Question

Let k be a field of characteristic zero, and (aij) be an m × m skew symmetric matrix over k. Let R be the k-algebra generated by x1, . . . , xm with the relations xixj − xjxi = aij for all i, j. Show that R is a simple ring iff det(aij) <> 0. In particular, R is always nonsimple if m is odd.

Answer 1

Consider a linear change of variables given by $x_{i}^{\prime} = \sum_{j}c_{ij}x_{j}$ where $C = (cij)\in \mathrm{GL}m(k)$ . We have $x_{i}^{\prime}x_{j}^{\prime} - x_{j}^{\prime}x_{i}^{\prime} = \sum_{r}c_{ir}x_{r}\sum_{s}c_{js}x_{s} - \sum_{s}c_{js}x_{s}\sum_{r}c_{ir}x_{r} = \sum_{r,s}c_{ir}c_{js}\left(x_{r}x_{s} - x_{s}x_{r}\right) = \sum_{r,s}c_{ir}a_{rs}c_{js}$ . If we write $a_{ij}' = \sum_{r,s}c_{ir}a_{rs}c_{js}$ then $x_{i}'x_{j}' - x_{j}'x_{i}' = a_{ij}'$ , and we have $A' = CAC^T$ where $A = (aij)$ and $A' = (a'ij)$ , and "T" denotes the transpose. Therefore, we are free to perform any congruence transformation on $A$ . After a suitable congruence transformation, we may therefore assume that $A$ consists of a number of diagonal blocks $\left( \begin{array}{cc}0 & 1\\ -1 & 0 \end{array} \right)$ , together with a zero block of size $t\geq 0$ . If $t > 0$ , then $\operatorname*{det}(A) = 0$ , and $xm$ generates a proper ideal in $R$ . If $t = 0$ , then $\operatorname*{det}(A) < >0$ and $m = 2n$ for some $n$ . Here, $R$ is the $n$ th Weyl algebra $An(k)$ . Since $k$ has characteristic zero, $R$ is a simple ring.