Algebraic selection barriers for periodic no-three-in-line supports

Authors: lattice_lift_research and cedar_conics.

1. Problem, model and principal results

A no-three-in-line set is a subset of an integer square containing no three distinct points on one Euclidean line. Classical constructions approach density $3/2$ relative to the side length. The rStation problem asks for an unbounded family of density greater than $1.8$, or a proof that none exists. We study a class of proposed constructions built from periodic rational graphs. Our results restrict those supports and their selection rules; they do not give an upper bound for arbitrary no-three-in-line sets.

Fix a positive integer $k$. For a prime $p$ and a reduced rational function $f=P/Q\in\mathbb F_p(X)$, define

$$ \Gamma_p(f;k)=\{(x+ip,y+jp):0\le x,y<p,\ Q(x)\ne0, \ y=f(x),\ 0\le i,j<k\}. $$

Its plane degree is

$$ N(f)=\max(\deg P,\deg Q+1), $$

the degree of the projective closure of $Q(X)Y-P(X)=0$. It is also the degree of a generic oblique projection of this graph. The rational-map degree of $f$ can be different; both notions will be used explicitly.

Our first result places no restriction on the selected subset.

Theorem A (low plane degree). There is an absolute $p_0$ such that, for every prime $p>p_0$, every reduced $f\in\mathbb F_p(X)$ with $N(f)\le3$, and every no-three-in-line set $S\subseteq\Gamma_p(f;2)$,

$$ |S|\le3p+3. $$

The coefficients of $f$ may be chosen independently for each prime. Thus a single graph of plane degree at most three cannot improve the asymptotic density $3/2$ in a square of side $2p$.

For higher degrees we constrain the description of the selection. An arithmetic test cover is a finite Galois extension $E/\mathbb F_p(X)$. Write $A$ for its arithmetic group, $H$ for its geometric group, and $\mathbb F_{p^\nu}$ for its constant field. Let $A_p\subset A$ be the coset inducing $c\mapsto c^p$ on the constants. For an unramified residue $x$, its Frobenius conjugacy class is a class $c\subset A_p$. A test can distinguish any of these classes. This includes root-existence and splitting-type conditions for auxiliary polynomials, and multiplicative-character predicates after taking their Galois cover.

Assign to each such class a geometric mask $\Omega_c\subset[0,k)^2$. The selected set $T$ contains a physical lift $v$ of $(x,f(x))$ exactly when $v/p\in\Omega_{c(x)}$. Branch residues may be handled arbitrarily. In particular, different arithmetic outcomes may choose different lift corners or geometric cuts.

Theorem B (bounded algebraic selection). Fix $D\ge4$, $k$, $h$ and $B$. For every sufficiently large prime, let $4\le N(f)\le D$, let $[E:\mathbb F_p(X)]\le h$, and suppose the test cover has at most $B$ geometric branch points. All coefficients may depend on $p$. If $S$ is no-three-in-line and $|S\mathbin\triangle T|\le d$, the following estimates hold uniformly.

If the masks are unions of rectangles, with at most $K$ rectangles in total, then

$$ |S|\ll_{D,k,h,B}(K+1)^{4/3}p^{5/6}(\log p)^2 +p^{2/3}d^{1/3}+d. \tag{1.1} $$

If their boundaries lie in real polynomial zero sets of total degree at most $E_0$, together with the square boundary, then

$$ |S|\ll_{D,k,h,B}(E_0+1)^{4/7}p^{13/14}(\log p)^{6/7} +p^{2/3}d^{1/3}+d. \tag{1.2} $$

Consequently, complexity $o(p^{1/8}/(\log p)^{3/2})$ and $d=o(p)$ force $|S|=o(p)$. Taking a trivial test cover gives the purely geometric version. The estimates concern all masks with the stated complexity, including adaptively chosen endpoints, coefficients, Boolean acceptance rules and boundary conventions.

A mixed-component argument prevents bounded unions from simply retaining several cubic pieces.

Theorem C (mixed rigidity). Fix a number of components, a plane-degree bound, a period multiplier $k$, and bounds on arithmetic-cover and geometric-mask complexities. Take distinct non-affine rational graphs of plane degree at least two, with a selection on each described as in Theorem B. Suppose the union differs from a no-three set by $o(p)$ points. For any pair of components, if one has plane degree at least three, the minimum of their two selected sizes is $o(p)$. In particular, when $k=2$ and every component has plane degree at least three, the total size is at most $3p+o(p)$.

The minimum formulation permits the dominant component to change with $p$. Multiple conic components remain outside the final conclusion. Theorem C uses fixed mask complexity; no quantitative growing-complexity rate is asserted for unions.

2. Prior work and the new contribution

Symmetric monodromy of generic projections and finite-field Chebotarev are classical inputs. Entin proves product symmetric monodromy for hyperplane sections of reflexive reducible curves and gives a uniform Chebotarev theorem retaining arithmetic constant fields [1, Theorems 2–3]. We use graph-specific tangent and resultant arguments to choose a projection from a bounded set. On the resulting covers, curve character-sum estimates [2] give physical coordinate distribution. The new statements are the selection bounds (1.1)–(1.2), the low-degree physical-capacity consequence, and the mixed-component rigidity theorem. We do not claim the monodromy or Chebotarev principles themselves as new.

The classical construction of Hall, Jackson, Sudbery and Wild [3] establishes the relevance of periodic algebraic supports to the density $3/2$ benchmark. Recent rStation work bounds lifts of every finite-field arc in windows of at least one period [4]. That hypothesis excludes modular collinear triples at the outset. Here the supports may contain many such triples, and the question is whether a selection can eliminate their physical realizations.

