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Can anyone give an example of an "non-artificial" algebraic structure that fails to be a ring only because of a lack of one- and two-sided distributive property?

John Smith
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    There are a bunch of good examples here: http://en.wikipedia.org/wiki/Near-ring – Casteels Oct 03 '13 at 08:54
  • @Casteels: This is an answer, not just a comment. – Martin Brandenburg Oct 03 '13 at 09:22
  • This is an example of structure that satisfies all axioms of a ring except distributivity (and it is commutative): https://mathoverflow.net/questions/424372/anti-dual-numbers-and-what-are-their-properties – Anixx Jun 23 '22 at 00:56

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Casteels has rightly pointed you to the Wikipedia article, but since you want only one distributive law not to hold (but addition to be commutative), let us see precisely this.

Consider a multiplicatively written abelian group $G$ (more later), and the set $N$ of maps $G \to G$, written as exponents.

Define operations $+$ and $\cdot$ on $N$ by $$ g^{m + n} = g^{m} g^{n}, $$ and $\cdot$ is composition $$ g^{m n} = (g^{m})^{n}. $$ Since $G$ is abelian, addition is commutative.

As to the distributive properties, $m (n + k) = m n + m k$ holds by the definition of $+$, but $$(m + n) k = m k + n k\tag{distr}$$ will fail if $k$ is not an endomorphism. To see this, take any two elements $a, b \in G$, consider any two maps $m, n$ such that $m : 1 \mapsto a$ and $n : 1 \mapsto b$, then if (distr) holds you have $$ (a b)^{k} =(1^{m} 1^{n})^{k} = (1^{m + n})^{k}= 1^{(m + n) k} = 1^{m k + n k}= 1^{m k} 1^{n k} = a^{k} b^{k}. $$

So if $G$ has at least two elements, (distr) fails.

Andreas Caranti
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