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Is there an abelian group $R$ with multiplication operator with this properties ?

(i) $a(bc)=(ab)c$
(ii) $a(b+c)=ab+ac$ , $(b+c)a=ba+ca$

And a unique element $e$ s.t
(iii) $ea=a\quad \forall a\in R $

But there exists $x\in R$ s.t $xe\neq x$ .

Najib Idrissi
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  • You are looking for a structure with only a left identity and no right identity. There are structures which have more than one left identity, however this implies that there exists no right identity. – Edward Evans Mar 16 '16 at 11:33
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    @Ed_4434 It's not possible in noncommutative rings? What is your proof? –  Mar 16 '16 at 11:33
  • In any ring the additive identity and (if there is one) the multiplicative identity are unique. – Edward Evans Mar 16 '16 at 11:39
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    @Ed_4434 $e$ is not multiplicative identity . –  Mar 16 '16 at 11:41
  • Then I'm not quite sure what you're looking for. You've asked if there is an Abelian group $R$ under multiplication with element $e$ such that $ea = a$, yet $e$ is not a multiplicative identity? – Edward Evans Mar 16 '16 at 11:44
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    @Ed_4434 $(R,+)$ is an abelian group, but $(R,\times)$ is not a group. –  Mar 16 '16 at 11:50
  • @NajibIdrissi Thanks, but I'm looking for the structure with only one left identity, all the answers have more than one left identity. –  Mar 16 '16 at 12:22
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    Ah. It wasn't very clear from your question. Then read this: http://math.stackexchange.com/q/1392057/10014 – Najib Idrissi Mar 16 '16 at 12:24
  • @NajibIdrissi thanks . My question is duplicate! –  Mar 16 '16 at 12:32
  • It is indeed. Since I had already cast a vote for the (wrong) previous one I cannot cast one again, but hopefully someone will. – Najib Idrissi Mar 16 '16 at 12:35

1 Answers1

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My solution was inspired by this question. Consider the set $R$ of matrices with integer entries of the form $$ \begin{pmatrix} a & a\\ b & b \end{pmatrix} $$ Then $R$ is an abelian group under addition and a semigroup under the multiplication of matrices. Distributivity also holds. Now, the formula $$ \begin{pmatrix} a & a\\ b & b \end{pmatrix}\begin{pmatrix} x & x\\ y & y\end{pmatrix} = \begin{pmatrix} a(x+y) & a(x+y)\\ b(x+y) & b(x+y) \end{pmatrix} $$ shows that every matrix $\begin{pmatrix} x & x\\ y & y\end{pmatrix}$ such that $x + y = 1$ is a right identity. However, there is no left identity. You can of course take the dual operation if you wish to have a left identity and no right identity, like in your question.

Community
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J.-E. Pin
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    Thanks , but i'm looking for the structure that have only one left identity. your example is very good but has infinite many identity. –  Mar 16 '16 at 12:18