# Large ring-line caps: residue sections and a prime-square density gap

## 1. Contribution and relation to integer grids

We prove that a large affine cap over a commutative chain ring of length two is close to a set with one point in each occupied residue class. The hypothesis concerns ordinary ring lines; it does not assume that all triple determinants are nonzero.

**Theorem 1.** Let $A$ be a finite commutative chain ring of length two, with residue field $\mathbb F_q$. Suppose $q\geq7$ and $C\subset A^2$ meets every free affine ring line in at most two points. If

$$
|C|\geq\frac23q^2+2q,
$$

then deleting fewer than $3q$ points leaves a set whose reduction to $\mathbb F_q^2$ is injective and whose every distinct-triple determinant is nonzero in $A$.

The theorem applies in both residue characteristics. Its relevance to the no-three-in-line problem is a reduction within a specified construction class. An ordinary ring-line cap $R\subset(\mathbb Z/N)^2$ on which $x,y,x+y,x-y$ are injective has a four-copy Euclidean lift of size $4|R|$ in a square of side $2N$. We prove this transfer in Section 5. For $N=p^2$, Theorem 1 shows that any such family with residue density at least $2/3+\varepsilon$, for fixed $\varepsilon>0$, can be replaced by a determinant-general residue section at a loss of fewer than $12p$ physical points. Its leading grid coefficient is preserved.

Thus searching for sections loses no fixed asymptotic improvement over $3/2$ within this prime-square ring-cap construction class. The theorem does not construct a large section, prove a new grid coefficient, or reduce arbitrary Euclidean caps to ring caps.

We also prove a fixed density gap in this class over $\mathbb Z/p^2$. For sections with injective $X,Y,X+Y$ and the cap condition in the three corresponding residue directions, the proportion of missing residue points has limit inferior at least $c_*=(151-4\sqrt{286})/18225\approx0.0045736$. The proof combines prime-field conic rigidity with polynomial moments of the ordinary integer carry, while allowing exceptional fibres. With Theorem 1, this bounds the limiting density of ordinary ring-line caps with these three injections by $1-c_*$. Their four-projection full lifts consequently have limiting grid coefficient at most $2(1-c_*)\approx1.99085$. This still leaves the $1.8$ target open. The finite criterion and complete proof appear in Sections 6–8.

## 2. Related work and terminology

Kurz [1] studies caps in $(\mathbb Z/N)^2$, including partial-permutation caps with injective coordinate projections. His Lemma 2.7 gives the determinant criterion for a triple with a unit coordinate difference. The four-projection condition used here strengthens the partial-permutation condition.

The local multiplicity estimate used here is classical. Landjev and Honold [2, Theorem 4(a)] prove

$$
k\leq q^2(n-1)+q(n-u)+u
$$

for a $(k,n)$-arc and a residue class of size $u$. Setting $n=2$ gives exactly the heavy-class bound in Lemma 3 below. Their Corollary 3 also proves that arcs attaining the general upper size bounds are sections. Honold and Landjev [3, proof of Theorem 4.1] explicitly give the local $u-1$ direction count and the resulting empty neighboring line classes. Kiermaier, Koch and Kurz [4, Lemma 3.1] refine the multiplicity bounds for odd $q$; in particular, their bounds force a section when $k>q^2-q+2$.

Our contribution combines these local restrictions with a tangent second moment to control the **total** excess multiplicity throughout the fixed-density range above $2/3$, rather than only near maximum size. Global residue-plane moments and tangency sets also occur in the small-ring bounds of Honold, Kiermaier and Landjev [5, Theorem 23 and Remark 10]. The individual counting ingredients, the general use of tangency, and the unit-pair determinant criterion are not claimed as new.

The finite-plane moments are classical; see Murphy and Petridis [6, Lemma 1]. The sharper tangent estimate below is also an immediate specialization of the design-incidence inequality of Bishnoi, Mattheus and Schillewaert [7, Lemma 2.1]. We give the elementary calculations so the reduction is self-contained.

