
> 
> **Definition.** $$\mathbb{J} = \{1,2,3,\ldots\}.$$
> 
> 
> We can refer to the elements of $\mathbb{J}$ as "joiners."
> 
> 
> * The product of joiners is inherited from $\mathbb{Z}$.
> * The sum of joiners will be defined by $$a \oplus b = a+b-1.$$
> 
> 
> 


The motivation is that if we're thinking of the elements of $\mathbb{J}$ as non-empty finite totally ordered sets, then we can think of $a \oplus b$ as the result of "identifying" or *joining* the greatest element of $a$ to the least element of $b.$


This makes the $\mathbb{J}$ into a commutative monoid in two different ways. The good thing is that the identity elements of both these monoids are the same (namely $1$). The bad thing is that the distributivity law doesn't hold. Nonetheless, we *do* have the "variant distributivity" law:


$$(a(b \oplus c)) \oplus a = ab \oplus ac$$


*Proof.*


$$\mathrm{LHS} = a(b+c-1) \oplus a = (ab+ac-a) \oplus a = ab+ac-a+a-1 = ab+ac-1= \mathrm{RHS}$$


From this, we can deduce further "variant distributivity" identities, one for each natural number. For example, the $n=3$ case is $$a(b \oplus c \oplus d) \oplus a \oplus a = ab \oplus ac \oplus ad$$


This motivates the following definition.



> 
> **Definition.** Define that a *flask* is a pair of monoids on a common set, one denoted concatenatively, the other denoted with $\oplus$, such that:
> 
> 
> * $\oplus$ is commutative
> * their identity elements are equal and denoted $1$, and
> * the variant distributivity law holds (on both sides).
> 
> 
> 


It makes sense to speak of "commutative flasks" of course, the prototypical example being $\mathbb{J}$. There's lots of other flasks, though; for example, every ring $R$ can be viewed as a flask in the obvious way, and if $R$ is commutative, the corresponding flask will obviously be commutative. Note that we can view $\mathbb{J}$ as a subflask of $\mathbb{Z}$. This implies that subrings can have subflasks that aren't themselves subrings (because they don't contain $0$ and aren't closed under subtraction.)



> 
> **Question.** Has anything interesting ever been done with the algebraic structure I'm denoting $\mathbb{J}$ and/or the class of algebraic structures I'm referring to as flasks?
> 
> 
> If so, is there a book or article that provides a good reference?
> 
> 
> 



Has anything ever been done with the set $\{1,2,3,4,\ldots\}$ equipped with the operation $a \oplus b = a+b-1$ and the usual notion of multiplication?
I believe this could be related to an algebraic structure useful for studying set-theoretical solutions of the Yang-Baxter equation, the so-called ***braces***. At least the funny distributive property appears in the theory of set-theoretical solutions of the Yang-Baxter equation.


**Definition.** An abelian group $(A,+)$ is a **left brace** if there is a multiplication $m\colon A\times A\to A$, $(a,b)\mapsto ab$, turning $A$ into a group such that 
$$a(b+c)+a=ab+ac$$ 
holds for all $a,b,c\in A$.


Similarly one defines right braces.


The connection to the Yang-Baxter equation is as follows. First, for each left brace $A$ 
one has a canonical solution $(A,r\_A)$ of the Yang-Baxter equation: 
$$r\_A\colon A\times A\to A\times A,\quad r\_A(a,b)=(ab-a,(ab-a)^{-1}ab).$$


One also has the following.


**Theorem.** Let $X$ be a set, $$r\colon X\times X\to X\times X,\quad r(x,y)=(\sigma\_x(y),\tau\_y(x))$$ be a 
non-degenerate involutive solution of the Yang–Baxter equation. (Non-degenerate means that $\sigma\_x$ and $\tau\_x$ are permutations for each $x\in X$, and involutive means that $r^2=\mathrm{id}\_{X\times X}$.) Then there exists a unique left brace structure over the group 
$$G=G(X,r)=\langle X:xy=\sigma\_x(y)\tau\_y(x)\rangle$$
such that its associated solution $r\_G$ satisfies
$r\_G(\iota\times \iota) = (\iota\times \iota)r$, where $\iota\colon X\to G(X,r)$ is the canonical map.
Furthermore, if $B$ is a left brace and $f\colon X\to B$ is a map such that $(f\times f)r=r\_B(f\times f)$,
then there exists a unique brace homomorphism $\varphi\colon G(X,r)\to B$ such that $f = \varphi\iota$.
and $(\varphi\times\varphi)r\_G = r\_B(\varphi\times\varphi)$. 


The theorem is implicit in the work of Rump where braces were discovered:


* Rump, Wolfgang. [Braces, radical rings, and the quantum Yang-Baxter equation](https://doi.org/10.1016/j.jalgebra.2006.03.040). J. Algebra 307 (2007), no. 1, 153--170. [MR2278047](https://mathscinet.ams.org/mathscinet-getitem?mr=2278047)


You can find some interesting examples of braces here:


* Bachiller, David. [Classification of braces of order $p^3$](https://doi.org/10.1016/j.jpaa.2014.12.013). J. Pure Appl. Algebra 219 (2015), no. 8, 3568--3603. [MR3320237](https://mathscinet.ams.org/mathscinet-getitem?mr=3320237), [arXiv: 1407.5224](https://arxiv.org/abs/1407.5224)


See also this survey:


* Rump, Wolfgang. The brace of a classical group. Note Mat. 34 (2014), no. 1, 115--144. [MR3291816](https://mathscinet.ams.org/mathscinet-getitem?mr=3291816)


There is a non-commutative version of braces useful to study 
non-involutive solutions, see [this paper](http://arxiv.org/abs/1511.03171).


**Added.** Rump proved that braces on both sides are in bijective correspondence with radical rings. The correspondence is as follows. If $R$ is a non-zero radical ring (for all $x\in R$ there is $y\in R$ such that $x+y+xy=0$) then $R$ with $a\circ b=ab+a+b$ is a two-sided brace. Conversely, if $A$ is a two-sided brace, then $A$ with $a\*b=ab-a-b$ is a radical ring. 


