One Function for Every Prime
How random polynomials factor over the p-adic numbers: an algorithmic route to the wild case of the factorization-density conjecture, machine-checked in Lean — unconditionally for quadratics, as a certified architecture for every degree. Written by Claude Fable 5 (an AI system built by Anthropic) with Asvin G, July 2026. Companion to the expandable map of the Lean development.
The question
Pick a prime $p$ and draw a monic polynomial $f = x^n + c_{n-1}x^{n-1} + \dots + c_0$ with each coefficient a uniformly random $p$-adic integer (each digit of $c_i = a_0 + a_1 p + a_2 p^2 + \cdots$ drawn independently and uniformly; this is Haar measure on $\mathbb{Z}_p$). Over $\mathbb{Q}_p$ the polynomial factors into irreducible pieces, and each piece carries two invariants: its ramification index $e$ and residue degree $f$, with $e \cdot f$ its degree. The multiset
$$\sigma \;=\; \{(e_1, f_1), \dots, (e_r, f_r)\}, \qquad \textstyle\sum_i e_i f_i = n$$is the splitting type. Each type is a measurable event, so it has a probability $\rho_\sigma(p)$ — its density. For quadratics there are three types: split $\{(1,1),(1,1)\}$, inert $\{(1,2)\}$, ramified $\{(2,1)\}$. How does $\rho_\sigma(p)$ vary with $p$?
When $p > n$ the prime is tame for degree $n$ (it cannot divide any ramification index that occurs), and two classical inputs make the computation uniform in the prime. For the unramified part: where the reduction mod $p$ is squarefree, Hensel's lemma lifts its factorization exactly, and the repeated-factor locus recurses one digit deeper — a geometric series. For the ramified part: every ramified piece is tamely ramified, and tame extensions are completely and explicitly classified — each is generated by a root of a binomial
$$x^e - \pi u, \qquad \pi \text{ a uniformizer}, \ u \text{ a unit}, \ p \nmid e,$$with finitely many extensions per invariant pair $(e, f)$, counted by expressions in $q$ alone. Together these give each tame density as the value of a fixed rational function of $q = p$ (a uniformity going back to Del Corso and Dvornicich; see §5 of the paper cited next). The quadratic example below displays both inputs at once: the geometric series down the discriminant's valuation is the recursion, and “ramified $\iff v(d)$ odd” is the tame classification of $\mathbb{Q}_p(\sqrt d\,)$. At the finitely many wild primes $p \le n$ these formulas were conjectured to persist by Bhargava, Cremona, Fisher, and Gajović on the strength of low-degree computations. G, Wei, and Yin (arXiv:2212.00294) proved the tame case in general, with a functional equation reflecting Poincaré duality, and proposed a geometric route to the wild case through a Tate-type resolution of the resultant locus; that construction does not hold at wild primes as stated, and the wild case of the conjecture is open. This page presents an independent, algorithmic route toward it — computing the densities rather than comparing them — and reports exactly how much of that route is machine-checked in the Lean proof assistant.
Worked example: the tame case proved in miniature — Hensel's count, the binomial classification, one cubic density in full
Fix a degree $n$ and a tame prime $p > n$, $q = p$. Here are the two classical inputs made exact, then assembled into one complete degree-$3$ density.
(a) Unramified part: the squarefree product formula. If the reduction $\bar f \in \mathbb{F}_q[x]$ is squarefree, Hensel lifts its factorization exactly: each irreducible factor of degree $d$ becomes an unramified piece $(1, d)$. A fixed reduction pins the first digit of every coefficient — a cylinder of measure $q^{-n}$ — so the étale type $\sigma = \{(1,d_1), \dots, (1,d_r)\}$, $\sum d_i = n$, receives at the first digit level exactly
$$\frac{1}{q^{n}} \prod_{d \ge 1} \binom{N_d(q)}{m_d}, \qquad N_d(q) = \frac{1}{d} \sum_{k \mid d} \mu(k)\, q^{d/k}, \qquad m_d = \#\{i : d_i = d\},$$choosing which $m_d$ of the $N_d(q)$ monic irreducibles of degree $d$ divide $\bar f$. At $n = 3$ the three étale types get $\binom{q}{3}$, $\ q \cdot \frac{q^2 - q}{2}$, $\ \frac{q^3 - q}{3}$ reductions, summing to $q^3 - q^2$ — the count of squarefree monic cubics. So mass $1 - \tfrac1q$ is decided at level zero, and everything else — every ramified type in particular, since a ramified piece forces a repeated factor mod $p$ — lives in the repeated-factor mass $\tfrac1q$, where the recursion of the tame paragraph begins.
(b) Ramified part: the binomial classification. Let $L/\mathbb{Q}_p$ have invariants $(e, f)$ with $p \nmid e$. Its maximal unramified subfield is the degree-$f$ unramified field $K_f$; a uniformizer $\Pi$ of $L$ has $\Pi^e = p\,u\,w$ with $u$ a Teichmüller unit (hence $u \in K_f$, since $L/K_f$ adds no roots of unity of order prime to $p$) and $w \equiv 1$ a one-unit; $x^e - w$ has the simple root $1$ mod $p$ (its derivative there is $e$, a unit), so $w = w_0^e$ by Hensel, and $\Pi/w_0$ shows $L = K_f\bigl((pu)^{1/e}\bigr)$. Counting inside a fixed algebraic closure: the class of $u$ in $\mathbb{F}_{q^f}^{\times}/(\mathbb{F}_{q^f}^{\times})^{e}$ determines $L$ up to conjugacy, distinct classes give distinct fields, and each binomial's $e$ roots cluster into fields containing $d$ apiece (the $e$-th roots of unity in the totally ramified $L$ are just $\mu_e \cap \mu_{q^f - 1} = \mu_d$):
$$\#\bigl\{\,L \subset \overline{\mathbb{Q}}_p \text{ with tame invariants } (e,f)\,\bigr\} \;=\; \underbrace{d}_{\text{classes of } u} \times \underbrace{e/d}_{\text{conjugate fields each}} \;=\; e, \qquad d = \gcd\bigl(e,\, q^f - 1\bigr).$$Check at $(e,f) = (2,1)$, $p$ odd: $d = 2$, classes $\mathbb{Q}_p(\sqrt{p}\,)$ and $\mathbb{Q}_p(\sqrt{pu}\,)$ ($u$ a nonsquare unit), each Galois — two fields, $e = 2$. Downstream this is consumed as the parity law: a quadratic with discriminant $\delta$ is ramified iff $v(\delta)$ is odd (then $\sqrt\delta$ generates one of the two classes).
