Question

Show that, if group G is not {1} can be right ordered, then, for any domain k, A = kG is J-semisimple.

Answer 1

Consider a product $\alpha\beta$ where

$$\begin{array}{l} \alpha = a_1 g_1 + \cdots + a_m g_m, \; g_1 < \cdots < g_m, \; a_i \neq 0 \ (1 \leq i \leq m), \\ \beta = b_1 h_1 + \cdots + b_n h_n, \; h_1 < \cdots < h_n, \; b_j \neq 0 \ (1 \leq j \leq n). \end{array}$$

Choose $i_0, j_0$ such that $g_{i0}h_{j0}$ is least among $\{g_i h_j\}$.

Then $i_0 = 1$ (for otherwise $g_1 < g_{i0} \Rightarrow g_1 h_{j0} < g_{i0} h_{j0}$).

In particular, $g_{i0}h_{j0} = g_i h_j$ implies $i = i_0 = 1$ and hence $j = j_0$.

This shows that, in the product $\alpha\beta$, $a_1 b_{j0} g_1 h_{j0}$ cannot be “canceled out” by any other term, so $\alpha\beta \neq 0$.

Thus, it has to be a domain, and for the $J$-semisimplicity of $A$ (when $G$ in non $\{1\}$) we are done since any domain is $J$-semisimple.