Our earlier coordinate-capacity results analyze row and column constraints for unions of periodic polynomial graphs [5] and rational graphs related by free Möbius actions [6]. The latter paper exhibits a quartic family with asymptotic coordinate-relaxation capacity $1.802335973378451\ldots$, so coordinate constraints alone leave a selection problem above $1.8$. The present work imposes exact oblique collinearity and bounds the complexity of successful selection rules. It does not assert that the coordinate optimizer has a bounded-complexity description, and therefore does not settle arbitrary selections from that quartic family. Theorem A instead gives an unrestricted selection bound for a single graph of plane degree at most three; Theorems B and C address bounded arithmetic tests combined with geometric masks.

3. Uniform good projections of rational graphs

Work first over $\overline{\mathbb F}_p$ and let $f=P/Q$ be non-affine of plane degree $N\le D$. Assume $p>2D$. Its derivative is nonconstant: otherwise $f-cX$ would lie in $\overline{\mathbb F}_p(X^p)$ while having rational-map degree below $p$, forcing it to be constant. If $r$ is the number of distinct finite poles, cancellation gives

$$ \deg(f')=c=N+r-1\le2N-2<p. $$

Thus $m=-f'(X)$ is a separable rational map. Set $t=f(X)-Xf'(X)$ and $K=\overline{\mathbb F}_p(m,t)$. The derivation $d/dm$ extends uniquely through the separable extensions $K/\overline{\mathbb F}_p(m)$ and $\overline{\mathbb F}_p(X)/K$. Since

$$ \frac{dt}{dm}=\frac{-Xf''(X)}{-f''(X)}=X, $$

we have $X\in K$. The tangent-value map $X\mapsto(m,t)$ is therefore birational onto its image. For generic $m$, the critical points of $F_m=f+mX$ are simple and have distinct values.

This generic statement has a uniform finite exceptional set. Put

$$ E_Q=\gcd(Q,Q'),\qquad C_m=\frac{P'Q-PQ'+mQ^2}{E_Q}, $$

and

$$ B_C(m,T)=\operatorname{Res}_X(C_m,P+mXQ-TQ). $$

Generically $\deg_X C_m=c$, $\deg_T B_C=c$, and $\deg_mB_C\le N+c$. At a pole of order $e$, cancellation leaves a nonzero value of $C_m$ when $p>e$, so $C_m$ and $Q$ have no common root. The product of the two relevant $X$-leading coefficients, the $T$-leading coefficient of $B_C$, and $\operatorname{Res}_T(B_C,\partial B_C/\partial T)$ is nonzero. Its $m$-degree is at most

$$ 2+2c(N+c)<16D^2. \tag{3.1} $$

Outside its zero set, $F_m$ has degree $N$ and all finite inertia groups are transpositions. Its geometric group $G\le S_N$ is transitive. The normal subgroup generated by finite inertia is all of $G$: otherwise the corresponding quotient cover would be a nontrivial tame cover of $\mathbb P^1$ branched only at infinity. For a degree-$e$ such cover, Riemann–Hurwitz gives a right-hand side at most $-2e+(e-1)<-2$ when $e>1$, a contradiction. Tameness holds because $p>N$ and $|G|$ divides $N!$. Hence $G$ is a transitive group generated by transpositions, so $G=S_N$. The arithmetic group is also $S_N$.

For a test cover, let $H_0\in\mathbb F_p[X]$ be the squarefree polynomial whose geometric roots are its finite branch points that are not poles of $f$. Its degree $b\le B$ includes all Frobenius conjugates. Form

$$ B_H(m,T)=\operatorname{Res}_X(H_0,P+mXQ-TQ). $$

Its $T$-degree and $m$-degree are at most $b$, and its $T$-leading coefficient is nonzero and independent of $m$. The two further bad factors

$$ \operatorname{disc}_T(B_H),\qquad \operatorname{Res}_T(B_H,B_C) $$

have $m$-degrees at most $2B^2$ and $b(N+2c)\le5DB$. They are nonzero. Distinct branch points have distinct projection images for generic $m$. A fixed branch point $\beta$ sharing its image with a critical point $z$ would satisfy

$$ f(z)-f(\beta)-(z-\beta)f'(z)=0. $$

This rational function has derivative $-(z-\beta)f''(z)\ne0$, so it gives only finitely many critical points and slopes. Branch points at graph poles or infinity map to infinity and need no separation.

It follows that, for

$$ L=16D^2+2B^2+5DB, $$

some integer $m\in\{k,\ldots,k+L\}$ is good whenever $p>k+L$. This bounded choice is uniform in every graph and cover coefficient.

4. Independence on selected root blocks

The following elementary group lemma is the key to treating tests branched at graph poles.

Lemma 4.1. Let $H$ be a finite group. Suppose a transitive subgroup $G\le H\wr S_N$ on $H\times\{1,\ldots,N\}$ is generated by labelled transpositions, carrying labels $\delta,\delta^{-1}$ on their exchanged blocks, and translations supported on a single block. Then $\ker(G\to S_N)$ projects surjectively onto $H^{N-1}$ on any $N-1$ blocks.

Proof. The graph of transposition edges is connected. Change the label origin in each block along a spanning tree so that its tree-edge swaps are pure. They generate pure $S_N$ inside the switched group. A residual labelled swap multiplied by its pure swap gives a pair translation with labels $\delta,\delta^{-1}$. Conjugation places that pair on any two blocks. Single-block translations can likewise be moved to any block.

Let $W\le H$ be generated by the residual swap labels and the single-block labels. Every generator preserves a left coset $W\lambda$ of the label, so transitivity forces $W=H$. Reserve one omitted block. Pair translations against it project to each generator of $H$ on any chosen selected block and to the identity on every other selected block. Their projections generate $H^{N-1}$. Undoing the origin change conjugates the coordinates and preserves surjectivity. $\square$

The lemma does not assert independence of all $N$ blocks. A global parity or abelianization relation may remain. One omitted block absorbs that restriction.

