Solution to Bishnoi's conjecture on minimal t-fold blocking sets of maximal size

TL;DR

This paper proves Bishnoi's conjecture that minimal t-fold blocking sets of maximal size in projective planes are either the entire plane minus one point, the complement of a Baer subplane, or a unital.

Contribution

The paper provides a proof confirming Bishnoi's conjecture, classifying all minimal t-fold blocking sets of maximal size in projective planes.

Findings

Confirmed Bishnoi's conjecture.

Classified maximal minimal t-fold blocking sets.

Established conditions for such sets in projective planes.

Abstract

Bishnoi conjectured that if a minimal t-fold blocking set in a projective plane of prime power order has maximal size then it is either a projective plane minus one point, the complement of a Baer subplane or a unital. In this note we prove this conjecture.

Equations

$$\frac{1}{2}n\sqrt{4tn-(3t+1)(t-1)}+\frac{1}{2}(t-1)n+t$$

$$p^h(p^h+\alpha p^l-1)-\alpha p^l+{\alpha }^2{p}^{2l}=\alpha p^{l}q$$

$$(p^h+\alpha p^h-1)-\alpha +{\alpha }^2{p}^h=\alpha q$$

Full text

A $t$-fold blocking set in a projective plane $\pi$ of order $n$ is a set of points $\mathcal{K}$ such that each line of $\pi$ intersects $\mathcal{K}$ in at least $t$ points and some line of $\pi$ intersects $\mathcal{K}$ in exactly $t$ points. It is called minimal if every point of $\mathcal{K}$ is contained in a line intersecting $\mathcal{K}$ in exactly $t$ points.

The case $t=1$ is well-studied going back to Bruen and Thas [4]. They have shown that in a finite projective plane of order $n$ the size of a (1-fold) minimal blocking set is bounded above by $n\sqrt{n}+1$. This bound is sharp in the case when $n$ is a square, and $S$ is a unital in $\pi$, i.e. a set of $n\sqrt{n}+1$ points in $\pi$ such that each line of $\pi$ intersects $S$ in 1 or $\sqrt{n}+1$ points. An example of a unital in a Desarguesian plane of square order $\mathrm{PG}(2,q^{2})$ is the point set $\mathcal{K}$ of a non-degenerate Hermitian variety, i.e. the points $(x,y,z)\in\mathrm{PG}(2,q^{2})$ satisfying $x^{q+1}+y^{q+1}+z^{q+1}=0$.

Multiple blocking sets were introduced by Bruen in [3] and lower bounds were obtained by Ball in [1]. In [2] Bishnoi proved the first general upper bound on the size of minimal $t$-fold blocking sets extending the result of Bruen and Thas to the case of general $t$. More precisely he proved the following

[TABLE]

If the size of $S$ is equal to this upper bound, then every line of $\pi$ intersects $S$ in exactly $t$ or $b+1:=\frac{1}{2}(\sqrt{4tn-(3t+1)(t-1)}+t-1)+1$ points.

In particular for the case $t=1$ he recovers the result of Bruen and Thas, for $t=n$ the bound is equal to $n^{2}+n$, and $S$ is a projective plane minus a point, and when $n$ is a square for $t=n-\sqrt{n}$ we obtain the upper bound $n^{2}-\sqrt{n}$, in which case $S$ is the complement of a Baer subplane. A Baer subplane $\mathcal{B}$ in $\pi$ is a set of $n+\sqrt{n}+1$ points of $\pi$ such that each line of $\pi$ intersects $\mathcal{B}$ in 1 or $\sqrt{n}+1$ points. For example, in a Desarguesian plane of square order $\mathrm{PG}(2,q^{2})$ the subplane $\mathrm{PG}(2,q)$ is a Baer subplane.

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$t=1$* when $q$ is a square, in this case $S$ is a unital in $\pi$.*

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$t=q-\sqrt{q}$* when $q$ is a square, in this case $S$ is the complement of a Baer subplane in $\pi$.*

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$t=q$* for any $q$, in this case $S$ is the plane $\pi$ with one point removed.*

[TABLE]

By $(\dagger)$ $t$ is a divisor of $b(b+1)=(p^{h}-1+t)(p^{h}+t)$, hence $t=\alpha p^{l}$ is a divisor of $p^{h}(p^{h}-1)$, so $l\leq h$ and $\alpha$ divides $p^{h}-1$ $(\ast)$. We will distinguish four cases.

Case I $l>0,h>0,l<h$: dividing (1) by $p^{l}$ implies that $p$ divides $\alpha$, a contradiction.

Case II $l>0,h>0,l=h$: Then after division of (1) by $p^{h}$ we obtain

[TABLE]

So $\alpha=\beta p^{h}-1$ and by (*) $\alpha=p^{h}-1$, and (2) yields after simplification that $p^{2h}=q$, hence $q$ is a square, $t=q-\sqrt{q}$ and $b=q-1$, and $S$ is the complement of a Baer subplane.

Case III $l=0,h>0$: Then from (1) we get $p^{h}(p^{h}+\alpha-1)-\alpha+\alpha^{2}=\alpha q$ which implies that $\alpha^{2}-\alpha$ is a multiple of $p^{h}$. Since $(\alpha,p)=1$ we must have $\alpha=\beta p^{h}+1$. By $(\ast)$ $\beta=0$, so $\alpha=1$ implying $p^{2h}=q$ and hence $q$ is a square, $t=1$ and $b=\sqrt{q}$, and $S$ is a unital.

Case IV $l=h=0$: Then we have $b=t=q$, and $S$ is the projective plane with one point removed.

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