This is a development of a previous post of mine, now with a more focused question at the end.
Let $R$ be a commutative ring (throughout this post, by "ring" I'll mean not necessarily an "associative ring"). If we denote by $+$ the addition component-wise, we get that $(R^n,+)$ is an abelian group ($n$ integer, $n \ge 2$) with additive unit $\bar 0:=(0,0,\dots,0)$. If we further request:
- $R$ with unit $1$;
- $(R^n,+)$ left $R$-module
then $\forall \bar u \in R^n, \bar u = u^i\bar e_i$ (summation convention), where $\bar e_i:=(0,\dots,0,1,0,\dots,0)$ (the only $1$ is in the i-th entry).
Now, suppose that we want to build up in $R^n$ a multiplication, $\times$, such that:
- $(R^n,+,\times)$ is a ring, namely $\times$ is distributive with $+$;
- $\times$ is bilinear on $R$.
Then, $\bar e_i \times \bar e_j=\alpha_{ij}^k\bar e_k$ completely defines the multiplication under construction.
Within this general formulation, we can get at least ($\delta_m^l$ is Kronecker's symbol):
Direct product of $n$ copies of $R$. This is the case for: $$\alpha_{ij}^k=\delta_i^k\delta_j^k \tag 1$$ The element $\bar e \in R^n$ is multiplicative unit if and only if: \begin{alignat}{1} \bar e_j=\bar e \times \bar e_j=e^i\bar e_i \times \bar e_j=e^i\alpha_{ij}^k\bar e_k, \forall j &\Leftrightarrow e^i\alpha_{ij}^k=\delta_j^k, \forall j,k \\ &\Leftrightarrow e^i\delta_i^k\delta_j^k=\delta_j^k, \forall j,k \\ &\Leftrightarrow e^i\delta_i^j=1, \forall j \\ &\Leftrightarrow e^j=1, \forall j \end{alignat} Thus, $\bar e=(1,1,\dots,1)$ is multiplicative unit.
[Any name for this?]. This is the case for:
$$\alpha_{ij}^k=\delta_{i+j-1}^k \tag 2$$
The element $\bar e \in R^n$ is multiplicative unit if and only if:
\begin{alignat}{1} \bar e_j=\bar e \times \bar e_j=e^i\bar e_i \times \bar e_j=e^i\alpha_{ij}^k\bar e_k, \forall j &\Leftrightarrow e^i\delta_{i+j-1}^k=\delta_j^k, \forall j,k \\ &\Leftrightarrow e^{k-j+1}=\delta_j^k, \forall j,k, j \le k \\ &\Leftrightarrow (e^{k-j+1}=0, 1 \le j < k \le n) \wedge (e^1=1) \\ &\Leftrightarrow (e^{k-j+1}=0, 1 \le k-j \le n-1) \wedge (e^1=1) \\ &\Leftrightarrow (e^m=0, 2 \le m \le n) \wedge (e^1=1) \\ \end{alignat}
Thus, $\bar e=(1,0,\dots,0)$ is multiplicative unit.
My question is: Given $R$ and $n$, how particular/exaustive are $(1)$ and $(2)$ in defining all the possible ring classes(?) $(R^n,+,\times)$?