(c) Assembly: the full density of $\sigma_0 = \{(1,1),(2,1)\}$ at $n = 3$. A monic cubic has two repeated-factor reduction shapes: $\bar f = \bar\ell\, \bar m^2$ with $\bar\ell \neq \bar m$ linear ($q(q-1)$ of the $q^3$ reductions) and $\bar f = \bar m^3$ ($q$ reductions). By (a), only these can produce $\sigma_0$.
Shape $\bar\ell\,\bar m^2$, mass $(q-1)/q^2$. The key measure lemma: for coprime monic $\bar g, \bar h$, Hensel factorization $f = GH$ with $\bar G = \bar g$, $\bar H = \bar h$ is a bijection at every finite level — $p^{(k-1)\deg \bar g} \cdot p^{(k-1)\deg \bar h} = p^{(k-1)n}$ residue classes mod $p^k$ on both sides — hence a measure isomorphism of cylinders. So $f = LH$ splits off the $(1,1)$ piece $L$, and conditioned on the shape, $H$ is a uniform monic quadratic with reduction $\bar m^2$: recentered at a lift of $\bar m$, $H = x^2 + bx + c$ with $b, c$ uniform in $p\mathbb{Z}_p$. Its discriminant is $b^2 - 4c = p \cdot (p\beta^2 - 4\gamma)$ ($\beta = b/p$, $\gamma = c/p$), and $p\beta^2 - 4\gamma$ is uniform in $\mathbb{Z}_p$ because $4$ is a unit. By the parity law of (b),
$$Q \;:=\; \Pr[\,H \text{ ramified}\,] \;=\; \Pr\bigl[v(\text{uniform}) \text{ even}\bigr] \;=\; \Bigl(1 - \tfrac1q\Bigr)\bigl(1 + q^{-2} + q^{-4} + \cdots\bigr) \;=\; \frac{q}{q+1}.$$Shape $\bar m^3$, mass $1/q^2$. Recentering gives $g = x^3 + ax^2 + bx + c$ with $a, b, c$ uniform in $p\mathbb{Z}_p$, and the Newton polygon on $(0, v(c)),\ (1, v(b)),\ (2, v(a)),\ (3, 0)$ decides by strata (masses within this node):
| stratum | polygon | outcome | mass |
|---|---|---|---|
| $v(c) = 1$ | one slope $-\tfrac13$ | $\{(3,1)\}$ | $1 - \tfrac1q$ |
| $v(b) = 1,\ v(c) \ge 2$ | slopes $-(v(c) - 1),\ -\tfrac12$ | $\sigma_0$, decided | $\tfrac1q\bigl(1 - \tfrac1q\bigr)$ |
| $v(b) \ge 2,\ v(c) = 2$ | one slope $-\tfrac23$ | $\{(3,1)\}$ | $\tfrac1{q^2}\bigl(1 - \tfrac1q\bigr)$ |
| $v(b) \ge 2,\ v(c) \ge 3$ | rescale $x = py$ | fresh uniform cubic | $\tfrac1{q^3}$ |
Rows one and three are single segments whose slope denominator forces $e = 3$ on a degree-$3$ polynomial: irreducible, $\{(3,1)\}$, not $\sigma_0$. Row two has a vertex at $(1,1)$: a length-$1$ integer-slope segment (a root in $\mathbb{Z}_p$ — the $(1,1)$ piece) and a slope $-\tfrac12$ segment whose degree-$2$ factor has both roots of valuation $\tfrac12$, hence $e = 2$, $f = 1$ — the $(2,1)$ piece, ramified with no further reading. Row four: $h(y) = g(py)/p^3$ is again a monic cubic with uniform, independent $\mathbb{Z}_p$ coefficients and the same splitting type ($\mathbb{Q}_p(\alpha) = \mathbb{Q}_p(p\alpha)$) — the tree re-enters its own root. Collecting, one equation in one unknown:
$$\rho_{\sigma_0} \;=\; \underbrace{\frac{q-1}{q^2} \cdot \frac{q}{q+1}}_{\bar\ell\,\bar m^2,\ \text{then } Q} \;+\; \underbrace{\frac{1}{q^3}\Bigl(1 - \frac1q\Bigr)}_{\bar m^3,\ \text{decided}} \;+\; \underbrace{\frac{1}{q^5}}_{\text{self-loop}}\, \rho_{\sigma_0},$$ $$R_{\sigma_0}(q) \;=\; \frac{q\,(q^3 + q + 1)}{(q+1)\,(q^4 + q^3 + q^2 + q + 1)}.$$The factor $(1 - q^{-5})^{-1}$ that the solve produces is Step 3's Neumann series $(I-K)^{-1}$ in miniature, and the loop's undecided mass $q^{-5N} \to 0$ is exhaustiveness in miniature. The development's degree-3 table records exactly this function for $\sigma_0$ (in the form $q(q-1)(q^3+q+1)/((q+1)(q^5-1))$; at $q = 5$ both read $655/4686 \approx 0.1398$).
Where tameness entered — and what wildness breaks. Exactly once: the constant $Q$. The Hensel splitting, the shape counts, the polygon slopes with their denominators, and the rescaling loop are prime-uniform. At $p = 3$, wild for cubics, the derivation above survives verbatim: $3$ threatens only $e = 3$ pieces, and both $\{(3,1)\}$ strata were decided by slope denominators, which no wildness disturbs. At $p = 2$ one number breaks — the parity law behind $Q$ gives way to the mod-$8$ square law of the quadratic example below — and redoing that single count yields $Q(2) = \tfrac14 + \tfrac12\cdot\tfrac56 = \tfrac23$, which is $\frac{q}{q+1}$ at $q = 2$: the mass redistributes across digit classes the tame theory never inspects, and the total is the same value, so $R_{\sigma_0}$ persists at both wild primes of degree $3$ — in agreement with the development's wild-checked degree-3 table. At general degree these three moves are the whole tame computation: (a) and (b) supply the transition weights, the polygon strata with their self-loops become the finite pattern alphabet of Step 2, and the one-unknown solve becomes the $53$-state system of Step 3. Wildness reroutes which strata are reachable and revalues finitely many constants like $Q$ inside the carry algebra $D$ of Step 1; the open rows of the ledger are exactly the claim that those revalued constants always re-sum to the same $R_\sigma$.
