Is there any nontrivial commutative ring without multiplicative identity that satisfies alternating property ($x \cdot x = 0$ for all $x$ where $\cdot$ is multiplication operator and $x \cdot y \neq 0$ when $x \neq y$ and $x$ and $y$ are nonzero except in the case of $x \cdot (x \cdot y)=0$)?

The commutative ring may be a commutative ring that does not have multiplicative identity.

The total cardinality of distinct elements of the ring must be infinite. (So this eliminates the case like $Z/7Z$.)

And what happens if we eliminate the additive inverse axiom from commutative ring mentioned above? I know that it will no longer be a ring, but that is fine. So this would be commutative semiring without requirement of multiplicative identity.

so I am basically asking for two cases: commutative semiring or commutative ring - with both not requiring multiplicative identity.

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    $\begingroup$ $\,3^2=0\,$ in the ring $\,R:=\Bbb Z/9\Bbb Z\,$ ... $\endgroup$ – DonAntonio Apr 10 '13 at 2:20
  • $\begingroup$ @DonAntonio, maybe he means that $x^2 = 0$ for all $x$? $\endgroup$ – Paul Gustafson Apr 10 '13 at 2:35
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    $\begingroup$ Yes @user66345, I think you're right. Then take any abelian group (written additively, to avoid confusion) and define $\,x\cdot y=0\,\,\,,\,\,\forall\,x,y$ in the group $\endgroup$ – DonAntonio Apr 10 '13 at 2:51
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    $\begingroup$ Let $x$ and $y$ be distinct and nonzero. Then, $x\cdot y\neq 0$. If $x$ and $x\cdot y$ were distinct, we would thus have $x\cdot \left(x\cdot y\right)\neq 0$, contradicting $x\cdot \left(x\cdot y\right) = \left(\underbrace{x\cdot x}_{=0}\right)\cdot y = 0$. Hence, $x = x\cdot y$. Similarly, $y = x \cdot y$. Hence, $x = y$, contradicting $x\neq y$. Thus, the ring can have at most one element $\neq 0$. $\endgroup$ – darij grinberg Apr 10 '13 at 3:52
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    $\begingroup$ Instead ofadding strange exceptions, it might be more useful to explain what you are trying to do. $\endgroup$ – Mariano Suárez-Álvarez Apr 10 '13 at 4:12

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