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Please note: Since this topic is an ongoing research, there are some terminology that are created solely for describing the topic. A relevant context or mathematical application need to be found before the investigation will arose interest of the mathematics community and hence having a chance to go mainstream via journal publication
To begin, we first define a special type of ring like structures which characterise the general type of algebraic structures that the associative subset of these algebras belong: Given a ring, semiring, near ring and near semiring, the reversed ring,reversed semiring, reversed near ring and reversed near semiring are distributive structures defined like the usual rings and generalisations, except that given the binary operator $*$ for multiplication and $+$ for addition.
Because of this "distributive law the wrong way around" of these associative distributive structures (also known as ringnoids, where it is deliberately misspelt with an extra n so as not to confuse with ringoids in the category of groups), we commonly use $*$ in place of $+$ and vise versa as if we are studying ordinary rings.
Note because of the "reversed" nature of these algebras, e.g. for semirings, you will expect additive annihilators instead of multiplicative ones, though both can potentially exists at the same time
The key concepts in zero term algebra are closely related to the theory of semigroups. This is because idempotents are very common in these algebras thus tools and theorems in semigroups are often used.
Zero term: Given a zero term algebra $S$ and some element $a \in S$ where a zero element $0 \in S$, a left zero term is a nonzero product of the form $0a^n$ while a right zero term is a nonzero product of the form $a^n0$, for some $n \in {\mathbb{Z}}$, known as the order of the zero term. If it is both a left and right zero term, then it is a two-sided zero term $0^m0$ for some $m \in \mathbb{Z}$.
Zero inverse: Given a zero term algebra $S$ with an identity (may be one sided) $1$, a left zero inverse is an element $q \in S$ such that $q0=1$. A right zero inverse is an element $r \in S$ such that $0r=1$. If it is both a left and right zero inverse, then it is a two sided zero inverse. Like all inverses, the two sided zero inverse is unique.
Division by zero algebra: A division by zero algebra $\%_D$ is a zero term algebra where zero inverses exists. In other words, an algebra that allow legitimate division by zero.
Zero divisor: Given a ringnoid, a zero divisor is a nonzero element $a,b \neq 0$ such that $ab=0$. Here $a$ is a left zero divisor and $b$ is a right zero divisor. If given some nonzero $x$, $c$ is a two sided zero divisor when $cx=0$ and $xc=0$. If $a=b$, hence $a^2=0$ then $a$ is a nilpotent element.
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Divisible zero divisor (Under refinement) Given a zero term algebra $S$, a right divisible zero divisor for a left zero divisor $a$ where $\exists x\in S, ax=0$ (can be one or two sided) such that $\exists b \in S, 0b=x$. Alternately $axb=x$. Similarly a left divisible zero divisor for a right zero divisor $c$ where $\exists y\in S, yc=0$ such that $\exists d \in S, d0=y$. Alternately $dyc=y$.
Unit zero divisor Given a zero term algebra $S$ with a (possibly one sided) zero divisor $a$ and (possibly one sided) identity $1$, a right unit zero divisor is there exists an element $b$ such that it is a right inverse of $a$ i.e. $ab=1$. Similarly a left unit zero divisor is there exists an element $c$ such that it is a left inverse of $a$ i.e. $ca=1$. It is a two sided unit zero divisor if it is both left and right sided.
Note unit zero divisors are impossible in ordinary rings and their generalisations as 0 is the absorbing element in those structures since e.g. $ab=0$, $a=0b^{-1}=0$ which is a contradiction since $a\neq 0$ and similarly for $b$.
Theorem 1: Idempotence: All left zero terms are idempotent in a left distributive zero term algebra, while all right zero terms are idempotent in a right distributive zero term algebra.
Proof: For all $a \in S$ and suppose it is left distributive. Then
$$a0+a0=a(0+0)=a0$$
The proof for the right distributive case is similar.
$$a0+a0=a(0+0)=a0$$
The proof for the right distributive case is similar.
Definition 1: Dominance Given a magma $M$ with elements $a,b \in M$. $a$ dominates $b$ on the left if $ba=a$ and $a$ dominates $b$ on the right if $ab=a$.
We can notate this in short hand using a partial ordering $\leq_D$ or $\geq_D$ where $D$ stands for left (L) or right (R).
Therefore
$$ab=a\equiv b \leq_L a$$
$$ab=b\equiv a \leq_R b$$
$$ba=a\equiv b \leq_R a$$
$$ba=b\equiv a \leq_L b$$
Therefore
$$ab=a\equiv b \leq_L a$$
$$ab=b\equiv a \leq_R b$$
$$ba=a\equiv b \leq_R a$$
$$ba=b\equiv a \leq_L b$$
If $a$ left and right dominates $b, then $a$ dominates $b$. The one sided partial orderings $a \geq_L b$ and $a \geq_R b$ becomes simply $$a \geq b$$
This dominance relationship is very important for associative zero term algebras as the zero terms mostly control the addition structure in this fashion. Outside of zero term algebras, they are also important in commutative magmas such as Rock Paper Scissors (RPS) magmas, although rarely an ordering is specified explicitly in studies of those structures.
NB. The equality case for the above partial order means idempotence. $a+a=a$ for some idempotent $a$.
(Amendment to zero terms): The element $a$ is known as the zero term coefficient or just "coefficient" of the zero term.
Theorem 2: Coefficient dominance All coefficients left dominates the left zero terms with the same coefficient in a left distributive zero term algebra with a right additive identity $0$ and right multiplicative identity $1$, while all coefficients right dominates the right zero terms with the same coefficient in a right distributive zero term algebra with a left additive identity $0$ and left multiplicative identity $1$.
Proof: Let $a \in S$ be a coefficient. Suppose $0,1$ are right additive and multiplicative identities respectively and $S$ is left distributive at least for $a$. Then:
$$a+a0=a(1+0)=a1=a$$
The proof for right zero terms are similar.
$$a+a0=a(1+0)=a1=a$$
The proof for right zero terms are similar.
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Theorem 4: Squared zero is the least element of the zero terms of order 1 Given a right (left) distributive division by zero algebra $S$ with a left (right) additive identity $0$. Then $0^2$ is the right (left) least element of the set of all right (left) zero terms of order 1 in $S$. i.e. $\forall a \in S, 0^2 \leq_R a0$
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Theorem 5: Given a zero term algebra that is a commutative reversed semiring $R'$ (i.e. both $+$ and $*$ are commutative monoids, $*$ left right distributes over $+$ and there exists a unique additive annihilator/absorber $a$), Given $a$. If $0^2=a$, then $a0=a=0a$.
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Theorem 6: Division by zero no-go theorem (commutative reversed semiring). Zero inverses does not exist in commutative reversed semirings.
Proof Suppose a zero inverse $q$ exists, i.e. $0q=1$. Now since this is a commutative reversed semiring, b Theorem 5, $\exists a \in R', a0=00$. Now multiply $q$ on the right both sides
$$(a0)q=(00)q$$
Associativity results in
$$a(0q)=0(0q)$$
$$a1=01$$
$$a=0$$
A contradiction. Since this holds for any $a \in R'$, commutative reversed semirings do not form a division by zero algebra.
$$(a0)q=(00)q$$
Associativity results in
$$a(0q)=0(0q)$$
$$a1=01$$
$$a=0$$
A contradiction. Since this holds for any $a \in R'$, commutative reversed semirings do not form a division by zero algebra.
note since in a commutative reversed semiring, $*$ is abelian, thus $0q=q0=1$ and the proof proceed as usual.
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Dec '1626
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Zero term algebra
All discussions on the ongoing project of algebraic structures...