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I defined an "aneloid" to be a set endowed with two operations, adition and multiplication, with multiplication being distributive BOTH sides in relation to adition.

I tried to find an example of "plain" aneloid, a such one that adtion and multiplication doesn't have any property of an abelian group, i. e. both operations should be not commutative, not associative, not have a neutral element (I'm not taking account of another properties like cancellation and idempotents).

I tried many finite and infinite examples but seems to me that the exigence of both sides distributivity implies at least one property of group (if you want distributivity by one side only is very easy to find examples).

I tried to proof by absurd if there exists a plain aneloid then one side of distributivity is violated, but I do not have sucess.

My question is: do you know an example of plain aneloid? Or a plain aneloid violates the both side distributivity?

Red Banana
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Rodrigo Tavares
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1 Answers1

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$$\begin{array}{c|cccc} \diamond & 0 & a & b & 1\\ \hline 0 & 0 & 1 & a & b \\ a & b & a & 1 & 0\\ b & 1 & 0 & b & a \\ 1 & a & b & 0 & 1\\ \end{array}$$

$\diamond$ is not commutative: $a\diamond b =1\neq 0 = b\diamond a$

$\diamond$ is not associative: $(a\diamond a)\diamond b= a \diamond b = 1\neq 0 = a\diamond 1 = a\diamond(a\diamond b)$

Take any finite field $\mathbf{F}$ of order $p^n\geq 3$ and a generator $a\in F$ such that $a\neq0,1$. Define an operation $\diamond$ on $F$ by $x\diamond y=a\cdot x + (1-a)\cdot y$. According to Theorem 2.6 in Jezek and Kepka's paper "Atoms in the lattice of varieties of distributive groupoids," $\diamond$ distributes over itself.

Therefore if $A=\{0,1,a,b\}$, then $\mathbf{A}=(A,\diamond,\diamond)$ is a 'plain aneloid.'

Also, the word plain has already been used for algebraic structures: see here.

Eran
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