For some reason, I assumed that for any ring $R$, the subset $R_{\neq 0}$ is a monoid under multiplication. However, that is only true if there are no zero divisors, i.e. in a domain! On the other hand, including the zero just gives us a semigroup, since $1=0$ implies that 1 cannot be an identity. What a pity, I was hoping to deal with a monoid… Pretend that the last sentence has never been written, and continue with me.
However, on second sight, multiplication by $1$ and $0$ is not too different, after all:
This motivates the following definition:
DEF A semigroup $S$ equipped with a system of subsets $S_0\subsetneq \cdots \subsetneq S_n\subseteq S$ is called an $n$-graded semigroup if the following holds: Let $T_0 := S_0$ and $T_{i+1}:=S_{i+1}\setminus S_i$. For every $0\leq i \leq n$ there is an element $e_i\in T_i$ such that
Note that the requirement to be an identity on $T_i$ does not imply that the latter is a subsemigroup – for that, we'll see an example soon. However, It is not necessary to ensure uniqueness of the identities:
LEM Let $S$ be a semigroup and $S_0\subsetneq \cdots \subsetneq S_n\subseteq S$. If Elements $e_i$ satisfying the identity axiom (1) exist, they are unique.
Proof. Assume there were identities $e_i and f_i\in T_i$. Then $e_i=e_i f_i = f_i$.
LEM $S_i\leq S$
Proof Let $s,t\in S_i$ and let $j\leq i$ such that $t\in T_j$. Then $st = s(te_j)=(st)e_j\in S_j\subseteq S_i$.
LEM Let $(S, {(S_i)}_i$ be a graded semigroup with identities $e_i$, and $\lambda_x,\rho_x$ the left and right multiplication with $x\in S$, respectively. Then $S_i = \bigcup_{j\leq i} \mathrm{Im} \lambda_{e_i} = \bigcup_{j\leq i} \mathrm{Im} \lambda_{e_i}$.
Proof.
EX Consider $S={e_0,e_1,e_2}$ with the multiplications necessary to make it into a grading, i.e. $e_0\cdot - = - \cdot e_0 = e_0$, $e_1^2 = e_1$, $e_2^2 = e_2$. To see what other restrictions we need, note that $bc,cb\in {a,b}$