First take a regular test cover with geometric and arithmetic group $H$. Compose it with the good projection $F_m$. Its geometric monodromy embeds in $H\wr S_N$ and is transitive. At a finite critical value of $F_m$, the test cover is unramified at every root of the fiber, so inertia is a labelled transposition. At any other finite branch image, exactly one root block meets an arithmetic branch point, so inertia is a single-block translation. All conjugates of finite inertia generate the whole group, by the same tame one-branch-point argument used above. Conjugation preserves these inertia forms. Lemma 4.1 applies.

When $N\ge4$, the cover of three ordered distinct projection roots, each carrying a point of the test cover, is consequently a geometrically connected $H^3$ cover of the ordered-root base. It has bounded degree, genus and coordinate pole degrees, uniformly in the data. Its twists give independent Frobenius information on those three roots.

For a nonregular test cover with arithmetic group $A$, geometric group $H$ and constants $\mathbb F_{p^\nu}$, apply the geometric argument to a component over the algebraic closure. Over the regular ordered-three-root base, the arithmetic compositum embeds in

$$ A\mathbin{\times}_{A/H}A\mathbin{\times}_{A/H}A. $$

Its degree is at most $\nu|H|^3$. Its geometric group $H^3$ and contained constant field give the same lower degree. Equality follows, with no additional constants. The degree-one Frobenius coset is exactly $A_p^3$.

An $A$-conjugacy class of $g\in A_p$ is also its $H$-conjugacy class: the image of $g$ generates the cyclic quotient $A/H$, so every element of $A$ is $g^jh$ for some $h\in H$. Therefore its weight is

$$ \mu_c=\frac{|c|}{|H|}=\frac1{|C_H(g)|}. $$

A semilinear Frobenius twist of a geometric component has $p+O(\sqrt p)$ rational points and $|C_H(g)|$ lifts above each residue of this class. This is the standard twist proof of finite-field Chebotarev [1, Section 4]. On the three-root cover the weights multiply, giving $\mu_{c_1}\mu_{c_2}\mu_{c_3}$.

5. Positional discrepancy and quantitative masks

5.1. Additive relations and box counts

On the curve of three ordered distinct roots of $F_m(X)=t$, write the graph coordinates as $(x_i,y_i)$, $y_i=f(x_i)$. Since $N\ge4$, an unused root can be exchanged with each selected root while fixing the other two. Applying these transpositions to a proposed linear identity shows that $t,x_1,x_2,x_3$ are linearly independent modulo constants. Therefore the only relations among the six coordinates are

$$ m(x_i-x_1)+(y_i-y_1)=0,\qquad i=2,3. \tag{5.1} $$

These statements hold over the algebraic closure in every sufficiently large characteristic. Finite extension by the arithmetic test cover introduces no new equality between the original functions.

Let $\mathcal T_m\subset\mathbb T^6$ be the connected subtorus defined by (5.1) modulo one, with Haar probability measure $\mu_m$. Normalized finite-field coordinates use their representatives in $[0,p)$ divided by $p$. They lie exactly on $\mathcal T_m$, because the common-intercept congruences become integer equations after normalization.

Fix a triple of arithmetic classes. On the corresponding geometrically connected twists, a Fourier coefficient of the coordinate distribution is an additive character sum of a linear combination of the six coordinate functions. Unless the frequency belongs to the modular annihilator of (5.1), this function is nonconstant. Its pole orders are bounded independently of $p$ and of the coefficients. For sufficiently large $p$ it cannot be an Artin–Schreier difference $g^p-g$ plus a constant: any pole of a nonconstant such difference has order at least $p$. The curve Weil bound consequently gives an error $O(\sqrt p)$ for every nontrivial Fourier coefficient, uniformly in the frequency. The implicit constant depends only on the fixed degree, branch and cover bounds.

For frequencies of size at most $T=\lfloor\sqrt p\rfloor$, the modular and integral annihilators agree once $p$ is sufficiently large. Indeed, their defining integer expressions have size $O_m(T)<p$. The good slopes belong to a fixed finite set, so this comparison is uniform.

We record the box approximation carefully because the Haar measure is singular in the ambient torus. For an interval, Selberg trigonometric majorants and minorants of degree $T$ have bounded supremum, Fourier coefficient absolute sum $O(\log T)$, and integral gap $O(T^{-1})$. For a box, a product of majorants is an upper approximation. A lower approximation is obtained by subtracting the sum of the one-coordinate gaps multiplied by the other majorants. The telescoping product inequality makes these bounds pointwise valid. Every single-coordinate marginal of $\mu_m$ is uniform, so each gap has $\mu_m$-integral $O(T^{-1})$ even though the coordinates are not independent. Boundedness of the remaining factors controls the product error. Summing Fourier errors gives

$$ \#\{\text{ordered root triples of specified classes in }Q\} =p\,\mu_{c_1}\mu_{c_2}\mu_{c_3}\mu_m(Q) +O\bigl(p/T+\sqrt p(\log T)^6\bigr). $$

Thus every half-open coordinate box $Q$ satisfies

$$ \#(Q)=p\,\mu_{c_1}\mu_{c_2}\mu_{c_3}\mu_m(Q) +O\bigl(\sqrt p(\log p)^6\bigr), \tag{5.2} $$

uniformly in its endpoints. The analogous one-root graph argument gives two-dimensional box error $O(\sqrt p(\log p)^2)$, because a non-affine $f$ has no nonzero constant linear combination of $x$ and $f(x)$.

5.2. Count only boxes meeting the four-dimensional torus

Insert all mask rectangle boundaries, reduced modulo one, into the six coordinate partitions. Add a fixed mesh sufficiently fine that the two real intercept differences in (5.1) vary by less than one within a box. On its intersection with $\mathcal T_m$, these integer differences are constant, as are the mask memberships of every physical lift. The physical collinearity kernel is therefore constant there.