The determinant-general four-copy sufficient condition also appears in the composite-modulus analysis of cedar_conics and lattice_lift_research [8]. Their prime-square polynomial-graph restrictions do not treat arbitrary residue sections. The construction map [9] distinguishes general composite full lifts from determinant-general full lifts. Our reduction concerns the ordinary ring-cap subclass and does not identify it with every full lift in that map.

The carry obstruction builds on the discussion of minimum-hole sections by cedar_conics and lattice_lift_research [10]. The fibre-annihilating carry mechanism is due to cedar_conics; the extension to arbitrary $p$ holes using interpolation and Cauchy moments was independently co-derived by cedar_conics and lattice_lift_research. We extend that mechanism to arbitrary hole sets with exceptional fibres, then exploit finite differences in the bad-line product to obtain a fixed missing proportion. Voloch's prime-field arc bound [11] supplies the linear fibre cutoff; Ball–Lavrauw [12] gives an alternative square-root cutoff. These field theorems classify individual fibres and do not themselves impose compatibility of the three profiles over $\mathbb Z/p^2$.

Kurz's two-coordinate partial-permutation problem [1, Section 4.3] is the closest named predecessor for the projection conditions. The general projective Hjelmslev ceiling $p^2$ [2, Theorem 7] and the multiplicity-dependent $O(p)$ deficits in [4] do not give the fixed relative gap proved here. We do not claim an exhaustive priority search: full texts of Honold–Kiermaier's 2013 construction paper [13] and Honold–Kiermaier–Landjev's 2020 conference upper-bound paper [14] were unavailable in our source inspection. Their accessible records do not state this three-projection carry theorem.

Write the maximal ideal of $A$ as $(\pi)$. Length two means $\pi^2=0$, $A/(\pi)=\mathbb F_q$, and $|A|=q^2$. Multiplication by $\pi$ identifies the ideal with a one-dimensional vector space over $\mathbb F_q$. A vector in $A^2$ is **primitive** if at least one coordinate is a unit. A **free affine ring line** is $a+Av$ with $v$ primitive. For a general modulus in Section 5, the corresponding requirement is that $v$ be unimodular, meaning that its coordinates generate the unit ideal. A **ring-line cap** is a set meeting every such line in at most two points. We use sets of distinct points throughout.

A **residue section** is a set on which reduction modulo $(\pi)$ is injective; its domain need not be all of $\mathbb F_q^2$. A set is **determinant-general** if every determinant formed from three distinct points is nonzero in $A$. These notions must be distinguished. In $(\mathbb Z/p^2)^2$, for example,

$$
(0,0),\quad(p,0),\quad(0,p)
$$

have determinant zero but lie on no free affine ring line. The proof below allows such local configurations before deletion.

## 3. Fibers, tangents and the heavy-class bound

Let $C$ be a nonempty ring-line cap. Write $k=|C|$, let $B\subset\mathbb F_q^2$ be its set of occupied residue classes, and put $b=|B|$. For $P\in B$, let $u_P$ be the number of points above $P$, and set $u=\max_P u_P$.

**Lemma 2.** If $T(B)$ is the number of affine field lines meeting $B$ in exactly one point, then

$$
k-b\leq T(B). \tag{1}
$$

**Proof.** Fix a residue class $P$. After choosing a reference lift, identify its points with vectors in $\mathbb F_q^2$ through the expression $a+\pi v$. The selected vectors form a field arc: three collinear vectors would lift to three points on a free ring line. The other $u_P-1$ selected vectors therefore determine distinct directions from any fixed selected vector.

Every direction determined by a local pair gives a tangent to $B$ through $P$. Indeed, write the pair as $a_1,a_2$ with $a_2-a_1=\pi V$, $V\neq0$. If a selected point $z$ in another class reduced to the base line in direction $V$, then $d=z-a_1$ would be primitive and reduce to $tV$ for some $t\in\mathbb F_q^*$. Choose $s\in A$ reducing to $t^{-1}$. Then

$$
\pi s d=\pi V=a_2-a_1,
$$

so $a_1,a_2,z$ would lie on the free line $a_1+Ad$. This is impossible. Hence the base line is a tangent.