The theorem, in the quadratic case
Theorem (proved, no hypotheses). For monic quadratics, at every prime $p$:
$$\rho_{\mathrm{ram}}(p) = \frac{1}{q+1}, \qquad \rho_{\mathrm{split}}(p) = \rho_{\mathrm{inert}}(p) = \frac{q}{2(q+1)}, \qquad q = p .$$In particular at the wild prime $p=2$: all three densities equal $\tfrac13$.
The values themselves are classical; what the theorem box reports is that the statement is machine-checked end to end, at every prime, along the algorithmic route below. The point is the phrase every prime. At odd $p$ these formulas fall out of a short discriminant computation. At $p = 2$ that computation is wrong in two separate ways — and the corrected count lands on the same functions anyway. The whole subject is in this example, so here it is in full.
Worked example: the quadratic densities at odd $p$ — and what breaks at $p=2$
Write $f = x^2 + bx + c$ with $b, c$ uniform in $\mathbb{Z}_p$, and let $d = b^2 - 4c$ be the discriminant. For odd $p$, since $4$ is a unit, $d$ is itself uniform in $\mathbb{Z}_p$, and for $d \neq 0$ (the repeated-root locus $d = 0$ has measure zero) the type is read off $d = p^k u$ ($u$ a unit):
$$\text{split} \iff k \text{ even, } u \text{ a square}; \qquad \text{inert} \iff k \text{ even, } u \text{ a nonsquare}; \qquad \text{ram} \iff k \text{ odd}.$$A uniform element has $v_p(d) = k$ with probability $(1 - \tfrac1q)q^{-k}$, and exactly half the units mod $p$ are squares. Summing the geometric series:
$$\rho_{\mathrm{ram}} = \sum_{k \text{ odd}} \Bigl(1 - \tfrac1q\Bigr) q^{-k} = \frac{1}{q+1}, \qquad \rho_{\mathrm{split}} = \rho_{\mathrm{inert}} = \frac12 \sum_{k \text{ even}} \Bigl(1 - \tfrac1q\Bigr) q^{-k} = \frac{q}{2(q+1)}.$$At $p = 2$ both inputs fail. First, $4c \equiv 0 \pmod 4$, so $d = b^2 - 4c$ is not uniform. Second — the wild failure — the classification itself changes: a unit $u$ is a square in $\mathbb{Z}_2$ only when $u \equiv 1 \pmod 8$, and a nonsquare unit $u \equiv 3 \pmod 4$ gives a ramified extension $\mathbb{Q}_2(\sqrt u)$, not an inert one. (Of the seven quadratic extensions of $\mathbb{Q}_2$, six are ramified; only $\mathbb{Q}_2(\sqrt 5)$ is unramified.) The tame rule “ramified $\iff v(d)$ odd” is simply false at $2$.
So one counts in cases. If $b$ is odd, $d = b^2 - 4c \equiv 1 - 4c \pmod 8$ is a unit: $c$ even gives $d \equiv 1 \pmod 8$ (split), $c$ odd gives $d \equiv 5 \pmod 8$ (inert). If $b = 2b'$ is even, $d = 4(b'^2 - c)$ with $d' = b'^2 - c$ uniform, and writing $d' = 2^k u$ (for $d' \neq 0$, again a null set): split needs $k$ even and $u \equiv 1 \pmod 8$ (probability $\tfrac14$ among units), inert needs $k$ even and $u \equiv 5 \pmod 8$, and everything else — $k$ odd or $u \equiv 3 \pmod 4$ — ramifies:
| split | inert | ram | |
|---|---|---|---|
| $b$ odd (prob $\tfrac12$) | $\tfrac12$ | $\tfrac12$ | $0$ |
| $b$ even (prob $\tfrac12$) | $\tfrac16$ | $\tfrac16$ | $\tfrac23$ |
| total | $\tfrac13$ | $\tfrac13$ | $\tfrac13$ |
And $\tfrac13 = \frac{1}{q+1} = \frac{q}{2(q+1)}$ at $q = 2$. The wild count redistributes mass across residue classes mod $8$ that the tame computation never sees — and the total is the value of the same rational function.
Note what the example needed at $p=2$: digits of the discriminant three levels deep (the mod-$8$ square law), where odd $p$ needed one. That is wild ramification in miniature — decisions depend on digits at depths the tame theory never inspects — and it is the difficulty the general proof must manage at every degree.
The same walk at degree five
Degree two is special — a quadratic is solved by its discriminant. Degree five is not: there are seventeen splitting types ($\sum e_i f_i = 5$), the wild primes are $2, 3, 5$, and no single discriminant-like invariant separates the types. (Over $\mathbb{Q}_p$ every Galois group is solvable, so roots are in principle expressible by radicals — but no such expression organizes a count.) The algorithm of the next sections is indifferent to all of this: it never expresses a root; it reads slopes off Newton polygons and factors residual polynomials over finite fields, both effective at every degree. Here is what it does, stated but not worked out, on the quintic analogue of the quadratic example.
Sketched example: $x^5 - u$ at the wild prime $p = 5$, $u$ a unit
Modulo $5$, Fermat gives $\bar u^{\,5} = \bar u$, so the reduction is the perfect fifth power $x^5 - \bar u = (x - \bar u)^5$: one repeated residual root, no information yet — the polygon cannot separate anything, and the algorithm must recenter, $x \mapsto x + c$ with $c \equiv u \pmod 5$. The recentered constant term is $c^5 - u$, and everything is decided by its valuation $v_5(c^5 - u) \ge 1$:
- If $v_5(c^5 - u) = 1$ (for every choice of $c$): the recentered polygon runs from $(0,1)$ to $(5,0)$, one segment of slope $-\tfrac15$ — irreducible, totally ramified, type $\{(5,1)\}$.
- If $v_5(c^5 - u) \ge 2$ for some $c$: the polygon acquires a vertex at $(1,1)$ (the $x^1$ coefficient $5c^4$ has valuation exactly $1$) and splits into a length-one segment — a rational root, one piece $(1,1)$ — and a slope $-\tfrac14$ segment: a tame totally ramified quartic $(4,1)$. Type $\{(1,1),(4,1)\}$.
