Exponential restricts to special linear matrices

Question

Let's consider a field $k$ of characteristic $p$ and a matrix $M \in \mathfrak{sl}_n$ (the Lie algebra of trace $0$ matrices). Assume $M^r = 0$ for some $r < p$ so that the exponential $$\exp(M) = I + M + \frac12M^2 + \cdots + \frac{1}{(r-1)!}M^{r - 1}$$ is well defined. Then it's true that $\det(\exp(M)) = 1$.

The easiest way to see this is to note that $\exp$ commutes with conjugation so you just have to prove it for matrices in Jordan normal form, and there it's obvious because all the eigenvalues are zero.

My question is how can this be proven without using Jordan normal form? I ask because I would like to be able to show that this is still true when the entries in $M$ lie in some $\mathbb F_p$-algebra which is not necessarily a field.

Answer 1

Let $R$ be a commutative ring with unity and $M\in M_n(R)$ s.t. $M^r=0$ ; thus $e^M$ can be defined. Moreover a result due to Almkvist says that $tr(M)^{(r-1)n+1}=0$ and, consequently, $e^{tr(M)}$ can be defined. A consequence of this result is the required equality $\det(e^M)=e^{tr(M)}$ where $M$ is a nilpotent matrix over $R$. cf. footnote $9$ in

http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimPDF/gert.pdf

EDIT: a necessary condition when $r\geq 2$ : $((r-1)n)!$ must be invertible in $R$.

Answer 2

The following argument works when the scalars are a field (or, I believe, a UFD). I'm unfortunately too ignorant of generalizations of linear algebra to rings to say if it will help in your setting:

Let $N = I-\exp(M)$. Then $N^r = 0$, so $N$ is also nilpotent and has characteristic polynomial $x^n$. Therefore $$x^n = \det(Ix-N)$$ and substituting $x=1$ gives $\det(\exp(M)) =1$.