Traces of nilpotent matrices

Posted on 18 November 2022 by Remy

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

Posted on 22 February 2022 by Remy

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
$x_0x_1^qx_2^{q^2}+x_1x_2^qx_0^{q^2}+x_2x_0^qx_1^{q^2}-x_0^{q^2}x_1^qx_2-x_1^{q^2}x_2^qx_0-x_2^{q^2}x_0^qx_1$
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

Posted on 1 April 2021 by Remy

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. $\qedsymbol$