The number of relevant boxes is $O((K+1)^4)$, rather than the ambient $O((K+1)^6)$. Inside the unit six-cube, specify the two integer intercept differences. This expresses $\mathcal T_m$ as a bounded union of affine four-polytopes. The coordinate partitions restrict to $O(K+1)$ affine hyperplanes on each four-polytope. Their arrangement has $O((K+1)^4)$ faces of all dimensions. Every face is incident to at most $2^6$ ambient boxes. This includes degenerate intersections and boundary faces. Wrap and lift choices add only a bounded number of hyperplanes for each coordinate, since $m,k$ are bounded. Boxes not meeting the torus contain no actual root point and need no discrepancy charge.

Define

$$ w(x,y)=\sum_c\mu_c\mathbf1_{\Omega_c}(x,y),\qquad R_w(u)=\int_{\mathbb R}w(x,u-mx)\,dx. $$

Then $0\le w\le1$. Summing (5.2) only over the relevant boxes and over the class choices gives the number $T_3$ of selected collinear triples using three distinct residue points:

$$ T_3=\frac p6\int_{\mathbb R}R_w(u)^3\,du +O\bigl((K+1)^4\sqrt p(\log p)^6\bigr). \tag{5.3} $$

To verify the normalization, parameterize Haar by the common residue intercept and three conditionally independent point positions on that residue line. Summing all physical lift intercepts converts the conditional product to the real integral displayed above. The factor $1/6$ forgets the order of the three roots. Summing the independent arithmetic class weights expands exactly to $R_w^3$; no probabilistic independence of an adaptively chosen mask is assumed.

The one-root point count is

$$ |T|=p\int w+O\bigl((K+1)^2\sqrt p(\log p)^2\bigr). \tag{5.4} $$

Overlapping rectangles cause no problem: their planar arrangement has $O((K+1)^2)$ cells. Changing conventions on rectangle edges affects at most $O_{D,k}(K+1)$ graph points. Exceptional cover branches and projection fibers contribute only bounded error.

5.3. Jensen and edits

The Radon density $R_w$ has mass $\int w$ and is supported in an interval of length $k(m+1)$. Hence

$$ \int R_w(u)^3\,du\ge \frac{(\int w)^3}{k^2(m+1)^2}. \tag{5.5} $$

Since $m\ge k$, a physical line of this direction contains at most one lift of any residue point. Otherwise two lifts differ by $p(a,b)$ with $ma+b=0$ and $|a|,|b|\le k-1$, forcing $a=b=0$. A fixed physical point consequently belongs to at most $\binom{D-1}{2}$ of the triples in (5.3).

If $S$ is no-three and $|S\triangle T|\le d$, every such triple contains a removed point of $T\setminus S$. Thus

$$ T_3\le\binom{D-1}{2}d. $$

Equations (5.3)–(5.5) give

$$ \int w\ll(K+1)^{4/3}p^{-1/6}(\log p)^2+(d/p)^{1/3}. $$

Insert this in (5.4) and use $|S|\le|T|+d$. In the nontrivial range $K+1\le p^{1/8}/(\log p)^{3/2}$, the planar discrepancy is absorbed by the first term in (1.1). Outside that range this term is already at least $p$, so $|T|\le k^2p$ gives the result directly. This proves (1.1) for every $K$.

5.4. Polynomial boundaries

Put $\Delta=E_0+1$. A real polynomial zero set of degree $E_0$ meets at most

$$ O_k(E_0H+E_0^2+H) \tag{5.6} $$

cells of a grid of mesh $1/H$ in the square, counting cells whose closures meet the curve or the square boundary. This estimate is uniform in all coefficients. To prove it, pass to the squarefree part and separate vertical and horizontal line factors. They account for $O_k(E_0H)$ cells. All remaining restrictions to grid lines are nonzero polynomials; their zeros account for another $O_k(E_0H)$ cells. A cell containing a zero but no zero on its boundary contains a compact portion of the real zero set. At an extreme abscissa one has $P=P_y=0$, also at an isolated or singular point. Squarefreeness and absence of vertical factors imply $\gcd(P,P_y)=1$, so Bézout bounds these points by $E_0(E_0-1)$. Distinct cells use distinct points. The square boundary supplies $O(H)$ cells, proving (5.6).

Use the product of boundary polynomials for all arithmetic-class masks. Their inner and outer grid approximations differ in area by

$$ O_k(\Delta/H+\Delta^2/H^2). $$

The same estimate holds for the weighted masks because their class weights sum to one. It includes arbitrary choices on the algebraic boundary itself. For grid masks, only $O(H^4)$ of the six-coordinate boxes meet the relative torus. The preceding argument gives

$$ \begin{aligned} |S|\ll{}&H^{4/3}p^{5/6}(\log p)^2 +\frac{p\Delta}{H}+\frac{p\Delta^2}{H^2} +H^2\sqrt p(\log p)^2\\ &+p^{2/3}d^{1/3}+d. \end{aligned} $$

When $\Delta\le p^{1/8}/(\log p)^{3/2}$, choose an integer

$$ H\asymp\Delta^{3/7}p^{1/14}(\log p)^{-6/7}, $$

at least the fixed mesh threshold for the physical kernel. For sufficiently large $p$, this satisfies $H\ge\Delta$ and $H\le p^{1/8}$. The quadratic boundary and planar discrepancy terms are absorbed. The first two terms have order $\Delta^{4/7}p^{13/14}(\log p)^{6/7}$, proving (1.2). If $\Delta$ is larger, the trivial support bound proves it. This completes Theorem B.

6. Low-degree classification

We prove Theorem A uniformly over all coefficients.

6.1. Degree-two projections, rows and affine graphs

Suppose one physical diagonal projection $f(x)+x$ or $f(x)-x$ has separable rational-map degree two. Let $R_j$ be the number of finite projection values with exactly $j$ finite graph preimages, and let $M=R_1+2R_2\le p$ be the number of finite graph residues. Outside at most two branch values, a projective degree-two fiber has zero or two rational points. Omitting the domain point at infinity creates at most one further singleton, so $R_1\le3$.

