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This question already has an answer here:

$R$ is a ring, containing more than one element, such that
$\forall 0\ne a\in R \exists ! b\in R; aba=a$.

Here's what I did in the very first place:
$(aba)a^{-1}=aa^-1 \text{ and } a^{-1}(aba)=a^{-1}a\Longrightarrow \left\{ \begin{array}{rl} ab=1 & \forall0\ne a\in R \exists !b\in R \\ ba=1 & \forall0\ne a\in R \exists !b\in R \end{array} \right.$
Therefore, since $ab\in R\Longrightarrow 1\in R$. So $R$ is a ring with identity.

But then I realized, how one can say $aa^{-1}=1$ when one doesn't know whether $R$ is a ring with identity.

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marked as duplicate by rschwieb abstract-algebra Feb 12 '15 at 16:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ In a ring, if you don't know it has an identity, you definitely don't know it has multiplicative inverses - $a^{-1}$ is meaningless. $\endgroup$ – Thomas Andrews May 7 '14 at 21:58
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    $\begingroup$ @Brenin I think OP just means the ring contains more than one element (it is non-trivial.) The "such that" modifies the ring, not the elements. $\endgroup$ – Thomas Andrews May 7 '14 at 22:00
  • $\begingroup$ Silly me! right, I'll delete my comment $\endgroup$ – Brenin May 7 '14 at 22:01
  • $\begingroup$ I'll add commas to make it clearer. $\endgroup$ – Thomas Andrews May 7 '14 at 22:02
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Try to first show $R$ has no zero divisors (see below), and having that, it follows that $bab=b$.

With this in hand, let $c\in R$. Then $aba=a$ implies $caba=ca$, so cancelling the nonzero $a$, $cab=c$.

On the other hand, $bab=b$ implies $bc=babc$, so cancelling the nonzero $b$ yields $c=abc=cab$. That is, $c=(ab)c=c(ab)$ for any $c\in R$. So $ab$ satisfies the properties of $1$, so $ab=1$.


To see there are no zero divisors, assume $a\neq 0$. Let $b$ be the unique $b$ such that $aba=a$. Suppose $c$ is such that either $ca=0$ or $ac=0$. Then $a(b+c)a=aba+aca=a$ so by uniqueness, $b=b+c$, or $c=0$.

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  • $\begingroup$ Any reason for the downvote? $\endgroup$ – Ben West May 7 '14 at 22:11
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    $\begingroup$ I can see no reason. $\endgroup$ – egreg May 7 '14 at 23:22
  • $\begingroup$ I'm a bit confused here @BenWest . In the second line, how can we cancel the element $a$ when we don't know whether a $1_R$ exists such that $caba=ca\Rightarrow caba(a^{-1})=ca(a^{-1}) \Rightarrow cab(1)=c(1)$. Do I miss anything? $\endgroup$ – Mill May 8 '14 at 8:38
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    $\begingroup$ @Milad You're cancelling by the fact that $R$ has no zero divisors, not by assuming an inverse of $a$. The reasoning is $caba=ca$ implies $caba-ca=0$ implies $(cab-c)a=0$. Since $a\neq 0$ and there are no zero divisors, necessarily $cab-c=0$, so $c=cab$. $\endgroup$ – Ben West May 8 '14 at 18:01
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    $\begingroup$ That's right, @BenWest . $R$'s being an integral domain is what I miss. Problem solved and thank you. $\endgroup$ – Mill May 9 '14 at 6:54
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$a = aba = abababa = ... = (ab)^n a \quad \forall n$

Therefore $a=(ab)a$ and $a=a(bab)a$.

But $\exists ! b$. So $b=bab$. So if $a\mapsto b$ by that rule, $b\mapsto a$.

Now $ab$=$abab$=$(ab)^n \forall n$

So $\forall c \in R, \quad ab c=(ab)^2c=(ab)^n c$

If $c \neq (ab)c$, then since $(ab)\neq 0$

$(ab)c\neq (ab)^2 c$. But this is false, so $c=(ab)c$ for all $c$, therefore $ab=1$

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  • $\begingroup$ I'd say in the last step, "therefore $ab$ is a left-identity." $=1$ is meaningless. $\endgroup$ – Thomas Andrews May 7 '14 at 22:25
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aba = a

aba - a = a - a

aba - a = 0

EDIT:

aba - aba = 0

baba - baba = 0

baba - ba = 0

a(bab - b) = 0

b = bab

Now since b = bab let c be any element in R

since bc = babc we see that c = (ab)c

we also have ca = caba

thus c = c(ab)

Therefore ab must be a an identity in R

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  • $\begingroup$ Where does $c$ comes from? $\endgroup$ – egreg May 7 '14 at 23:21
  • $\begingroup$ when you factor out the a its quite obvious that the c has to be 1 but for generality I'm assuming that some ac = a there. $\endgroup$ – meh May 7 '14 at 23:51
  • $\begingroup$ I can't see what you mean with $aba-a=a(ba-c)$: you are assuming the existence of $1$, in order to do it. Where are you using the uniqueness of the pseudoinverse $b$, which is necessary for concluding the existence of $1$? There are examples of von Neumann regular rings without $1$. $\endgroup$ – egreg May 7 '14 at 23:53
  • $\begingroup$ I'm not assuming the existence of 1. I will edit my post for clarity. $\endgroup$ – meh May 7 '14 at 23:57
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    $\begingroup$ @Milad It comes from the fact that there are no Zero divisors. Since a != 0 and c != 0 we have the equation bc = babc which is equivalent to saying b(abc - c) = 0. This shows that abc = c and hence why (ab) = 1. $\endgroup$ – meh May 9 '14 at 2:33

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