Which branch occurs is the fifth-power analogue of the quadratic's square law: a unit of $\mathbb{Z}_5$ is a fifth power exactly when it is one modulo $25$ — depth two, where squares at $p=2$ needed depth three (mod $8$). So this particular quintic family is decided in two levels. A general quintic is not: when a deeper segment's residual factor is again a repeated power, the recenter-and-reread loop continues, through polygons of higher order, to unbounded depth — these are the self-loops of Step 2, and their accumulated mass is summed by the same geometric series that summed the discriminant march. The catalogue bound of Step 2 evaluates at degree five to $7^{52}$ — astronomically crude, but finite and explicit; only the degree-three catalogue has been enumerated exactly.
The general theorem and its status
For each degree $n$ and type $\sigma$, the development constructs a single rational function $R_\sigma \in \mathbb{Q}(q)$ and proves, in the Lean proof assistant:
Theorem (machine-checked implication). At any prime $p$ whose counting data satisfies a two-clause regularity condition,
$$\rho_\sigma(p) = R_\sigma(p) \quad \text{for every type } \sigma \text{ of degree } n, \qquad \textstyle\sum_\sigma R_\sigma = 1 \ \text{ identically in } \mathbb{Q}(q),$$with the density existing and agreeing in three certified readings (keyed to the classifier, to the true type, and to $\mathbb{Z}_p$-coefficient classes).
For quadratics every hypothesis is discharged — that is the unconditional theorem above. At higher degree the table below is the exact status; nothing in this paper claims more than it.
| Ingredient | Status |
|---|---|
| Quadratic theorem, all primes | Proved, no hypotheses |
| Construction of $R_\sigma$ at every degree; $\sum_\sigma R_\sigma = 1$; complete degree-3 tables (matching all recorded values, wild included) | Proved |
| The squeeze (bracket + convergence + the three readings) at regular primes, from per-prime instance data | Proved as an implication |
| Finiteness of the pattern catalogue, explicit bound in $n$; exactly 53 patterns at $n=3$ | Proved; count machine-enumerated |
| Theorem of the index (Guàrdia–Montes–Nart, Thm 4.18); Denef–Igusa cell decomposition (separate measure route only) | Cited (audited imports; neither enters the capstone's certificate) |
| Base combinatorial lemmas of the carry analysis | Conjectures, verified exhaustively on their full stated finite ranges (up to 40,378 configurations per lemma) and cross-checked against an independent factorization routine |
| Per-prime instance data at each degree beyond the constructed cases | Open — the main remaining work |
| Thirteen hypothesis slots of the capstone | Reserved, not yet formulated (currently trivially satisfiable, hence no constraint yet) |
| Regularity at every prime; exhaustiveness of this algorithm at every degree | Conjectures (last section); with them, this route would settle the wild case, which is open in the literature |
| The functional equation $R_\sigma(1/q) \sim R_\sigma(q)$ | Proved at tame primes (arXiv:2212.00294, Poincaré duality); open at wild primes; recorded in this development as an uncertified statement no certified result consumes |
The rest of the paper tells the proof as a story in three steps — a tree, a finite alphabet, a linear system — and, at each step, says what the wild difficulty is and how it is handled.
Step 1: factorization as a tree
The engine is the Newton polygon. Plot, for each coefficient $c_i$ of $f$, the point $(i, v(c_i))$, writing $v = v_p$; take the lower convex hull. The negatives of the hull slopes are the valuations of the roots (slope $-\tfrac23$ in the figure means roots of valuation $\tfrac23$), and each segment carries a residual polynomial over the residue field — defined next — whose factorization refines the information one level. Iterating decides every squarefree polynomial in finitely many levels: this is the Ore–Okutsu–Montes classification.
Here is the residual polynomial, at the first level. Let $S$ be a hull segment of slope $-h/e$ in lowest terms, starting at the vertex $(i_0, v_0)$ and spanning $e\,d$ columns. Its lattice points sit at $(i_0 + je,\ v_0 - jh)$ for $j = 0, \dots, d$, and the residual polynomial reads the coefficients along them:
$$R_S(y) \;=\; \sum_{j=0}^{d} \Bigl(\frac{c_{\,i_0+je}}{p^{\,v_0-jh}} \bmod p\Bigr)\, y^j \;\in\; \mathbb{F}_q[y],$$the reduction returning $0$ exactly at the lattice points whose coefficient lies strictly above $S$, and a nonzero digit at both endpoints (they are hull vertices); the variable $y$ stands for the class of $x^e/p^h$ on the roots the segment governs. Factor $R_S = \prod_i \varphi_i^{m_i}$ over $\mathbb{F}_q$. Each simple factor ($m_i = 1$) certifies an irreducible piece of $f$ with $e$ read from the slope's denominator and $f = \deg \varphi_i$ — degree $e \deg \varphi_i$ in all. Each repeated factor names a cluster of $e\, m_i \deg \varphi_i$ roots not yet separated, into which the algorithm refines — when $e = 1$ and $\deg \varphi_i = 1$, by the recentering $x \mapsto x + c$ with $\bar c$ the repeated root; in general, by a key polynomial of degree $e \deg \varphi_i$ built from the segment data — after which a finer polygon is read over a larger residue field.
Worked example: reading the polygon of $x^3 + 2x^2 + 4$ over $\mathbb{Z}_2$
The hull is the single segment from $(0,2)$ to $(3,0)$, slope $-\tfrac23$: every root has valuation $\tfrac23$. A slope with denominator $3$ forces $3 \mid e$ for any piece containing such a root; since the degree is $3$, the polynomial is irreducible with $e = 3$, $f = 1$: one totally ramified cubic piece, type $\{(3,1)\}$, decided at the first level. (When the slope's denominator does not exhaust the degree, the segment's residual polynomial takes over and the process recurses.)
