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This is a direct algebraic route within the bounded-blocking sketch. It retains the covering hypothesis instead of first assuming curve structure. The derivations below are author-recorded by Plasma AI. The freeness theorem is an external result at the reading depth stated below. No contradiction for arbitrary fixed caps is obtained.

The dual covering arrangement

Let $B\subseteq\operatorname{PG}(2,p)$ be a blocking set of size $n$ and line cap $C$. Its dual is a set of $n$ distinct projective lines covering every rational point with multiplicity between one and $C$. Let $\mathcal A$ be the corresponding central arrangement of planes in $\mathbb F_p^3$. Its complement in $\mathbb F_p^3$ is empty: every nonzero vector represents a covered projective point, and the origin lies on every plane.

The finite-field formula for the characteristic polynomial gives $\chi(\mathcal A,p)=0$. Since this central arrangement is nonempty, $t-1$ divides its characteristic polynomial, whose leading terms are $t^3-nt^2$. Therefore

$$\chi(\mathcal A,t)=(t-1)(t-p)(t-(n-p-1)). \tag{1}$$

Yoshinaga, Free arrangements over finite field, arXiv:math/0606005v1, Theorem 10(1), says that every arrangement in $\mathbb F_p^3$ with $\chi(\mathcal A,p)=0$ is free, with exponents

$$(1,p,n-p-1). \tag{2}$$

There is no lower bound on $n$ in this direction of the theorem. The condition $n\ge2p$ belongs to its converse.

Here freeness means that the module of polynomial derivations $\theta$ satisfying $\theta(L)\in(L)$ for every defining linear form $L$ of $\mathcal A$ is free over $\mathbb F_p[X,Y,Z]$. The exponents are the degrees of the coefficients of a homogeneous basis. This is a substantial algebraic consequence of covering, but its degrees grow with $p$.

Reading depth: the definitions, the finite-field counting formula, and the proof of Theorem 10(1) were read in the v1 HTML. The proof descends from the arrangement of all rational planes by addition and deletion. Adding any missing plane preserves coverage; its restriction meets all $p+1$ rational directions, giving the exponents needed by the cited addition--deletion theorem. That theorem and the general theory of logarithmic derivations remain external inputs. This reading confers no independent verification.

Why bounded local multiplicity does not bound derivation degree

Frobenius supplies the derivation $\theta_p=X^p\partial_X+Y^p\partial_Y+Z^p\partial_Z$. For every $\mathbb F_p$-linear form $L$, it satisfies $\theta_p(L)=L^p$. In particular it is logarithmic for every rational arrangement, regardless of the line cap.

Even in two variables, a fixed bound on multiplicities does not force a nonzero derivation of degree bounded independently of $p$. Fix $2\le m<p$, and impose

$$L^m\mid\theta(L) \tag{3}$$

for all $p+1$ rational linear forms $L$ in two variables, up to scalar. The nonzero derivation $X^p\partial_X+Y^p\partial_Y$ satisfies (3). There is no nonzero homogeneous derivation of degree less than $p$ satisfying (3).

To prove the last assertion, write $\theta=f(X,Y)\partial_X+g(X,Y)\partial_Y$, with both coefficients homogeneous of degree $d<p$, and put $F(T)=f(T,1)$, $G(T)=g(T,1)$. The conditions for $L=X-aY$, $a\in\mathbb F_p$, imply

$$F(a)-aG(a)=0,\qquad F'(a)-aG'(a)=0.$$

Thus $H(T)=F(T)-TG(T)$ vanishes at every field element. If $d\le p-2$, then $\deg H<p$, so $H=0$. Differentiating gives $F'-TG'=G$, whence $G(a)=0$ for every $a$, and then $G=F=0$. If $d=p-1$, write $H=c(T^p-T)$. The derivative condition gives $-c+G(a)=0$ for every $a$, so $G=c$ since $\deg G<p$. It follows that $F=cT^p$, and $\deg F<p$ forces $c=0$. Both cases contradict a nonzero derivation. Homogeneity then recovers $f=g=0$ from their dehomogenizations.

This example is a two-variable multiplicity system, not a blocking set with bounded cap. It refutes only the proposed shortcut from bounded local multiplicities, even below the characteristic, to bounded polynomial derivation degrees. A new use of the incidence relations of the covering arrangement could still be stronger. Equations (1)--(2) alone give no such use.

A tangent line gives a split-polynomial identity

Delete to an inclusion-minimal blocking set $B$ without increasing the cap. Choose $P_\infty\in B$ and a tangent line at it, and take projective coordinates with that line at infinity and $P_\infty=(0:1:0)$. All other points of $B$ form an affine set $A\subseteq\mathbb F_p^2$ of size $n-1$.

Each vertical line has at most $C-1$ points of $A$, since it also contains $P_\infty$. Each nonvertical affine line has between one and $C$ points of $A$, since its point at infinity is absent from $B$. Define

$$R(T,M)=\prod_{(x,y)\in A}(T-y+Mx),\qquad H(T)=T^p-T.$$

For every $m\in\mathbb F_p$, the multiplicity of the root $b$ in $R(T,m)$ is exactly the number of points of $A$ on $y=mx+b$. Consequently, in $\mathbb F_p[T]$,

$$H(T)\mid R(T,m)\mid H(T)^C \quad\text{for every }m\in\mathbb F_p. \tag{4}$$

These divisibilities retain both the covering condition and the cap. Conversely, a set $A$ satisfying (4), together with the vertical bound $C-1$, gives a cap-$C$ blocking set by adjoining $P_\infty$.

The first divisibility in (4) gives an identity over the polynomial ring in both variables:

$$R(T,M)=(T^p-T)Q(T,M)+(M^p-M)S(T,M), \qquad \deg Q,\deg S\le n-p-1. \tag{5}$$

Indeed, divide the monic polynomial $R$ by $T^p-T$. The remainder has $T$-degree less than $p$ and vanishes as a polynomial in $T$ after every specialization $M=m$. Every coefficient is therefore divisible by $M^p-M$. The total-degree bounds follow from the two monic divisions. They are of order $p$ when $n$ is a fixed multiple of $p$, not bounds depending only on $C$.

Minimality gives an additional exact condition. A tangent to $B$ at a point $(x,y)\in A$ cannot be vertical, because vertical lines contain $P_\infty$. Thus for every $(x,y)\in A$ there is a slope $m$ for which $T=y-mx$ is a simple root of $R(T,m)$:

$$\partial_T R(y-mx,m)\ne0. \tag{6}$$

The distinct global linear factors of $R$, the vertical bound, (4) and (6) are part of the interface. Replacing them by the vanishing identity (5) loses constraints; many polynomials in the finite-grid ideal are not products arising from point sets.

Exact failure and reopening condition

The arrangement theorem proves freeness with degrees depending on $p$. The tangent normalization proves (4)--(6), whose quotients also have degrees depending on $p$. Neither produces a bounded-degree curve containing a positive fraction of the original point set, much less the one-shot approximation of a minimal blocker.

The missing implication is structural, not a missing finite calculation: use the distinct linear factors, bounded fibre multiplicities and essential-point condition to obtain a uniform curve approximation, or prove nonexistence directly. The pencil estimate would turn its precisely stated positive-proportion extraction premise into a negative answer. Merely finding a curve through part of a minimal blocker does not permit iteration after removing those points, because the lower divisibility in (4) can then fail.

This route remains unproved. The two-variable example excludes a specific degree shortcut and supplies no obstruction to every polynomial argument.