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A ring can be defined as a near-ring satisfying two-sided distributivity, whose underlying additive group is Abelian. Negating this second stipulation, we obtain the following definition. A silly-ring is a near-ring satisfying two-sided distributivity, whose underlying additive group is non-Abelian.

Do silly-rings exist?

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Fantastic title. –  Dylan Yott Dec 9 '13 at 4:05
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@PedroTamaroff, I did not know that. Proof? –  user18921 Dec 9 '13 at 4:16
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@user18921 Distribute $(x + 1)(y + 1)$ both ways. –  Ryan Reich Dec 9 '13 at 4:26
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Alternative to Ryan: $(1+1)(a+b)$ in both ways. –  Pedro Tamaroff Dec 9 '13 at 4:33
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Less cleverly, if you just distribute $(a+b)(c+d)$ both ways, you get $ad+bc=bc+ad$. Well, $RR=R$ is weaker that "$R$ has a $1$", isn't it? (I wouldn't know, not an algebraist.) –  bof Dec 9 '13 at 4:50
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up vote 6 down vote accepted

Yes, they exist. Let $(R,+)$ be any non-Abelian group. Define $xy=0$ for all $x,y\in R$. (Ask a silly question, . . .)

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The comments show that if near-rings are assumed to have unity, then silly-rings do not exist. Here's a slightly more general proposition, following along the lines of bof's comment.

Proposition. Consider a structure $(R,+,\cdot)$ such that

  1. Two sided distributivity holds.
  2. $+$ is associative
  3. Both cancellation laws hold.

Then for all $x,y \in R$, we have that if $x$ factors as $ab$ and $y$ factors as $cd$, then $x+y=y+x$.

Proof. Distribute $(a+c)(b+d)$ both ways, and cancel.

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Of course the two-sided cancellation law is equivalent to the conjunction of the two one-sided cancellation law. –  bof Dec 9 '13 at 5:26
    
@bof, it certainly folkows from the conjunction of the one-sided cancellation laws. Are you saying the converse holds? –  user18921 Dec 9 '13 at 5:40
    
$a+x=a+y\Rightarrow a+x+b=a+y+b\Rightarrow x=y$. –  bof Dec 9 '13 at 5:52
    
@bof, yep good point. –  user18921 Dec 9 '13 at 6:32
    
On the other hand, if you take the usual axioms for a ring (without identity) and replace the two one-sided distributive laws by the "two-sided distributive law" $(a+b)(c+d)=ac+ad+bc+bd$, you get something strictly weaker. –  bof Dec 9 '13 at 7:20
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