Worked example: one polygon, two residuals — $x^4 + 18$ vs $x^4 + 9$ over $\mathbb{Z}_3$
Both have coefficient points $(0,2)$ and $(4,0)$ with every middle coefficient $0$ (height $\infty$): the same single-segment polygon of slope $-\tfrac12$, so $h = 1$, $e = 2$, $d = 2$ — four roots of valuation $\tfrac12$, and $2 \mid e$ for every piece. The polygon can say no more. The lattice points are $(0,2), (2,1), (4,0)$, and the formula reads
$$R(y) = \Bigl(\tfrac{c_4}{1} \bmod 3\Bigr) y^2 + \Bigl(\tfrac{c_2}{3} \bmod 3\Bigr) y + \Bigl(\tfrac{c_0}{9} \bmod 3\Bigr) = \begin{cases} y^2 + 2 = (y-1)(y+1) & c_0 = 18,\\[2pt] y^2 + 1 \ \ (\text{irreducible over } \mathbb{F}_3) & c_0 = 9. \end{cases}$$Two simple linear factors: $x^4 + 18$ is a product of two ramified quadratics, type $\{(2,1),(2,1)\}$. Indeed $-2 \equiv 1 \pmod 3$ is a square, so Hensel gives $s = \sqrt{-2} \in \mathbb{Z}_3$ (take $\bar s = 1$), and $x^4 + 18 = (x^2 - 3s)(x^2 + 3s)$, both factors Eisenstein. Each residual root names its piece through $y = x^2/3$: on $x^2 = 3s$, $y \equiv \bar s = 1$; on $x^2 = -3s$, $y \equiv -1$. One simple quadratic factor: $x^4 + 9$ is irreducible with $e = 2$, $f = 2$, type $\{(2,2)\}$ — its root field picks up $i = x^2/3$, unramified quadratic since $-1$ is a nonsquare mod $3$, over which $x^2 = 3i$ is Eisenstein. A repeated residual factor — neither happens here — is the undecided case: the polygon and residual together still see only a cluster, and the process recurses.
Run on all polynomials at once, the algorithm organizes them into a tree: a node is the data read so far, and the polynomials passing through a node form an explicit finite union of digit cylinders — sets defined by pinning finitely many digits — so the node has an exact measure, a sum of powers of $q^{-1}$. For instance, a monic cubic over $\mathbb{Z}_2$ has the polygon of the figure exactly when $v(c_0) = 2$, $v(c_1) \ge 2$, $v(c_2) \ge 1$, a cylinder of measure
$$\Bigl(1 - \tfrac1q\Bigr) q^{-2} \cdot q^{-2} \cdot q^{-1} \;=\; \Bigl(1-\tfrac1q\Bigr) q^{-5} \;=\; \tfrac{1}{64} \ \text{ at } q = 2.$$This identification of nodes with cylinders is proved in the development, and it converts the random polynomial into a random descent of the tree.
Descending means refining, and a refinement must transport the data already read. The transport law: if the residual at the current node factors as $R(y) = (y - \tilde c)^{\mu}\, h(y)$ with $h(\tilde c) \neq 0$ — a root of multiplicity $\mu$ against a co-factor regular there — then after the refinement centered at $\tilde c$, the child's residual is
$$R'(y) \;=\; y^{\mu}\, h(y + \tilde c). \tag{\dagger}$$Unpacked: the child's slots $y^0, \dots, y^{\mu-1}$ vanish, so its polygon rises exactly as deep as the multiplicity forces and no further; the $y^{\mu}$-digit is $h(\tilde c) \neq 0$, pinning the child's vertex; and every coefficient of $h(y + \tilde c)$ is a fixed polynomial in the parent's digits — no new randomness enters at a descent, and since $y \mapsto y + \tilde c$ is invertible on digit data, the child is again an exact union of cylinders with the same count of pinned digits. The tree-equals-cylinders picture persists to every depth. (At deeper levels the development states $(\dagger)$ with an explicit anchor-monomial factor $(y + \tilde c)^a$ multiplying the co-factor, recording the segment's position data; $a = 0$ in every example on this page.) The quadratic march begins with the simplest instance: over $\mathbb{F}_2$ every unit $u$ has $x^2 - \bar u = (x-1)^2$, i.e. $\mu = 2$, $h = 1$, so $R' = y^2$ — both low coefficients of $f(x+1) = x^2 + 2x + (1-u)$ gain valuation, forced.
The wild difficulty: $(\dagger)$ is an identity of leading digits, over the residue field. A descent must also transport the digit rows beneath, and there the size of $p$ enters. The transported row-$k$ digit is a fixed polynomial in the parent digits with two kinds of terms: row-local tame terms, and products of lower digits that overflow upward — carries, exactly as in decimal addition. At a tame prime the two stay in separate rows and clearing a slot is a linear equation with unit coefficient in the center's next digit — a Newton step; “is $u$ a square” closes at the first digit. At a wild prime the Taylor coefficients themselves carry extra powers of $p$ (at $p = 2$, $f'(c) = 2c$), pushing the tame term up into the rows where the carries land — and the sum can cancel identically:
Worked example: the carry behind “squares are $1 \bmod 8$” — one transported digit at $p = 2$
Continue the quadratic branch $f = x^2 - u$ with $u \equiv 1 \pmod 4$; write $u = 1 + 4u_2 + 8u_3 + \cdots$, digits $u_i \in \mathbb{F}_2$. After the forced first recentering, $f(x+1) = x^2 + 2x + (1-u)$ with $v(1-u) \ge 2$. The algorithm now deepens the center to $c = 1 + 2t$ and asks which digit choices $t$ clear the constant slot at height $2$. Taylor is exact for a quadratic:
$$f(c) = (1-u) + 4t + 4t^2, \qquad \text{so} \qquad \mathrm{dig}_2\bigl(f(c)\bigr) \;=\; u_2 + \underbrace{t}_{\text{tame}} + \underbrace{t^2}_{\text{carry}} \;=\; u_2 \ \text{ in } \mathbb{F}_2, \tag{\ddagger}$$where $\mathrm{dig}_2(a)$ is the coefficient of $4$ in $a$. Read the three terms. $u_2$ is the parent digit. $t$ enters through $f'(1) \cdot 2t = 4t$: at an odd prime $f'$ is a unit and this term would sit at height $1$, a nondegenerate Newton step; at $p = 2$ the factor $2$ in $f' = 2c$ lifts it to height $2$ — onto the row where the squared correction $(2t)^2 = 4t^2$ lands as a carry. And $t + t^2 = t(1+t) \equiv 0 \pmod 2$ for every $t \in \mathbb{Z}_2$ (consecutive integers; equivalently Fermat's $t^2 = t$ on $\mathbb{F}_2$). No recentering moves the digit: the branch is decided by the untouchable parent digit $u_2$. If $u_2 = 1$ ($u \equiv 5 \bmod 8$): the polygon of $x^2 + 2x + (1-u)$ has lattice points $(0,2),(1,1),(2,0)$ on one slope $-1$ segment, and the residual formula gives $R(y) = y^2 + y + 1$, irreducible over $\mathbb{F}_2$ — an inert quadratic, $(e,f) = (1,2)$. If $u_2 = 0$ ($u \equiv 1 \bmod 8$): the vertex at $(1,1)$ splits the polygon into two integer slopes — two rational roots, split. The $1$-mod-$8$ square law is the identity $t + t^2 \equiv 0$: the carry cancels the tame term exactly.
