  In this paper, we study the complexity of computing the determinant of a
matrix over a non-commutative algebra. In particular, we ask the question,
"over which algebras, is the determinant easier to compute than the permanent?"
Towards resolving this question, we show the following hardness and easiness of
noncommutative determinant computation.
  * [Hardness] Computing the determinant of an n \times n matrix whose entries
are themselves 2 \times 2 matrices over a field is as hard as computing the
permanent over the field. This extends the recent result of Arvind and
Srinivasan, who proved a similar result which however required the entries to
be of linear dimension.
  * [Easiness] Determinant of an n \times n matrix whose entries are themselves
d \times d upper triangular matrices can be computed in poly(n^d) time.
  Combining the above with the decomposition theorem of finite dimensional
algebras (in particular exploiting the simple structure of 2 \times 2 matrix
algebras), we can extend the above hardness and easiness statements to more
general algebras as follows. Let A be a finite dimensional algebra over a
finite field with radical R(A).
  * [Hardness] If the quotient A/R(A) is non-commutative, then computing the
determinant over the algebra A is as hard as computing the permanent.
  * [Easiness] If the quotient A/R(A) is commutative and furthermore, R(A) has
nilpotency index d (i.e., the smallest d such that R(A)d = 0), then there
exists a poly(n^d)-time algorithm that computes determinants over the algebra
A.
  In particular, for any constant dimensional algebra A over a finite field,
since the nilpotency index of R(A) is at most a constant, we have the following
dichotomy theorem: if A/R(A) is commutative, then efficient determinant
computation is feasible and otherwise determinant is as hard as permanent.
