \in R$. Now $ta=a$ so $xta=xa \, \forall x \in R$ so $(xt-x)a=0$ so ($a \not =0$) $xt=x \, \forall x\in R$. Similarly $ux=x \, \forall x \in R$. Finally $u=ut=t$ so $u=t$ is the (multiplicative) identity.

Answer 2

Let $a\in R$ be nonzero and let $c\in R$ be such that $ac=0$. There exists a unique element $b\in R$ such that $aba=a$, so from the identities $$a(b+c)a=(ab+ac)a=(ab)a=aba=a,$$ it follows that $b+c=b$, and hence that $c=0$. This shows that $a$ is not a zero divisor