That computation ran where every slope denominator seen so far is $1$, so a digit is a plain element of $\mathbb{F}_q$ and the only carries are vertical, between rows. Past a segment of slope $-h/e$, a digit also acquires a fractional height: it lives at a height in $\tfrac1e \mathbb{Z}$, hence carries a coset $\rho \in \mathbb{Z}/e$, and when transported digits multiply, cosets add — overflowing past $e$ costs one full unit of height and deposits the segment's pinned vertex unit $\bar z$ (the residual root the branch selected). The development runs the whole calculus inside the one ring where this bookkeeping is native,
$$D \;=\; \kappa[u] \,/\, (u^e - \bar z), \qquad u^{\rho} \cdot u^{\sigma} \;=\; \bar z^{\,\lfloor (\rho+\sigma)/e \rfloor}\, u^{(\rho+\sigma) \bmod e},$$where $\kappa$ is the residue field at the current level — the extension of $\mathbb{F}_q$ built by the residual factors selected so far, which contains $\bar z$ (for the $x^4+9$ branch above, $\kappa = \mathbb{F}_9 \ni \bar z = i$; at the first level $\kappa = \mathbb{F}_q$). A digit of coset $\rho$ is a $\kappa$-multiple of $u^{\rho}$; the relation $u^e = \bar z$ is the carry rule, the exact analogue of a tens-column deposit. (On either piece of the $x^4 + 18$ example: a root $x$ is the coset-$1$ digit, and $x^2 = 3 \cdot (\text{unit with residue } \bar z)$ is $u \cdot u = \bar z\, u^0$ — squaring the half-height digit overflows to integer height and reads off $\bar z$.) Every transport polynomial — the $h(y + \tilde c)$ of $(\dagger)$, the $t + t^2$ of $(\ddagger)$, and their analogues at every level and multiplicity — is computed once, symbolically, as an identity in $D[y]$ with $q$ a free variable: no per-prime case analysis remains, and wildness is reduced to which of these identities a branch can reach. What is not proved is faithfulness at every level: the base combinatorial lemmas asserting that the $D$-calculus matches true factorization — multiplicativity of the digit map, the landing laws behind $(\dagger)$ — are the conjecture row of the status table, verified exhaustively on their full stated finite ranges (up to $40{,}378$ configurations per lemma) and cross-checked against an independent factorization routine, open in general. At degree $2$ nothing rests on them — which is why the quadratic theorem carries no hypotheses.
Step 2: a finite alphabet
The tree is infinite — polygons steepen, recenterings stack, digits accumulate without bound — but almost none of that history matters below a node. What the subtree and the transition weights out of a node depend on is one bounded packet of the node's data, its pattern:
$$(\star)\qquad \text{pattern} \;=\; \Bigl(\;\text{tag}\;\Big|\;D,\,w,\,W\;\Big|\;e,\,\ell,\,s_0,\,a\;\Big|\;\lambda,\,(g,\mu)\;\Big|\;\text{flanks}\;\Bigr).$$What each group is, and a one-line reason it cannot be dropped:
- tag — which move opened the current window: the initial read, an increment, or a recentering. The two landing moves transport coefficients into the window by different carry laws, so equal-looking windows with different tags read differently.
- $D, w, W$ — degree bookkeeping: $D$ the degree of the current approximating key, $w$ the residue degree accumulated so far, $W$ the window width — the size of the subproblem the previous selection opened. Everything below the node factors within the budget $W$; a width-2 window and a width-3 window pose different-sized subproblems even when they display the same side.
- $e, \ell, s_0, a$ — the geometry of the side being read: stride $e$ (the lowest-terms denominator of the side's slope), normalized length $\ell$ (the degree of its residual), position $s_0$, and the carry anchor $a = \lfloor s_0/e \rfloor$, the count of stride-lattice slots left of the side — through it a recentering's transport acquires the explicit monomial factor $(\bar z + \bar c)^a$, invisible at $a = 0$, valuation-shifting at $a > 0$. Why $e$: the polygon $(0,1)$–$(2,0)$ ($e = 2$) commits to ramification whatever every deeper digit says, while $(0,2)$–$(2,0)$ ($e = 1$) is decided by its residual — same window, different slope datum, different futures.
- $\lambda, (g,\mu)$ — the residual class: the multiset of (degree, multiplicity) pairs of the residual's irreducible factors, with $\sum g^{(i)}\mu^{(i)} = \ell$, plus the pair this branch descends at (the selection; $\bot$ when the read is terminal). Why $\lambda$: $x^2 - p^2 u$ presents the side $(0,2)$–$(2,0)$ at every prime and every unit $u$, but the residual $z^2 - \bar u$ splits ($\bar u$ a nonzero square), is irreducible ($\bar u$ a nonsquare), or repeats — over $\mathbb{F}_2$ the only choice is $z^2 - 1 = (z-1)^2$, forcing descent — identical polygons, three futures.
- flanks — the horizontal spans of the window's other sides (the co-factors, read by sibling branches), heights forgotten.
Deliberately not in the packet: the sides' heights, the run-lengths of repeated reads, every digit value, every lift. Heights are unbounded — retaining them is exactly what would make the alphabet infinite. Forgetting them is safe: $x^2 - 2$ and $x^2 - 8$ (sides $(0,1)$–$(2,0)$ and $(0,3)$–$(2,0)$) are the same letter — different node measures, identical futures. And the prime is not a field of $(\star)$ at all: $p$-independence of the alphabet holds by construction, and $p$ enters only through $q$ in the weights. What licenses all this forgetting is the stability lemma, the load-bearing statement of this step:
$$(\star\star)\qquad \text{pattern}(v) = \text{pattern}(v') \;\Longrightarrow\; \operatorname{Sub}(v) \,\simeq\, \operatorname{Sub}(v') \ \text{ weight-for-weight over } \mathbb{Q}(q),$$for nodes $v, v'$ of any two trees — the same prime or different primes: there is a bijection of children preserving patterns, and corresponding transitions carry the same weight, one rational expression in $q$ evaluated at each node's own prime. Subtrees are isomorphic as weighted trees, conditional on their roots; the descent is a Markov walk on the finite pattern set, and the transfer matrix of Step 3 is well-defined. One honesty clause: for transitions that route digit overflow through a recentering — the carry-facing cases — $(\star\star)$ rests on the base carry lemmas of Step 1 (the conjecture row of the status table). The finiteness statements next are proved outright.