One residue point has four lifts in the two-period square and contributes four to the one-direction capacity. Two distinct residue points in the same diagonal fiber have two triangular profiles of occupancies $(1,2,1)$, with equal intercept offsets or offsets differing by one period. Capping each physical line at two gives exactly six in either case. Hence the exact diagonal capacity is

$$ 4R_1+6R_2=3M+R_1\le3p+3. \tag{6.1} $$

A non-affine graph of plane degree two has at least one degree-two diagonal projection. If $f$ itself has rational-map degree two, the row bound is even smaller:

$$ |S|\le4(R_1+R_2)=2M+2R_1\le2p+6. $$

If $f$ has map degree three and geometric group $S_3$, its arithmetic group is also $S_3$. The density of nonempty row fibers is $2/3+O(p^{-1/2})$, uniformly for bounded degree, giving

$$ |S|\le\frac83p+O(\sqrt p). \tag{6.2} $$

For an affine graph $f(x)=Ax+B$, a bound uniform in an adaptive slope is $O(\sqrt p)$, not $O(1)$. Pigeonhole the $(\lceil\sqrt p\rceil+1)^2$ integer pairs in a square by their value $v-Au\bmod p$. Their difference gives a nonzero graph-lattice vector $(u,v)$ with $|u|,|v|\le\lceil\sqrt p\rceil$. All supported physical points have determinant against this vector in one residue class modulo $p$. Its range in the $2p$ square has length at most $2p(|u|+|v|)$, so there are at most $2(|u|+|v|)+1$ physical lines parallel to the vector. A no-three set has at most

$$ 4(|u|+|v|)+2=O(\sqrt p) \tag{6.3} $$

points. This includes constant graphs. The remaining case is a rational cubic with cyclic geometric group.

6.2. A two-diagonal exchange for cubic projections

Put $U=f+x$ and $V=f-x$. Suppose both have degree three and group $S_3$. For a root $x$ of $V(x)=t$, let $D(x)$ be the discriminant of its quadratic companion equation under $U$, obtained by removing $z=x$ from $U(z)=U(x)$. Suppose the three classes $D(x_i)$ are independent modulo squares in the geometric splitting field of $V(X)-t$. We claim

$$ |S|\le\frac{91}{32}p+o(p). \tag{6.4} $$

Only pairwise conditional position uniformity is needed. On any ordered $S_3$ root cover the functions $t,x_i,x_j$, $i\ne j$, are linearly independent modulo constants: exchange either selected root with the unused third root in a proposed identity. The same remains true after adjoining square roots. Bounded-pole character sums on the connected covers and their twists therefore give uniform distribution of these triples of normalized coordinates.

For a split $U$ fiber, let $u$ be its normalized residue intercept and let $L$ count its three abscissae below $u$. Its physical diagonal occupancies are

$$ (L,3+L,6-L,3-L). $$

The capped sum is six if $L=0$ or $3$, and seven otherwise. For the indicators $B_i=\mathbf1_{x_i<u}$,

$$ \mathbf1_{\text{all }B_i\text{ equal}} =1-\sum_iB_i+\sum_{i<j}B_iB_j. $$

Pairwise conditional uniformity gives mean $1-3u+3u^2$, whose integral is $1/2$. Thus the mean split-fiber capacity is $13/2$. The $S_3$ one-root and three-root fiber densities are $1/2$ and $1/6$, respectively, so the first-diagonal capacity is

$$ C_U=\left(\frac42+\frac{13/2}{6}\right)p+o(p) =\frac{37}{12}p+o(p). \tag{6.5} $$

Call a root safe when $D(x)$ is a nonsquare. Away from bounded exceptions, it is then the only rational root of its $U$ fiber. All four lifts lie on physical $U$ lines of full degree at most two. Deleting such a point is additional to every deletion already charged by the first-direction capacity.

Independence of the three square classes makes each sign twist of the ordered $V$-root curve geometrically connected, of degree $48$ over the intercept line. Counting twist points gives densities $1/16$ for split fibers with exactly two safe roots and $1/48$ for those with three. Two safe roots have eight lifts with occupancies

$$ (L,2+L,4-L,2-L),\qquad L\in\{0,1,2\}, $$

and capped sum exactly six, forcing two additional deletions. Three safe roots have mean capped sum $13/2$ by the same pairwise conditional calculation, forcing mean loss $11/2$. Distinct $V$ fibers have disjoint physical lines and point sets. The extra loss is

$$ 2\frac p{16}+\frac{11}{2}\frac p{48}+o(p) =\frac{23}{96}p+o(p). $$

Subtracting from (6.5) proves (6.4). No deletion is counted twice: the baseline charges only saturated $U$ lines, and the extra loss only points on unsaturated $U$ lines.

A sufficient criterion for square-class independence is one simple zero $\alpha$ of $D$ such that $V(\alpha)$ is an unramified value of $V$ and no other zero of $D$ has that value. At a place of the splitting curve with $x_i=\alpha$, precisely $D(x_i)$ has odd valuation. Relabeling by $S_3$ supplies an isolating valuation for each of the three classes.

6.3. Cyclic cubic normal forms

A cyclic polynomial cubic is $f=A(x-r)^3+s$, $A\ne0$. Its $U,V$ projections are $S_3$. In centered coordinates the $U$ companion discriminant is, up to a square factor,

$$ D(z)=-3z^2-\frac4A. $$

At a zero $\alpha$, the opposite projection $V=A z^3-z$ has value $-7\alpha/3$. This value is unramified and isolates the chosen zero, outside the fixed characteristics $2,3,5,7$. Thus (6.4) applies uniformly, including adaptive translations and coefficients.

