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

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 q be a power of a prime p, and let x_0,\ldots,x_n \in \bar{\mathbf F}_q. Then x_0,\ldots,x_n satisfy a linear relation over \mathbf F_q if and only if

    \[\det \begin{pmatrix} x_0 & x_1 & \cdots & x_n \\ x_0^q & x_1^q & \cdots & x_n^q \\ \vdots & \vdots & \ddots & \vdots \\ x_0^{q^n} & x_1^{q^n} & \cdots & x_n^{q^n} \end{pmatrix} = 0.\]

Proof. If \sum_{i=0}^n c_ix_i = 0 for (c_0,\ldots,c_n) \in \mathbf F_q^n - \{0\}, then c_i^{q^j} = c_i for all i,j \in \{0,\ldots,n\} since c_i \in \mathbf F_q. As (-)^q \colon \bar{\mathbf F}_q \to \bar{\mathbf F}_q is a ring homomorphism, we find

    \[\begin{pmatrix} x_0 & x_1 & \cdots & x_n \\ x_0^q & x_1^q & \cdots & x_n^q \\ \vdots & \vdots & \ddots & \vdots \\ x_0^{q^n} & x_1^{q^n} & \cdots & x_n^{q^n} \end{pmatrix}\begin{pmatrix} c_0 \\ c_1 \\ \vdots \\ c_n \end{pmatrix} = 0,\]

so the determinant is zero. Conversely, the union of \mathbf F_q-rational hyperplanes H \subseteq \mathbf P^n_{\mathbf F_q} is a hypersurface Y of degree |\check{\mathbf P}^n(\mathbf F_q)| = q^n + \ldots + q + 1 (where \check{\mathbf P}^n denotes the dual projective space parametrising hyperplanes in \mathbf P^n). Since the determinant above is a polynomial of the same degree q^n + \ldots + q + 1 that vanishes on all \mathbf F_q-rational hyperplanes, we conclude that it is the polynomial cutting out Y, so any [x_0:\ldots:x_n] \in \mathbf P^n(\bar{\mathbf F_q}) for which the determinant vanishes lies on one of the hyperplanes. \qedsymbol

Of course when the determinant is zero, one immediately gets a vector (c_0,\ldots,c_n) \in \bar{\mathbf F}_q^{n+1} - \{0\} in the kernel. There may well be an immediate argument why this vector is proportional to an element of \mathbf F_q^{n+1}, but the above cleverly circumvents this problem.

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

  • n=0: a point x_0 \in \bar{\mathbf F}_q only satisfies a linear relation over \mathbf F_q if it is zero.
  • n=1: the polynomial x_0x_1^q-x_0^qx_1 cuts out the \mathbf F_q-rational points of \mathbf P^1.
  • n=2: the polynomial


    cuts out the union of \mathbf F_q-rational lines in \mathbf P^2. This is the case considered in the paper.

Algebraic closure of the field of two elements

I think I learned about this from a comment on MathOverflow.

Recall that the field of two elements \mathbf F_2 is the ring \mathbf Z/2\mathbf Z of integers modulo 2. In other words, it consists of the elements 0 and 1 with addition 1+1 = 0 and the obvious multiplication. Clearly every nonzero element is invertible, so \mathbf F_2 is a field.

Lemma. The field \mathbf F_2 is algebraically closed.

Proof. We need to show that every non-constant polynomial f \in \mathbf F_2[x] has a root. Suppose f does not have a root, so that f(0) \neq 0 and f(1) \neq 0. Then f(0) = f(1) = 1, so f is the constant polynomial 1. This contradicts the assumption that f is non-constant. \qedsymbol