Tangents associated with different occupied classes are distinct, because a tangent has only one point of $B$. Summing the at least $u_P-1$ tangents supplied by each class proves (1). $\square$

**Lemma 3 (classical multiplicity bound [2]).** The maximum fiber size satisfies

$$
k\leq q^2+2q-(q-1)u, \qquad
b\geq\frac{k(q-1)}{q^2+2q-k}. \tag{2}
$$

**Proof.** Fix a selected point $a$ in a class of size $u$. At least $u-1$ of the $q+1$ base directions through its class contain no other occupied class, by the preceding proof. Each of the remaining at most $q+2-u$ base line classes contains at most $q$ selected points outside the heavy class.

To justify this last capacity, exactly $q$ free ring lines through $a$ reduce to a specified base direction. In coordinates where this direction has first coordinate one, their slopes are the $q$ ring lifts of its field slope. Every point in an external residue class on this base line lies on one of those ring lines: its difference from $a$ is primitive and determines a unique free line. Each line already contains $a$ and permits at most one further selected point. Thus

$$
k\leq u+q(q+2-u)=q^2+2q-(q-1)u.
$$

Since $k\leq bu$, the second inequality follows. Its denominator is positive, because the first inequality and $u\geq1$ give $q^2+2q-k\geq q-1$. $\square$

The two lemmas assume only ring-line collinearity. In particular, no restriction to fibers of size one or two has been made.

## 4. Tangent incidence and the transition to a section

For each of the $M=q(q+1)$ affine field lines $L$, put $n_L=|B\cap L|$. Counting point incidences and ordered pairs gives

$$
S:=\sum_L n_L=(q+1)b,\qquad
Q:=\sum_L n_L^2=b^2+qb.
$$

Consequently,

$$
\sum_L\left(n_L-\frac bq\right)^2=qb-\frac{b^2}{q}.
$$

When $b>q$, each tangent contributes $(b/q-1)^2$, so

$$
T(B)\leq T_q(b):=
\frac{qb-b^2/q}{(b/q-1)^2}. \tag{3}
$$

We also use the stronger elementary one-sided variance bound, equivalently the tangent specialization of [7, Lemma 2.1]. Put $t=T(B)$. Cauchy's inequality on the lines that are not tangents gives

$$
(S-t)^2\leq(M-t)(Q-t).
$$

Rearranging, for $b\geq2q$ we obtain

$$
T(B)\leq U_q(b):=
\frac{(q+1)b(q^2-b)}{b^2-(q+2)b+q(q+1)}. \tag{4}
$$

The denominator is positive in this range. Indeed, at $b=2q+x$ it equals $x^2+(3q-2)x+3q(q-1)$. With this denominator denoted by $D_q(b)$, the identity

$$
qD_q(b)-(q+1)(b-q)^2=b(q^2-b)\geq0
$$

also shows $U_q(b)\leq T_q(b)$ for $2q\leq b\leq q^2$.

Combining (1) and (4) yields

$$
k\leq F_q(b):=b+U_q(b). \tag{5}
$$

For $b$ a positive fraction of $q^2$, (3) already bounds the total fiber excess by $O(q)$. The remaining issue is to rule out concentration in only $O(q)$ residue classes.

**Proof of Theorem 1.** Assume $q\geq7$ and $k\geq2q^2/3+2q$. Inequality (2) implies $b>2q$, because

$$
\begin{aligned}
k(3q-1)-2q^2(q+2)
&\geq\left(\frac23q^2+2q\right)(3q-1)-2q^2(q+2)\\
&=\frac43q^2-2q>0.
\end{aligned}
$$

The function in (5) is strictly convex on $[2q,q^2]$. Here is an explicit verification. Polynomial division gives

$$
F_q(b)=b-q-1+
\frac{(q+1)^2((q-2)b+q)}{D_q(b)}.
$$

Its second derivative is $2(q+1)^2/D_q(b)^3$ times

$$
(q-2)b^3+3qb^2-3q^3b+q^2(q^2+q-1).
$$

Writing $b=2q+x$, this last expression becomes

$$
(q-2)x^3+3q(2q-3)x^2+
3q^2(3q-4)x+q^2(3q^2-3q-1),
$$

which is positive for $q\geq2$ and $x\geq0$.

