I have a Clifford algebra defined over a field of characteristic not equal to $2$. Is there a formula for its discriminant in terms of the corresponding symmetric bilinear form (or in terms of its quadratic form)?
In particular, let $k$ be a field with $\text{char} \, k \neq 2$, let $n$ be an even positive integer, and let $$ W_n = k\left\langle x_1, x_2, \ldots, x_n \right\rangle / (x_i x_j + x_j x_i - 1 \text{ for } i \neq j). $$ As long as $n$ is even, then the center of this algebra is $C=k[x_1^2, \ldots, x_n^2]$, and $W_n$ is free and finitely generated over $C$. Therefore I can define the discriminant of $W_n$ over $C$ as the determinant of the appropriate matrix of traces; the discriminant will be an element of $C$.
I think that the answer should be as follows: let $M$ be the matrix with $(i,j)$ entry equal to $x_i x_j + x_j x_i$ (so $M$ has entries $2x_i^2$ down the diagonal, 1 off the diagonal). Then the discriminant should be $(\det M)^{2^{n-1}}$, up to a multiple by an element of $k$. Can anyone provide a reference?