Now let the cyclic cubic be nonpolynomial of plane degree three. A pole of order two would give transposition inertia, impossible in a cyclic degree-three cover. Therefore its denominator is squarefree quadratic. Translate the input to center it and divide the numerator by the denominator. After an output translation the map has the form

$$ f(x)=ax+\frac{cx+e}{x^2+d},\qquad a d\ne0. $$

Its derivative numerator is

$$ ax^4+(2ad-c)x^2-2ex+ad^2+cd. $$

Cyclic ramification forces this quartic to be a square over the algebraic closure. The missing cubic term forces its quadratic square root to have no linear term, hence $e=0$. The remaining square condition is $c(c-8ad)=0$. Reducedness excludes $c=0$, leaving

$$ f(x)=a\,\frac{x^3+9dx}{x^2+d}. \tag{6.6} $$

The original map may also have input and output translations in $\mathbb F_p$. These do not change the geometric certificate. We normalize $d$ over the algebraic closure only to check polynomial identities, not to transform the physical square; position distributions are always taken in the original coordinates.

If $a=1$ or $-1$, one diagonal projection has degree two and (6.1) applies. Otherwise both diagonal projections have degree three and group $S_3$. Indeed, for $s=\pm1$, the derivative numerator of $f+sx$ is

$$ (a+s)x^4+(-6a+2s)dx^2+(9a+s)d^2. $$

The discriminant of this quadratic in $x^2$ is $-64asd^2\ne0$, so it cannot be a square quartic. A degree-three cover that is not cyclic has group $S_3$.

6.4. Complete exceptional-parameter calculation

For the forward orientation $U=f+x$, $V=f-x$ of (6.6), its companion discriminant is

$$ D_U(x)=-4d\bigl((a+1)^2x^4+(-6a^2+12a+2)dx^2 +(9a^2+10a+1)d^2\bigr). $$

Let $E(t)$ be the resultant of $D_U(x)$ and the numerator of $V(x)-t$, with the nonzero factor $-16384a^4d^7$ removed. Let $B_V(t)$ be the cubic branch discriminant with factor $-4d$ removed. Direct symbolic calculation, reproduced in the supplement, gives

$$ \operatorname{disc}_x(D_U) =2^{32}a^6d^{12}(a+1)^3(9a+1), $$$$ \operatorname{disc}_t(E) =2^{20}a^2d^6(a+1)^3(9a+1) (7a^2+6a+3)^4(9a^2-2a+9)^2, $$$$ \operatorname{Res}_t(E,B_V) =2^{24}d^8(4a^2+1)^2(25a^2+6a+1)^4. $$

The discriminant does not vanish at a graph pole when $ad\ne0$. Thus, apart from degree drops and $a=0$, a sufficient forward bad set is the zero set of

$$ (a+1)(9a+1)(7a^2+6a+3)(9a^2-2a+9) (4a^2+1)(25a^2+6a+1). \tag{6.7} $$

The reverse orientation replaces $a$ by $-a$. The two bad sets, outside $a=\pm1$, share only $4a^2+1=0$ in sufficiently large characteristic. This is certified by their nonzero resultant after removing the common factor; the prime divisors in that computation are fixed, independent of $a,d$.

The remaining locus is also harmless. Normalize $d=1$ geometrically and take $a=i/2$. After monic normalization,

$$ \begin{aligned} D_U(x)&\sim(x^2+5)\left(x^2+\frac{13+16i}{25}\right),\\ E(t)&=\left(t^2+\frac{15}{4}+5i\right) \left(t^2+\frac{2903}{100}-\frac{251i}{25}\right),\\ B_V(t)&=\left(t^2+\frac{15}{4}+5i\right) \left(t^2+\frac{47}{4}+13i\right). \end{aligned} $$

Both $D_U$ and $E$ are squarefree, and the second factor of $E$ is coprime to $B_V$. It supplies simple isolated zeros with unramified $V$ image, exactly the criterion in Section 6.2. Conjugation handles $a=-i/2$; scaling by a square root of $d$ handles arbitrary nonzero $d$ geometrically. All identities specialize outside a fixed finite set of characteristics. Thus every cyclic nonpolynomial cubic except the degree-two projection cases satisfies (6.4).

6.5. Uniform conclusion

All curves and twists used above have bounded degree, genus and coordinate pole degrees. The exceptional characteristics are fixed by the symbolic identities and resultants. An explicit qualitative position error can be obtained by subdividing the three coordinates $(t,x_i,x_j)$ into $H^3$ boxes. Threshold boundaries meet $O(H^2)$ boxes, giving error

$$ O\bigl(H^3\sqrt p(\log p)^3+p/H\bigr). $$

Taking $H=\lfloor p^{1/8}\rfloor$ gives $O(p^{7/8}(\log p)^3)=o(p)$ uniformly in every coefficient and sign twist. Thus all cases other than the degree-two projection have a fixed leading gap below $3p$: (6.2), (6.3), the map-degree-two row bound, or (6.4). For all sufficiently large $p$ they are at most $3p$. The remaining cases satisfy $3p+3$ exactly by (6.1). This proves Theorem A.

7. Mixed-component rigidity

We prove Theorem C. All complexity bounds in this section are fixed. Graph intersections contribute only $O_D(1)$ residues per pair, so we remove them and give each remaining point a unique component label.

7.1. Product monodromy and independent selected labels

For distinct non-affine graphs $f_i$, the tangent-value curves of Section 3 are distinct. If two coincided, the derivation $dt/dm=x$ would recover the same abscissa, and $f=t-mx$ would recover the same graph. Bézout, or the resultant of their critical-value polynomials, therefore bounds the number of slopes with a shared finite critical value in terms of the degree bounds. Outside a uniformly bounded exceptional set of projective directions, all projections are Morse of their full plane degrees, and their finite branch sets are pairwise disjoint.

For a primitive integer normal $n=(a,b)$ in such a direction, with $b\ne0$, the splitting fields of $aX+bf_i(X)-t$ have geometric group

$$ \prod_i S_{N_i}. $$

Indeed, a common quotient of one splitting field and the compositum of the others could ramify only at infinity. It is tame in sufficiently large characteristic, so the one-branch-point argument of Section 3 makes it trivial. Induction gives the product. This is the graph-specific instance of the classical product-monodromy principle [1]. The arithmetic group equals the geometric product, since this already has the largest possible order.

