February | 2022 | Lovely little lemmas

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.

Lovely little lemmas

Or maybe ‘amusing observations’

Monthly Archives: February 2022

Union of hyperplanes over a finite field