Traces of nilpotent matrices

I needed the following well-known result in the course I’m teaching:

Lemma. Let k be a field and M \in M_n(k) a nilpotent matrix. Then \operatorname{tr} M = 0.

The classical proof uses the flag of subspaces

    \[0 \subseteq \ker M \subseteq \ker M^2 \subseteq \ldots \subseteq \ker M^n = k^n\]

to produce a basis for which M is upper triangular. Here is a slick basis-independent commutative algebra proof that shows something better:

Lemma. Let R be a commutative ring with nilradical \mathfrak r, and let M \in M_n(k) be a nilpotent matrix. Then the characteristic polynomial p_M(t) = \det(tI-M) satisfies

    \[p_M(t) \in t^n + \mathfrak r[t]_{< n}.\]

Here we write I[t]_{<n} \subseteq R for the polynomials in R of degree smaller than n whose coefficients lie in a given ideal I \subseteq R.

Note that the formulation should ring a bell: in the previous post we saw that R[t]^\times = R^\times + t\mathfrak r[t]. When R is a domain, this reduces to R[t]^\times = R^\times, and the lemma just says that p_M(t) = t^n.

This suggests that we shouldn’t work with \det(tI-M) but with its anadrome (or reciprocal) t^n\det(t^{-1}I-M) = \det(I-tM).

Proof of Lemma. We have to show that \det(I-tM) \in 1 + t\mathfrak r[t]. Since M is nilpotent, there exists r \in \mathbf N with M^{r+1} = 0, so (I-tM)(I+tM+\ldots+t^rM^r) = I. Thus \det(I-tM) \in R[t]^\times = R^\times + t\mathfrak r[t]. Evaluating at t=0 shows that the constant coefficient is 1. \qedsymbol