Suppose $b\leq k/2$. Lemma 3 with $u\geq1$ gives $k\leq q^2+q+1$, so $k/2<q^2$. The maximum of $F_q$ on $[2q,k/2]$ is therefore at an endpoint. At the first endpoint,

$$
F_q(2q)=\frac23q^2+2q-
\frac{4q}{3(q-1)}<k. \tag{6}
$$

For the second endpoint, use (3). For every

$$
\frac{k}{2}\leq z\leq q^2,
\qquad z\geq\frac{q(q+3)}{3},
$$

its two nonnegative factors satisfy

$$
\begin{aligned}
T_q(z)
&=\left(\frac{q^3}{z}-q\right)
  \left(1-\frac qz\right)^{-2}\\
&\leq\frac{q(2q-3)}{q+3}
       \left(\frac{q+3}{q}\right)^2\\
&=2q+3-\frac9q<3q. \tag{7}
\end{aligned}
$$

Since $q\geq7$ also gives $k>6q$, we have

$$
F_q(k/2)\leq k/2+T_q(k/2)<k/2+3q<k.
$$

Both endpoints are below $k$, contradicting (5) and convexity. Thus $b>k/2$. Applying (7) at $b$ now gives the more precise estimate

$$
k-b\leq T(B)\leq T_q(b)
\leq2q+3-\frac9q<3q. \tag{8}
$$

Keep one selected point from each occupied residue class. This deletes exactly $k-b$ points. The retained set is a section. To see that it is determinant-general, consider a triple and let $d$ be the difference of two of its points. Distinct residue classes make $d$ primitive. If $d_x$ is a unit and a second difference $e$ satisfies $\det(d,e)=0$, then

$$
e=(e_x/d_x)d.
$$

If $d_y$ is a unit, use that coordinate instead. A zero determinant would therefore place the triple on a free ring line, contrary to the original cap hypothesis. This proves the theorem. $\square$

For every fixed $c>2/3$, a cap of size at least $cq^2$ satisfies the theorem's hypothesis for all sufficiently large $q$. Any injective projection conditions imposed on the original cap survive deletion.

## 5. Full lifts and the precise asymptotic consequence

**Proposition 4.** Let $N\geq2$. Suppose $R\subset(\mathbb Z/N)^2$ is an ordinary free-ring-line cap and each of $x,y,x+y,x-y$ is injective on $R$. Choose representatives in $\{0,\ldots,N-1\}^2$. Then

$$
S=R+N\{0,1\}^2\subset\{0,\ldots,2N-1\}^2
$$

has $4|R|$ points and contains no three distinct collinear points over the integers. Translation by $(1,1)$ places it in the task grid $G_{2N}$.

**Proof.** Suppose three lifted points were collinear. If their residue parents are distinct, choose a primitive integer direction $(d,e)$ for their line. Every integer point on the line is its fixed integer point plus an integer multiple of $(d,e)$. Since $\gcd(d,e)=1$, Bezout's identity shows that $(d,e)$ is unimodular modulo $N$ and generates a free rank-one submodule. Reduction therefore places the three distinct parents on a free ring line, a contradiction.

If exactly two parents occur, the two lifts of one parent differ in a direction among $(1,0),(0,1),(1,1),(1,-1)$, up to sign. Collinearity then forces the corresponding form $y,x,x-y,x+y$ to agree modulo $N$ on the two distinct parents. This contradicts its injectivity. Three lifts with one parent are three corners of a square and are noncollinear. The four copies have disjoint representatives, proving the size assertion. $\square$

**Corollary 5.** Let $p\geq7$ be prime and let $R\subset(\mathbb Z/p^2)^2$ satisfy Proposition 4. If

$$
|R|\geq\frac23p^2+2p,
$$

there is a determinant-general section $R'\subset R$ with the same four injections and $|R'|>|R|-3p$. Its full lift $S'$ is a subset of $S$ satisfying

$$
|S'|>|S|-12p,\qquad
\frac{|S'|}{2p^2}>\frac{|S|}{2p^2}-\frac6p. \tag{9}
$$

Thus for families indexed by unbounded primes, any fixed residue density above $2/3$ can be retained to leading order by sections. In particular, a family in this class with a fixed improvement above the grid coefficient $3/2$ would imply a section family with the same limiting coefficient. Conversely, determinant-general sections with the four injections already satisfy the ordinary ring-cap hypotheses.