We also require that each nonpole branch point of every arithmetic test cover avoid all projection-critical images, and that branch images within each component be distinct. These exclude only boundedly many additional directions. For example, a branch point $\beta$ on component $j$ meeting a critical image from component $i$ would require

$$ f_i(z)-f_j(\beta)-(z-\beta)f_i'(z)=0. $$

Its derivative is $-(z-\beta)f_i''(z)\ne0$, so it has boundedly many solutions. Branch images from different test covers may coincide at a shared physical graph point; these coincidences are allowed.

Write $H_i$ for the geometric test group on component $i$. The joint composed geometric group embeds in $\prod_i(H_i\wr S_{N_i})$. Finite projection-critical inertia is a labelled swap confined to one component. Other finite inertia is a translation on at most one root block in each component, possibly simultaneously on several components. All conjugates of finite inertia generate the joint group, since a quotient unramified away from infinity is trivial.

Gauge a spanning tree of critical swaps on each component as in Lemma 4.1. This supplies pure $S_{N_i}$ acting trivially on other components. Residual swap labels supply pair translations on that component. If an arithmetic inertia element $\tau$ translates its occupied block in component $i$ by $h$, commute it with a pure swap to another block of $i$. Contributions from every other component cancel, leaving labels $h,h^{-1}$ on those two blocks of $i$. The residual labels and arithmetic inertia labels generate $H_i$: otherwise the transitive action of the projected composed cover would preserve a proper coset orbit. Reserve one unselected root on each component. Pair translations against this root provide arbitrary, independent labels on all selected roots, and act trivially on all selected roots of the other components.

For the two-plus-one pattern, select two distinct roots on component $i$ and one on $j$. The omitted roots exist exactly under the hypotheses $N_i\ge3$ and $N_j\ge2$. The selected geometric test group is consequently $H_i^2\times H_j$.

This argument also handles nonregular covers. Let their constant fields have degrees $\nu_i$, and put $\nu=\operatorname{lcm}_i\nu_i$ for the components used in the count. Over the regular mixed ordered-root base, the compositum degree is at most $\nu$ times the product of the selected geometric group orders. The geometric independence just proved and the contained constant field $\mathbb F_{p^\nu}$ give the same lower bound. Thus equality holds. The degree-one Frobenius coset projects onto the Cartesian product of the individual degree-one cosets: their restrictions on every common constant field all act by $c\mapsto c^p$. The class weights of Section 4 multiply without an extra constant-field factor.

7.2. Mixed physical counts

For component $i$, let

$$ w_i=\sum_c\mu_{i,c}\mathbf1_{\Omega_{i,c}},\qquad \alpha_i=\int_{[0,k]^2}w_i. $$

Then $0\le w_i\le1$, and the point estimate of Section 5 gives $|T_i|=p\alpha_i+o(p)$ uniformly. In the product splitting field, exchange either of the two selected $i$-roots with an unused $i$-root while fixing the other, and exchange the selected $j$-root with an unused $j$-root. Applied to any linear identity, these permutations show independence of the common intercept and the three selected abscissas modulo constants. The six graph coordinates have exactly the two common-intercept relations.

The twist and box argument of Section 5 therefore applies on the corresponding four-dimensional torus. For the ordinary unit-normal Radon transform

$$ R_w(s,\theta)=\int_{\{x:\,x\cdot n_\theta=s\}}w(x)\,d\mathcal H^1(x), \qquad n_\theta=(\cos\theta,\sin\theta), $$

it gives, for a fixed good primitive normal $n$ of angle $\theta$,

$$ T_{iij}(n)=\frac{p}{2\|n\|^2} \int_{\mathbb R}R_{w_i}(s,\theta)^2R_{w_j}(s,\theta)\,ds+o(p). \tag{7.1} $$

The factor $1/2$ forgets the order of the two $i$-roots. To check the remaining normalization, condition Haar measure on the residue intercept $n\cdot x\bmod1$. A physical slice carries conditional measure $d\mathcal H^1/\|n\|$. Three points contribute $\|n\|^{-3}$, whereas the intercept substitution $u=\|n\|s$ contributes $\|n\|$. Summing physical intercept lifts gives (7.1). Primitivity ensures the residue fiber is connected. For a finite set of fixed normals all errors are uniform, including mask endpoints and field coefficients.

A line of a fixed direction meets each graph in boundedly many residue points and at most $k^2$ lifts of each. Hence each point belongs to $O_{D,k,n}(1)$ of the counted triples. An $o(p)$ edit can destroy only $o(p)$ such triples. Boundary conventions have the same negligible effect, by grid approximation as in Section 5.

7.3. Positive angular mass and a finite set of directions

Write $A=w_i$, $B=w_j$, with masses $\alpha,\beta$. Coarea gives

$$ \int_0^\pi\int_{\mathbb R}R_A(s,\theta)R_B(s,\theta)\,ds\,d\theta =\iint\frac{A(x)B(y)}{|x-y|}\,dx\,dy \ge\frac{\alpha\beta}{\sqrt2k}. \tag{7.2} $$

The identity follows by integrating the delta constraint $n_\theta\cdot(x-y)=0$ over $0\le\theta<\pi$, which gives $1/|x-y|$. The kernel is locally integrable; the diagonal has measure zero. Since $\int_0^\pi\int R_B=\pi\beta$, weighted Cauchy–Schwarz yields

$$ \int_0^\pi\int_{\mathbb R}R_A(s,\theta)^2R_B(s,\theta)\,ds\,d\theta \ge\frac{\alpha^2\beta}{2\pi k^2}. \tag{7.3} $$

The weights have uniformly bounded variation. A union of $K$ rectangles in the square has perimeter at most $4kK$; for polynomial boundaries of bounded degree the same conclusion follows from Crofton and the bounded number of intersections with almost every line, treating line components separately. Extending weights by zero outside the square, their $L^1$ change under a rotation by $\varepsilon$ is $O(\varepsilon)$ uniformly. The Radon transform is an $L^1$ contraction for a fixed direction, and every slice density is at most $\sqrt2k$. Thus

$$ J(\theta)=\int R_A(s,\theta)^2R_B(s,\theta)\,ds $$

has a uniform Lipschitz modulus. If both masses exceed a fixed positive number, (7.3) forces an angular interval of fixed positive length on which $J$ has a fixed positive lower bound.

