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Consider an ordered set $(X,\geq)$ with a binary operation $*$ that satisfies the following axioms:

A1 (Closure) $\forall a,b\in X, a*b \in X$

A2 (Associativity) $\forall a,b,c\in X, (a*b)*c = a*(b*c)$

A3 (Identity) $\exists e\in X$ s.t. $\forall a\in X, a*e=a$

A4 (Commutativity) $\forall a,b\in X, a*b=b*a$

A5 (???) $\forall a,b\in X, a*b \geq a$

A6 (???) $\forall a,b\in X$ s.t. $a\geq b, \exists c\in X$ s.t $b*c=a$

So, it's a bit like an Abelian group with two modifications:

  • It's an ordered set and the "sum" is always bigger than its components (A5)
  • Invertability is replaced with a certain "divisibility" (A6)

Axiom A6 seems like a "natural" replacement for invertability given A5. (Note that A5 precludes the existence of an inverse)

Do axioms A5 and A6 have standard names? Are they familiar from other structures? Does this overall structure have a name?

exk
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  • I write $* = +$. Don't you want to add $a \geq b \Rightarrow a+c \geq b+c$ (monotony)? Then it would be an ordered commutative monoid consisting of non-negative elements. And A6 says that actually $\geq$ can be recovered by $+$. I've seen something like this in http://arxiv.org/abs/math/0212377 – Martin Brandenburg Aug 19 '13 at 15:42
  • Thanks Martin. Yes, you're right, it would be natural to add monotony. But, being a monoid does not require A6, does it? – exk Aug 19 '13 at 15:48

2 Answers2

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Let $a\ge b, \ a=b*c$. Then for every $x$ we have $a*x=b*c*x\ge b*x$. So $X$ is a commutative ordered monoid.

Addendum: In [G.Birkhoff, Lattice Theory (3ed, 1967) Chapt.XIV.2] ordered monoid in which $a\le b$ is equivalent to $b\in Xa\ \& \ b\in aX$, is called a divisibility monoid.

Boris Novikov
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  • Consider the following subset of positive integers {1,3,4,5,6,...}. This set, coupled with addition, is a commutative ordered monoid, but it does not satisfy A6. – exk Aug 19 '13 at 15:53
  • I don't think that these monoids have a name. I can add that a commutative ordered monoid with A6 embeds into a group (but again, converse is not true). – Boris Novikov Aug 19 '13 at 16:01
  • @exk: See the addendum to my answer. – Boris Novikov Aug 20 '13 at 07:47
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Aren't (A5) and (A6) actually $$ \forall a,b, \, a \leq b \iff b \in (a) \quad ?$$ (Where $(x)$ denoted the (two-sided) ideal generated by $x$.)

Then your structure is just a commutative monoid, because the order is always implicitly here.

Pece
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    The partial order defined by this condition is not an order in general; it doesn't have to be total or anti-symmetric. – Martin Brandenburg Aug 19 '13 at 16:06
  • As in the response to Boris' answer: being a commutative monoid does not guarantee A6. This "divisibility" axiom rules out commutative monoids like $\mathbb{N} \setminus {2}$. – exk Aug 19 '13 at 16:07
  • @MartinBrandenburg If I'm not mistaking, it is a preorder. But the axioms of the OP does not require $\leq$ to be total or antisymmetric. (Or is it implicit from the used symbol "$\leq$" for the OP ?) – Pece Aug 19 '13 at 16:23
  • @exk $\mathbb N \setminus {2}$ isn't closed under the law : 1+1 is not defined. Maybe you think of another law than addition ? – Pece Aug 19 '13 at 16:26
  • @Pece Sorry, that was a mistake. $\mathbb{N} \setminus {1}$ will do, however. That (under standard addition) satisfies all the axioms other than A6. – exk Aug 19 '13 at 21:01
  • @exk Because the common order on the natural integers is not the one I described in my answer : in the monoid $\mathbb N \setminus {1}$ with the order I gave, $2$ is not comparable to $3$. This was precisely my point : a (pre)order on a commutative monoid satisfying A5 and A6 is actually the preorder $a\leq b \iff b \in (a)$ ; so it does not have more structure than a commutative monoid. – Pece Aug 19 '13 at 22:31