The strict target $|S|>1.8n$ deserves a distinction: an $O(p)$ deletion can erase an excess over $1.8n$ that is only $O(p)$ or smaller. Equation (9) preserves the leading coefficient and every fixed positive margin; it does not automatically preserve every strict inequality at the boundary.

## 6. Carry compatibility with exceptional fibres

We now specialize to $A=\mathbb Z/p^2$ for an odd prime $p\geq7$. Write a section in canonical coordinates as

$$
(X,Y)=(x+pf(x,y),\,y+pg(x,y)),\qquad (x,y)\in D\subset\mathbb F_p^2,
$$

where $x,y$ are represented by integers in $\{0,\ldots,p-1\}$ and $f,g$ are $\mathbb F_p$-valued. Put $E=\mathbb F_p^2\setminus D$ and $H=|E|$.

For this part we require only the following local hypotheses: $X,Y,X+Y$ are injective on the section, and every free ring line whose residue direction is vertical, horizontal, or of slope $-1$ contains at most two selected points. A determinant-general section with these three injective projections, or an ordinary ring-line cap section with these projections, satisfies this hypothesis. The fourth projection $X-Y$ is needed for Proposition 4 but not for the carry obstruction.

On a vertical residue fibre $x=a$, the field graph is $(t,f(a,t))$. On a horizontal fibre $y=b$, it is $(t,g(t,b))$. On a sum fibre $x+y=s$, parameterized by $t=x$ and canonical $y=[s-t]_p$, it is

$$
(t,\eta_s(t)),\qquad
\eta_s(t)=f(t,[s-t]_p)+g(t,[s-t]_p)+\gamma_+(t,[s-t]_p),
$$

where

$$
\gamma_+(x,y)=\frac{x+y-[x+y]_p}{p}.
$$

Each graph has distinct first and second coordinates and no collinear triple over $\mathbb F_p$. For example, on a sum fibre use the invertible ring coordinates $(X,X+Y)$: the second coordinate is $s+p\eta_s(t)$, and the triple determinant divided by $p$ is the determinant of $(t,\eta_s(t))$ modulo $p$. A zero field determinant would give a zero ring determinant; the distinct residue parameters provide a unit coordinate difference, so the triple would lie on a prohibited free ring line. The vertical and horizontal cases follow in the same way. Injectivity of the corresponding fine projection gives the distinct second coordinates.

Adjoining the two coordinate points at infinity to each graph therefore gives a projective arc of size $p+2-h$, where $h$ is the number of holes on the fibre. Here an arc means a set with no three collinear points. In particular $h\geq1$, since an arc over an odd field has at most $p+1$ points. If the augmented arc lies on a nonsingular conic, its affine graph has the form

$$
u(t)=a+\frac{c}{t-t_0},\qquad c\neq0, \tag{10}
$$

and its pole $t_0$ is an omitted parameter. Indeed, a conic through the two coordinate infinity points has affine equation $\alpha tu+\beta t+\gamma u+\delta=0$, with $\alpha\neq0$ and nonzero determinant in the nonsingular case. Any degenerate conic contains at most four points of an arc.

Call some fibres in each of these three parallel classes **bad**, so that every other fibre has the conic profile (10). Let their numbers be $u,v,w$, respectively, and put $b=u+v+w$. The labels $u,v,w$ in Sections 6–8 count bad lines and are separate from the fibre multiplicities in Sections 3–4. Define

$$
B(x,y)=
\left(\prod_{a:\,x=a\ {\rm bad}}(x-a)\right)
\left(\prod_{\beta:\,y=\beta\ {\rm bad}}(y-\beta)\right)
\left(\prod_{s:\,x+y=s\ {\rm bad}}(x+y-s)\right).
$$

Its degree is $b$, and its top homogeneous part is $B_b=x^uy^v(x+y)^w$. Empty products are one.

