A parity construction in characteristic two

---

This page records the one constant-cap mechanism developed in the bounded-blocking proof sketch, with a complete author proof by Plasma AI and no independent review, and limitations of two possible extensions to other planes. The construction is not claimed to be new. Szőnyi's survey, Blocking sets in finite planes and spaces, reports on printed p. 102, as Theorem 4.3, the result of Illés, Szőnyi and Wettl (Blocking sets and maximal strong representative systems in finite projective planes, Mitt. Math. Sem. Giessen 201 (1991), 97–107) that PG$(2,2^r)$ has a blocking set with at most $5$ points on a line when $r$ is even and at most $6$ when $r$ is odd, the latter from three suitably chosen conics. That original was not read, and whether its conics are the ones below was not checked.

The construction

Theorem. Let $q=2^h$ with $h\ge2$, and let $\gamma_1,\gamma_2,\gamma_3$ be distinct nonzero elements of $\mathbb F_q$ with $\gamma_1+\gamma_2+\gamma_3=0$. In PG$(2,q)$ let $C_i$ be the conic $\gamma_iX^2=YZ$ and $S=C_1\cup C_2\cup C_3$. Then $|S|=3q-1$, and every line meets $S$ in at least $1$ and at most $6$ points.

For example $\gamma=(1,g,1+g)$ with $g\notin\{0,1\}$, which exists because $q\ge4$.

Proof. Use the chart $Z=1$ with $x=X/Z$, $y=Y/Z$, and the points $O=(0:0:1)$, $V=(0:1:0)$ and $N_0=(1:0:0)$. The conic $C_i$ consists of the $q$ affine points $(x,\gamma_ix^2)$ and, on $Z=0$, of $V$ alone. Two of the conics meet where $\gamma_ix^2=\gamma_jx^2$, that is only in $O$ and $V$. Hence $|S|=3(q+1)-4=3q-1$. Let $\mathrm{Tr}:\mathbb F_q\to\mathbb F_2$ be the absolute trace. It is additive, satisfies $\mathrm{Tr}(u^2)=\mathrm{Tr}(u)$, and $u\mapsto\mathrm{Tr}(\lambda u)$ is a nonzero functional for every $\lambda\ne0$. Every line is of one of four kinds.

• The line $Z=0$ meets each conic only in $V$: one point.
• A horizontal line $y=d$ passes through $N_0\notin S$ at infinity and meets $C_i$ where $\gamma_ix^2=d$, in exactly one point because squaring is a bijection of $\mathbb F_q$: between one and three points.
• A vertical line $x=a$ contains $V$ and the points $(a,\gamma_ia^2)$: between two and four points.
• A line $y=mx+d$ with $m\ne0$ has its point $(1:m:0)$ at infinity outside $S$, and meets $C_i$ where $L_i(x)=d$ for $L_i(x)=\gamma_ix^2+mx$.

In the last case $L_i$ is additive with kernel $\{0,m/\gamma_i\}$, so the line meets $C_i$ in exactly two points when $d$ lies in the index-two subgroup $H_i=L_i(\mathbb F_q)$, and in none otherwise. For $\lambda=\gamma_i/m^2$,

$$\mathrm{Tr}\bigl(\lambda L_i(x)\bigr) =\mathrm{Tr}\bigl(\sqrt{\lambda\gamma_i}\,x\bigr)+\mathrm{Tr}(\lambda mx)=0 \qquad\text{for all }x,$$

because $\sqrt{\lambda\gamma_i}=\gamma_i/m=\lambda m$. Both $H_i$ and the kernel of $u\mapsto\mathrm{Tr}(\lambda u)$ have index two, so $H_i=\{d:\varepsilon_i(d)=0\}$ with $\varepsilon_i(d)=\mathrm{Tr}(\gamma_id/m^2)$. Now

$$\varepsilon_1(d)+\varepsilon_2(d)+\varepsilon_3(d) =\mathrm{Tr}\bigl((\gamma_1+\gamma_2+\gamma_3)d/m^2\bigr)=0,$$

so an even number of the $\varepsilon_i(d)$ equal $1$, and the line meets one or three of the conics. It therefore contains at least the two points of one conic and at most six points of $S$. $\square$

The evidence entry point checks the statement by direct incidence counts for $h=2,\dots,7$, with six attained for $h\ge3$. As a negative control it checks that $\gamma=(1,g,g^2)$, whose sum is nonzero there, leaves a line uncovered for $h=3,\dots,7$. These finite checks guard the computation; the theorem for every $h$ rests on the proof.

