# Traces of nilpotent matrices

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

Lemma. Let be a field and a nilpotent matrix. Then .

The classical proof uses the flag of subspaces to produce a basis for which is upper triangular. Here is a slick basis-independent commutative algebra proof that shows something better:

Lemma. Let be a commutative ring with nilradical , and let be a nilpotent matrix. Then the characteristic polynomial satisfies Here we write for the polynomials in of degree smaller than whose coefficients lie in a given ideal .

Note that the formulation should ring a bell: in the previous post we saw that . When is a domain, this reduces to , and the lemma just says that .

This suggests that we shouldn’t work with but with its anadrome (or reciprocal) .

Proof of Lemma. We have to show that . Since is nilpotent, there exists with , so . Thus . Evaluating at shows that the constant coefficient is 1. # Union of hyperplanes over a finite field

The following lemma is a (presumably well-known) result that Raymond Cheng and I happened upon while writing our paper Unbounded negativity on rational surfaces in positive characteristic (arXiv, DOI). Well, Raymond probably knew what he was doing, but to me it was a pleasant surprise.

Lemma. Let be a power of a prime , and let . Then satisfy a linear relation over if and only if Proof. If for , then for all since . As is a ring homomorphism, we find so the determinant is zero. Conversely, the union of -rational hyperplanes is a hypersurface of degree (where denotes the dual projective space parametrising hyperplanes in ). Since the determinant above is a polynomial of the same degree that vanishes on all -rational hyperplanes, we conclude that it is the polynomial cutting out , so any for which the determinant vanishes lies on one of the hyperplanes. Of course when the determinant is zero, one immediately gets a vector in the kernel. There may well be an immediate argument why this vector is proportional to an element of , but the above cleverly circumvents this problem.

For concreteness, we can work out what this determinant is in small cases:

• : a point only satisfies a linear relation over if it is zero.
• : the polynomial cuts out the -rational points of .
• : the polynomial cuts out the union of -rational lines in . This is the case considered in the paper.

# Algebraic closure of the field of two elements

Recall that the field of two elements is the ring of integers modulo . In other words, it consists of the elements and with addition and the obvious multiplication. Clearly every nonzero element is invertible, so is a field.
Lemma. The field is algebraically closed.
Proof. We need to show that every non-constant polynomial has a root. Suppose does not have a root, so that and . Then , so is the constant polynomial . This contradicts the assumption that is non-constant. 