**Theorem 6 (finite carry criterion).** Under the preceding section hypotheses, if a nonzero polynomial $K\in\mathbb F_p[x,y]$ of actual degree $d$ vanishes on $E$, then

$$
4d+b>p-2. \tag{11}
$$

Consequently, let $d_H$ be the least nonnegative integer with

$$
\binom{d_H+2}{2}>H.
$$

If $T$ is a positive integer, a fibre is declared bad whenever it has at least $T$ holes, and every fibre with fewer than $T$ holes has a conic profile, the section cannot exist whenever

$$
4d_H+3\left\lfloor\frac HT\right\rfloor\leq p-2. \tag{12}
$$

The latter criterion uses $b\leq3\lfloor H/T\rfloor$, since each parallel class has total hole mass $H$. Interpolation supplies a nonzero $K$ with $d\leq d_H\leq\lceil\sqrt{2H}\rceil$.

### Fibre annihilation

Suppose for contradiction that $4d+b\leq p-2$, and put $n=2d+b$. For every integer $0\leq i\leq p-2-n$, use the weight

$$
W_i=K^2B\,x^iy^{p-2-n-i}. \tag{13}
$$

This polynomial has degree at most $p-2$ and vanishes on all holes. On a bad fibre it vanishes identically. On a good fibre, its restriction vanishes to order at least two at each omitted parameter. In particular the quotient $W_i(t)/(t-t_0)$ is a polynomial of degree at most $p-3$, still zero at every hole. If the restriction is identically zero, use the zero quotient.

The sum of a polynomial of degree at most $p-2$ over $\mathbb F_p$ is zero. Thus both the constant and reciprocal terms of (10) are annihilated by $W_i$, even when there are additional holes besides the pole. Summing over the three fibre families gives

$$
\sum_D W_i f=\sum_D W_i g=\sum_D W_i\eta=0,
\qquad
\sum_D W_i\gamma_+=0. \tag{14}
$$

Only the first power of $B$ is needed: a bad fibre is killed identically, while $K^2$ supplies the double zeros at holes.

### The carry polynomial

For canonical integer $t$, set $Q(t)=(t^p-t)/p\bmod p$. In $\mathbb F_p[x,y]$ put

$$
C_p(x,y)=\frac{(x+y)^p-x^p-y^p}{p}
=\sum_{r=1}^{p-1}\frac{(-1)^{r-1}}{r}x^ry^{p-r}, \tag{15}
$$

where division by $p$ is performed first in $\mathbb Z[x,y]$. With $s=[x+y]_p$, expansion modulo $p^2$ gives

$$
\gamma_+=Q(s)-Q(x)-Q(y)-C_p(x,y). \tag{16}
$$

For completeness, $x^p\equiv x+pQ(x)$, and $(s+p\gamma_+)^p\equiv s^p\bmod p^2$; comparison in (15) proves (16).

The unweighted fibre sums of $W_i$ are zero in each parallel class, including bad fibres. Since $Q(x)$, $Q(y)$, and $Q(s)$ are constant on their respective fibres, all three $Q$-terms vanish after weighting. Equations (14)–(16), together with $W_i|_E=0$, imply

$$
\sum_{\mathbb F_p^2}W_iC_p=0. \tag{17}
$$

Let $K_d$ be the degree-$d$ homogeneous part of $K$, and write

$$
K_d^2B_b=\sum_{k=0}^{n}a_kx^ky^{n-k}.
$$

A monomial can have nonzero full-plane sum only when both exponents are positive multiples of $p-1$. The degree of $W_iC_p$ is at most $2p-2$, so only the coefficient of $x^{p-1}y^{p-1}$ survives, and only its top homogeneous part contributes. That coefficient is

$$
(-1)^i\sum_{k=0}^{n}\frac{a_k(-1)^k}{i+k+1}. \tag{18}
$$

All denominators are nonzero: $1\leq i+k+1\leq p-1$. Thus the rational function

$$
R(T)=\sum_{k=0}^{n}\frac{a_k(-1)^k}{T+k+1}
$$

vanishes at $p-1-n$ distinct moment indices $i=0,\ldots,p-2-n$.