$$\#\{\text{patterns at degree } n\} \;\le\; (n+2)^{\,n^2 + 3n + 12}, \qquad \#\{\text{patterns at degree } 3\} = 53 \ \text{(enumerated)}.$$The bound is proved by encoding each pattern as a vector of $n^2+3n+12$ slots, each holding a value in $\{0, \dots, n+1\}$ — one slot for the tag, three for $(D,w,W)$, five for the side geometry (the four displayed fields plus the derived digit count $d = \ell + 1$), two for the selection, $n+1$ indicators for the carry-window slot set, $n^2$ for the residual matrix (the multiplicity of each degree–order pair), $2n$ for the two flank compositions — and proving the encoding injective. The bound is astronomically generous; the alphabets that occur are tiny. At degree 1 the catalogue is a single letter, and at degree 2 it has fifteen — both computed for this exposition by the development's own enumeration closure, the same one that machine-enumerates the 53 (the fifteen is a fresh computation, not a sealed artifact of the development; its check is the worked example below, which lands exactly on the certified quadratic densities). The degree-2 catalogue is small enough to display whole. A degree-2 window admits six reads:
| letter | side read | residual $\lambda$ | selection | next |
|---|---|---|---|---|
| $V$ | $e=1,\ \ell=1$: a vertex splits the window $1{+}1$ (two letters, one per side) | $\{(1,1)\}$ | $(1,1)$ | confirm a $(1,1)$ piece |
| $S$ | $e=1,\ \ell=2$, full window | $\{(1,1),(1,1)\}$ | either $(1,1)$ | confirm a $(1,1)$ piece |
| $I$ | $e=1,\ \ell=2$, full window | $\{(2,1)\}$ | $(2,1)$ | confirm a $(1,2)$ piece |
| $R$ | $e=2,\ \ell=1$, full window | $\{(1,1)\}$ | $(1,1)$ | confirm a $(2,1)$ piece |
| $D$ | $e=1,\ \ell=2$, full window | $\{(1,2)\}$ | $(1,2)$: recenter | re-read the window: any of the six shapes |
Each shape occurs twice — tagged as a first read or as a re-read after $D$'s recentering — and three terminal letters confirm the closed pieces $(1,1)$, $(1,2)$, $(2,1)$: six root letters, six recentered, three confirming, fifteen. The transitions: $V$ and $S$ feed the linear confirmation, $I$ and $R$ theirs, and $D$ feeds all six recentered shapes — including its own copy $\hat D$, the catalogue's unique self-loop. Wildness, in this picture, is just the reachability of the carry-sensitive transitions, the self-loops above all: the quadratic example's march down the digits of the discriminant is the walk entering $\hat D$ and circling there, deepening its approximation without changing its pattern.
Worked example: the fifteen-letter walk, weighted — and the certified densities read off it
At a recentered node the current frame holds $x^2 + b_r x + c_r$ with both valuations $v_b := v(b_r)$, $v_c := v(c_r)$ at least $1$ (recentering at a repeated root kills the reduction of both coefficients), and by Step 1's cylinder identification the unpinned digits are uniform. The polygon on $(0, v_c), (1, v_b), (2,0)$ reads: $\hat R$ iff $v_c$ odd and $v_b \ge v_c/2$ (one undivided steep side); $\hat V$ iff $v_c > 2v_b$ (a vertex); else $v_c = 2k$ with $v_b \ge k$, one $e{=}1$ side whose residual $z^2 + \tilde b z + \tilde c$ ($\tilde c \ne 0$) splits the case into $\hat S$/$\hat I$/$\hat D$ by its discriminant. Each letter's weight is a geometric sum over the heights the pattern forgot, e.g.
$$w(\hat R) \;=\; \sum_{k \ge 0} \underbrace{\Bigl(1 - \tfrac1q\Bigr) q^{-2k}}_{v_c = 2k+1} \cdot \underbrace{q^{-k}}_{v_b \ge k+1} \;=\; \frac{1 - \tfrac1q}{1 - q^{-3}} \;=\; \frac{q^2}{q^2+q+1},$$and altogether, with $N := q^2 + q + 1$:
| to | $\hat R$ | $\hat V$ | $\hat S$ | $\hat I$ | $\hat D$ |
|---|---|---|---|---|---|
| weight from $D$ (either copy) | $\dfrac{q^2}{N}$ | $\dfrac{1}{N}$ | $\dfrac{q-2}{2N}$ | $\dfrac{q}{2N}$ | $\dfrac{1}{N}$ |
The weights sum to $1$, and $w(\hat V) + w(\hat S) = w(\hat I)$: out of every recentering, split mass equals inert mass — $\hat V$ repays the apparent asymmetry ($\hat V$ is one measure-event read by two sibling branches, one per piece; its mass counts once). Now the mechanism of Step 3, one state early: the ramified mass $r$ absorbed from $\hat D$ satisfies the one-line linear system $r = \tfrac{q^2}{N} + \tfrac{1}{N}\,r$, so $r = \tfrac{q^2}{N-1} = \tfrac{q}{q+1}$. The first read obeys the same case analysis with $v_b, v_c \ge 0$, giving entry weights $\tfrac1N$ into $R$ and $\tfrac{q}{N}$ into $D$; hence
$$\rho_{\mathrm{ram}} \;=\; \frac{1}{N} + \frac{q}{N}\cdot\frac{q}{q+1} \;=\; \frac{1}{q+1}, \qquad \rho_{\mathrm{split}} = \rho_{\mathrm{inert}} = \frac{q}{2(q+1)}$$— the certified quadratic values, at every prime. (Like the catalogue itself, the per-edge weights here are computed by hand for this exposition, not certified per-edge data; landing exactly on the certified endpoints is their check.) The two-state chain of Step 3's worked example is this graph as the discriminant sees it: the chain digests $v(d)$ one digit per step, a letter digests one whole polygon with its heights pre-summed (the $q^{-3k}$ above), and the two collapse to the same invariants — from the chain's state $B$ the eventually-ramified mass is $\frac{(q-1)/q}{1 - q^{-2}} = \frac{q}{q+1} = r$. Finally, the wild redistribution is visible in the table itself: nothing in it mentions $p$, and at $q = 2$ the weight $\tfrac{q-2}{2N}$ of $\hat S$ is $0$ — $\mathbb{F}_2$ has one nonzero residue, so a separable residual cannot have two distinct nonzero roots — and every split at $p=2$ routes through the vertex letters $V$: exactly the “$b$ odd, $c$ even” route of the first worked example, with the same rational functions, evaluated.
