Hi isn't it easier to compute the matrix by seeing that if you divide it into 2x2 blocks, each of these blocks embed in a 2 dimensional algebra a + b.e where e² =1. This algebra is commutative. So the determinant of the original matrix corresponds to: (a +b.e)² - (c +d.e)² = (a + c + (b+d).e)(a - c + (b-d).e) Now, the determinant now corresponds to the determinant of this product (by reembedding it into the matrix algebra) which is much easier to do, and to factorize.
You can also just compute the final determinant through the norm (this is a finite dimension algebra extension after all, so the norm corresponds to the determinant of its embedding) (The only automorphism which fixes the base field is e -> -e.)
Hi isn't it easier to compute the matrix by seeing that if you divide it into 2x2 blocks, each of these blocks embed in a 2 dimensional algebra a + b.e where e² =1. This algebra is commutative. So the determinant of the original matrix corresponds to:
(a +b.e)² - (c +d.e)²
= (a + c + (b+d).e)(a - c + (b-d).e)
Now, the determinant now corresponds to the determinant of this product (by reembedding it into the matrix algebra) which is much easier to do, and to factorize.
You can also just compute the final determinant through the norm (this is a finite dimension algebra extension after all, so the norm corresponds to the determinant of its embedding)
(The only automorphism which fixes the base field is e -> -e.)