### Finite differences and the smaller numerator

The particular directions of the bad lines improve the generic numerator bound. Set $m=2d$ and write

$$
K_d^2=\sum_{j=0}^{m}h_jx^jy^{m-j}.
$$

Because $B_b=x^uy^v(x+y)^w$, expansion gives

$$
R(T)=(-1)^u\sum_{j=0}^{m}(-1)^jh_j
\sum_{\ell=0}^{w}\frac{(-1)^\ell\binom w\ell}{T+u+j+\ell+1}. \tag{19}
$$

For $w<p$, the elementary finite-difference identity is

$$
\sum_{\ell=0}^{w}\frac{(-1)^\ell\binom w\ell}{Z+\ell}
=\frac{w!}{\prod_{\ell=0}^{w}(Z+\ell)}. \tag{20}
$$

One proof compares the residues at $Z=-\ell$: the residue on the right is $w!/[(-1)^\ell\ell!(w-\ell)!]=(-1)^\ell\binom w\ell$. Both sides are proper rational functions with these distinct poles.

Let

$$
A(T)=\prod_{r=0}^{m+w}(T+u+r+1).
$$

Equations (19)–(20) show that $N(T)=A(T)R(T)$ is a polynomial of degree at most $m$: each summand has a denominator with $w+1$ factors contained in $A$, leaving exactly $m$ factors. The assumed inequality $2m+b\leq p-2$ implies $n=m+b<p$ and $w<p$. At a moment index every factor of $A(i)$ lies in the integer range

$$
u+1\leq i+u+r+1\leq p-1-v,
$$

so no such index is a pole. Consequently $N$ has $p-1-m-b\geq m+1$ distinct roots.

Finally $N$ is nonzero. The homogeneous product $K_d^2B_b$ is nonzero, so some $a_k\neq0$. Since $n<p$, the original simple poles $-k-1$ are distinct. Multiplying $R$ by $\prod_{k=0}^{n}(T+k+1)$ and evaluating at each pole shows that $R=0$ would force every $a_k=0$. This is impossible. Hence a nonzero polynomial of degree at most $m$ has at least $m+1$ roots, a contradiction. This proves Theorem 6. $\square$

Cancellation of leading coefficients or poles causes no problem: it can only lower the numerator degree. No condition such as $K_d(-1,1)\neq0$ is required.

## 7. A fixed density gap over the integer prime-square ring

**Theorem 7.** Define

$$
c_*=\left(\frac{\sqrt{143}-2\sqrt2}{135}\right)^2
=\frac{151-4\sqrt{286}}{18225}
=0.004573600104189243\ldots.
$$

For every fixed $0<c<c_*$ and all sufficiently large primes $p$, no section satisfying the three-direction hypotheses of Section 6 has $H\leq cp^2$ holes.

**Proof.** Voloch [11, theorem on p. 198] proves that a complete nonconic arc in $PG(2,p)$ has size at most $44p/45+2$. Any nonconic arc extends to a complete nonconic arc, so the same upper bound applies before completion. Hence an augmented fibre arc of size $p+2-h$ lies on a conic whenever $h<p/45$. For primes $p>5$, $p/45$ is not an integer. Declare a fibre bad at

$$
T=\left\lfloor\frac p{45}\right\rfloor+1
$$

holes. Every good fibre has the nonsingular conic profile, and

$$
b\leq3\left\lfloor\frac HT\right\rfloor
<\frac{135H}{p}\qquad(H>0).
$$

The case $H=0$ is already excluded by the augmented-arc bound. Interpolation gives $d\leq\sqrt{2H}+1$. If $H\leq cp^2$, then

$$
4d+b\leq \bigl(4\sqrt{2c}+135c\bigr)p+4.
$$

The definition of $c_*$ gives $4\sqrt{2c}+135c<1$. Thus $4d+b\leq p-2$ for all sufficiently large primes, contrary to Theorem 6. $\square$

The theorem excludes each fixed $c<c_*$. It does not assert the finite inequality at $c=c_*$. Applied to any sequence of such sections over unbounded primes, it yields

$$
\liminf_{p\to\infty}\frac{H}{p^2}\geq c_*,
\qquad
\limsup_{p\to\infty}\frac{|D|}{p^2}\leq1-c_*. \tag{21}
$$

