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.

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