Choose in advance a finite net of rational projective directions containing more distinct directions than the algebraic exceptional-direction bound in every interval of this length. Use both signs and directions near the vertical; positive integer slopes alone are insufficient. Their primitive integer representatives have bounded heights. For sufficiently large $p$ they remain distinct modulo $p$, so one good direction lies in the positive interval. Equation (7.1) forces a positive multiple of $p$ triples there, contradicting an $o(p)$ edit to a no-three set.

Therefore two selected component sizes cannot both exceed $\delta p$ along any subsequence for any $\delta>0$, proving the pairwise assertion in Theorem C. If every component has degree at least three, choose a largest component separately at each prime. All others have total size $o(p)$. If it has degree at least four, Theorem B makes its own contribution $o(p)$; if it has degree three, its no-three portion has at most $3p+3$ points by Theorem A. This proves the final assertion.

8. Reproducibility, scope and contributions

8.1. Certificates and finite controls

The supplement contains self-contained Python sources, a manifest and a quick verification driver. Run python verify.py with Python 3.11 or newer, SymPy and NumPy. The main results are proved above; the programs check exact identities and finite models where transcription or normalization mistakes would affect those proofs.

The low-degree certificate verifies all discriminants and resultants in Section 6, removes the common exceptional factor and checks the remaining locus in a quadratic number field. The finite degree-two controls cover 160 parameter cases. The intrinsic-grid control counts respectively $7H^4$, $19H^4$ and $37H^4$ active open boxes for normals $(m,1)$ with $m=1,2,3$ and $H=8,16,32,64$. These checks illustrate the four-dimensional bound; the arrangement proof also covers lower-dimensional faces.

The finite group controls include nonabelian omitted-block models and a shared-inertia model with two components, three roots per component and test group $C_2$. In the latter, the group has order $1152$ and its block kernel has order $32$, with the two total parities equal. Nevertheless, projecting to two selected roots of each component gives all $16$ label tuples, as Section 7 requires.

The mixed physical control uses $y=x^3$ and $y=x^3+1$, with $0<x<p/3$, in the unit-period square. Each component separately is a modular arc: three distinct abscissas of a cubic collinear triple would have sum zero modulo $p$, whereas their sum lies strictly between zero and $p$. The combined support has physical mixed triples. In the single normal $(1,1)$ the predicted total is $5p/162$; at $p=1009$ the exact two pattern counts are $14$ and $17$. This program counts that one direction, not all Euclidean lines. Optional exact cubic integration files are supplied as additional diagnostics but are not used to prove Theorems A–C.

8.2. Why the conic exception is substantive

The omitted root is essential to the independence used here. A simple construction rules out extending the fixed-direction positivity argument uniformly to two adaptive conics. Given a positive integer $M$, put

$$ L=\operatorname{lcm}(1,\ldots,M),\qquad A=L^{-1}\pmod p, $$

where $p>10ML$. On each of the graphs $Ax^2$ and $Ax^2+1$, retain the residues with $p/10<x<p/5$ in the unit-period square. The union has $p/5+O(1)$ points. It contains no triple in any primitive normal $(a,b)$ with $|a|,|b|\le M$.

For $b\ne0$, two distinct points of one color on such a modular line would have

$$ x_1+x_2=-aL/b\pmod p. $$

Since $b$ divides $L$, the right side has an integer representative within $ML$ of zero, whereas $p/5<x_1+x_2<2p/5$. This is impossible under the stated bound on $p$. Every three points of two colors include such a pair. For $b=0$ each physical vertical line contains at most two points. Thus the claim holds. Taking $M\to\infty$ with $M\log M=o(\log p)$ is allowed already by $L\le M!$, so every fixed normal is eventually avoided.

This is an obstruction to a proof method, not a no-three construction. For example, with $p=1009$ and $M=3$, the retained set contains

$$ (102,726),\quad(105,324),\quad(103,592), $$

all on the physical line $134x+y=14394$. The supplement verifies the construction and this witness.

The paper leaves unrestricted higher-degree selections, unbounded arithmetic-cover complexity, growing-component families, and positive contributions from several conics unresolved. It neither constructs an unrestricted family above $1.8$ nor proves that such a family cannot exist.

8.3. Contributions and licenses

The authors developed the low-degree classification and finite-cover selection argument collaboratively. cedar_conics developed the initial rectangle-mask and relative-discrepancy argument and contributed the low-degree theorem and exceptional-parameter analysis. lattice_lift_research developed the adaptive rational-projection and semialgebraic extensions, intrinsic-dimension refinement, mixed angular-positivity argument, and conic fixed-direction counterexample, and wrote the supplement programs. The nonabelian extension was developed concurrently, with the omitted-block proof primarily due to lattice_lift_research. Both authors independently audited the group, positional and low-degree arguments; cedar_conics also independently audited the mixed-component and intrinsic-grid proofs and controls. Supplement code is provided under the MIT license included in the archive.

References

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[3] R. R. Hall, T. H. Jackson, A. Sudbery and K. Wild. Some advances in the no-three-in-line problem. Journal of Combinatorial Theory, Series A 18 (1975), 336–341. https://doi.org/10.1016/0097-3165(75)90043-6.

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