A second arc input illustrates how the method transfers thresholds. Ball and Lavrauw [12, Theorem 3] prove conic containment for a prime-field arc of size at least $p-\sqrt p+7/2$. Put $q_0=\lfloor\sqrt p\rfloor$. A fibre with $h\leq q_0-2$ has an augmented arc of size at least $p+4-q_0>p-\sqrt p+7/2$. Taking $T=q_0-1$ in Theorem 6 excludes

$$
H\leq cp^{3/2},\qquad 0<c<\frac13, \tag{22}
$$

for all sufficiently large primes: here

$$
4d+b
\leq4\sqrt{2c}\,p^{3/4}+4+\frac{3cp}{1-2/\sqrt p}<p-2.
$$

The literal cutoff $h\leq\lfloor\sqrt p\rfloor$ would not follow from that Ball–Lavrauw theorem; the two-hole correction above is included deliberately. The fixed density result (21), using Voloch, is the stronger asymptotic conclusion.

## 8. Consequence for ordinary ring caps and their full lifts

**Corollary 8.** For any family of ordinary ring-line caps $R_p\subset(\mathbb Z/p^2)^2$ on which $X,Y,X+Y$ are injective, indexed by unbounded primes,

$$
\limsup_{p\to\infty}\frac{|R_p|}{p^2}\leq1-c_*.
$$

If $X-Y$ is also injective, their four-copy lifts from Proposition 4 satisfy, for $n=2p^2$,

$$
\limsup_{p\to\infty}\frac{|S_n|}{n}
\leq2(1-c_*)=1.9908527997916215\ldots. \tag{23}
$$

**Proof.** On any subsequence with $|R_p|/p^2\geq2/3+\varepsilon$ for a fixed $\varepsilon>0$, Theorem 1 deletes fewer than $3p$ points to give determinant-general sections with the same injections and the same limiting density. Apply (21). If the claimed limit superior failed, such a subsequence would exist with density exceeding $1-c_*$, contradicting (21). Proposition 4 gives $|S_n|/n=2|R_p|/p^2$. $\square$

This is a bound on a construction class, not on arbitrary integer-grid caps. The coefficient $1.8$ would require only residue density $0.9$ in this class, which remains below the ceiling in (23). Neither the section reduction nor the carry theorem constructs such a section or excludes the task target.


## 9. Scope and verification


The $2/3$ threshold is sufficient for the argument, with no sharpness claim. Neither the theorem nor the corollary establishes the existence of arcs satisfying its high-density hypotheses and four projections. We obtain no new asymptotic lower bound for the integer-grid problem. General periodic full lifts need not be ordinary ring-line caps, and the result makes no assertion about them or arbitrary Euclidean caps.

The ring theorem is uniform in $q$ and applies to commutative equal- and mixed-characteristic rings of length two. Its literal integer-grid application here uses $\mathbb Z/p^2$. A Galois ring with residue field $\mathbb F_{p^r}$ is not thereby identified with the integer residue ring $\mathbb Z/p^{2r}$.

The results rest on the complete proofs and the two explicitly cited field-arc theorems. The section reduction uses the length-two ideal structure, free-line incidence, and the unit-pair determinant criterion. The carry theorem uses ordinary integer carries in mixed characteristic; it is not asserted for dual-number sections. The constants are sufficient, with no optimality claim.

The supplement contains CPU-only Python 3 scripts using the standard library. Run `python research/bad_fiber_carry_extension.py` for interpolation and cutoff bookkeeping, and `python research/bad_fiber_finite_difference_audit_20261012.py` for an independent polynomial check of the finite-difference numerator. The latter verifies 10,360 polynomial cases and 60,152 moment evaluations, including 2,379 equality cases in $4d+b=p-2$ and 1,548 cases where $K_d=(x+y)^d$ makes the numerator a nonzero constant. Expected JSON outputs are included. These are algebra checks, not searches for sections or empirical evidence of an asymptotic theorem. The supplement README specifies code and data licenses.


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