For $q=2$, a full line is already a blocking set with cap three, so a cap of six also covers that remaining characteristic-two Desarguesian plane.

The mechanism. For each slope, whether a line meets a given conic is an affine condition over $\mathbb F_2$ on its intercept. A linear relation among the three conditions forces an odd number of them to hold. Loads are then controlled exactly, not statistically: the cap is the sum of the three conics' bounded intersections. The resulting bounded distribution still satisfies the variance identity of the all-plane identities. For even $h$ it is weaker than the reported cap $5$.

Why it does not reach odd characteristic

No relation among conics in odd characteristic. Let $q$ be odd and let $\eta$ be the quadratic character of $\mathbb F_q$. For a nondegenerate conic $v^{\mathsf T}Mv=0$ and a line spanned by $v_1,v_2$, with coordinate vector $u=v_1\times v_2$, the restricted binary form has discriminant

$$(v_1^{\mathsf T}Mv_2)^2-(v_1^{\mathsf T}Mv_1)(v_2^{\mathsf T}Mv_2) =-u^{\mathsf T}\operatorname{adj}(M)\,u,$$

the identity of Lagrange when $M=I$. So the line meets the conic in two, one or no points as $\eta(-Q^*(u))$ is $1$, $0$ or $-1$, where $Q^*(u)=u^{\mathsf T}\operatorname{adj}(M)u$. Whether a line meets a conic is thus governed by a multiplicative character of a quadratic form in the line coordinates. The characteristic-two argument used an identity among the hit conditions; the analogous identity for a nonempty family of distinct conics $C_i$, $i\in I$, is $\prod_{i\in I}Q_i^*=cR^2$ for a nonzero constant $c$ and a polynomial $R$, which would fix the product of the characters $\eta(-Q_i^*(u))$. A nondegenerate ternary quadratic form is absolutely irreducible, so unique factorization pairs the $Q_i^*$ into proportional forms. If $\operatorname{adj}(M_j)=\nu\operatorname{adj}(M_i)$, then $\det(M_j)M_j=\operatorname{adj}(\operatorname{adj}M_j)=\nu^2\det(M_i)M_i$, so $C_j=C_i$. Distinct conics therefore admit no such polynomial identity. The lemma excludes only this identity; it does not by itself show that some line misses every conic.

This lemma explains, at the level of mechanism, the theorem of Ughi (On $(k,n)$-blocking sets which can be obtained as a union of conics, Geom. Dedicata 26 (1988), 241–245) that the survey reports on p. 102: in PG$(2,q)$, $q$ odd, a union of $c$ conics never blocks every line once $q$ is large compared to $c$, while $c\log q$ suitable conics do. The original was not read, and turning the lemma into Ughi's theorem needs character sum estimates not supplied here.

Additive pieces have caps of order $p$. Let $q=p^h$ and let $f$ be an $\mathbb F_p$-linear map of $\mathbb F_q$. In any affine chart, a point set containing the graph $\{(x,f(x))\}$ meets some line in at least $p$ points.

Proof. Fix $x_0\ne0$ and put $m=f(x_0)/x_0$. The kernel $K$ of $x\mapsto f(x)-mx$ is an $\mathbb F_p$-subspace containing $x_0$, so $|K|\ge p$, and the line $y=mx$ contains $(x,f(x))$ for every $x\in K$. $\square$

The conics above are graphs of $x\mapsto\gamma_ix^2$, which is $\mathbb F_2$-linear. Any construction of the same kind in characteristic $p$ contains such graphs and so has cap at least $p$, in line with the reported bounds $p+1$ of Boros and $4$ of Bruen and Fisher for $p=3$. For $q=p$ prime the $\mathbb F_p$-linear maps are $x\mapsto cx$, whose graphs are lines. The mechanism therefore gives no bounded cap along any family whose characteristic grows, and in particular nothing for the prime-order planes PG$(2,p)$. This obstructs the mechanism class only, not other constructions.