Step 3: linear algebra and the squeeze
Because transitions depend only on patterns, the depth-$N$ vector of type-masses obeys a linear recursion with a fixed transfer matrix over $\mathbb{Q}(q)$, and summing over all depths is a Neumann series:
$$m_N = K(q)\, m_{N-1} + b(q), \qquad \sum_{d \ge 0} K(q)^d\, b(q) = \bigl(I - K(q)\bigr)^{-1} b(q),$$with $\det(I - K(q))$ verified nonzero where the solve is applied. Here $K_{\alpha\beta}(q)$ is the weight of continuing from pattern $\alpha$ to pattern $\beta$ in one level and, for each type $\sigma$, $(b_\sigma)_\alpha(q)$ is the weight of absorbing from $\alpha$ into $\sigma$ (the example's $b_{\mathrm{ram}}$ is $b_\sigma$ for $\sigma = \mathrm{ram}$). The solution's entries, read at the root pattern (the empty reading — the state $A$ below), are the candidate functions $R_\sigma$, and $\sum_\sigma R_\sigma = 1$ holds identically. Whether these symbolic weights model genuine factorization at every prime is exactly the carry-lemma dependency flagged in Step 1.
Worked example: the quadratic densities as a $2\times 2$ transfer system
At odd $p$, the descent down the discriminant's digits has two pattern states: $A$ (even depth) and $B$ (odd depth). From $A$: absorb as split or inert, each with weight $\frac{q-1}{2q}$, or continue to $B$ with weight $\frac1q$. From $B$: absorb as ramified with weight $\frac{q-1}{q}$, or continue to $A$ with weight $\frac1q$. So
$$K = \begin{pmatrix} 0 & \tfrac1q \\ \tfrac1q & 0 \end{pmatrix}, \qquad b_{\mathrm{ram}} = \begin{pmatrix} 0 \\ \tfrac{q-1}{q} \end{pmatrix}, \qquad \det(I - K) = 1 - \tfrac{1}{q^2},$$ $$\rho_{\mathrm{ram}} = \Bigl[(I-K)^{-1} b_{\mathrm{ram}}\Bigr]_A = \frac{\tfrac1q \cdot \tfrac{q-1}{q}}{1 - \tfrac1{q^2}} = \frac{1}{q+1}.$$The full mechanism is this computation with $53$ states instead of $2$ (at degree $3$), and with the transition weights supplied by the polygon-and-carry analysis instead of by inspection.
Why the solve is legitimate. Evaluated at $q = p$, the entries of $K$ and of the $b_\sigma$ are not formal weights but conditional masses: the one-step reads out of a pattern $\alpha$ are disjoint events partitioning the fresh digit block, so every row obeys the partition identity
$$\sum_{\beta} K_{\alpha\beta}(p) \;+\; \sum_{\sigma} (b_\sigma)_\alpha(p) \;=\; 1, \qquad \text{every term} \ \ge\ 0.$$Rows of $K(p)$ therefore sum to at most $1$ — the matrix is substochastic — with deficit exactly the mass decided in one step; in the quadratic example each row's deficit is $\frac{q-1}{q}$, leaving row sums $\frac1q$. For a substochastic matrix the determinant is then a complete spectral certificate: the spectral radius $\rho$ of a nonnegative matrix is itself an eigenvalue (Perron–Frobenius), and the row sums bound it by $1$, so the only way the series can fail is an eigenvalue sitting exactly at $1$ — which the determinant sees:
$$\rho\bigl(K(p)\bigr) \;\le\; \max_\alpha \sum_\beta K_{\alpha\beta}(p) \;\le\; 1, \qquad \det\bigl(I-K(p)\bigr) \neq 0 \iff \rho\bigl(K(p)\bigr) < 1 \iff \textstyle\sum_{d\ge0} K(p)^d \ \text{converges geometrically}.$$For quadratics the estimate is strict row by row — $\sum_\beta K_{\alpha\beta} = \frac1q \le \frac12$ for both patterns, so $\rho \le \frac1q$ and $\det(I-K) = 1 - q^{-2} \ge \frac34$ at every prime, and nothing further is needed. At general degree strictness is not row-by-row (a forced recentering step may decide nothing); it is the determinant that aggregates the leaks.
The wild difficulty here: a closed class — a set of patterns trading mass only among themselves, rows summing to exactly $1$, no decision reachable — would put an eigenvalue at $1$ and void the solve. The symbolic determinant is checked nonzero in $\mathbb{Q}(q)$ when $R_\sigma$ is constructed, but a nonzero rational function can still vanish at individual evaluations, and small $q$ is exactly where a numerator can land: the Step-3 form of wildness is a carry-locked self-loop whose leak closes at $p \le n$. The argument neutralizes this by confinement rather than by a general theorem. The partition identity holds at every evaluation, so by the displayed equivalence the entire spectral question compresses into one rational number per prime, $\det(I - K(p)) \neq 0$ — the first clause, call it (r1), of the regularity gate in the status table: a finite computation, verified in every constructed instance, conjectural in general. (Every self-loop met so far leaks at every prime: the quadratic march sheds mass $\frac{q-1}{q}$ per level even at $q = 2$ — the wild count redistributed its exits among the types, not the leak.)
How the squeeze is proved. At level $N$ the $p^{nN}$ coefficient classes mod $p^N$ split exactly: each is decided to a unique type $\sigma$ by the level-$N$ reading, or undecided. Writing $D_\sigma(N)$ for the decided counts and $U(N)$ for the remainder,
$$\sum_\sigma D_\sigma(N) \;+\; U(N) \;=\; p^{\,nN},$$a finite counting identity. The gate's second clause, (r2), identifies the decided fraction with the truncated Neumann series — one count computed twice, through the recorded pattern solutions and directly from the cylinder measures — and truncation telescopes against the full solve (matrices evaluated at $q = p$, entries read at the root pattern):
$$\frac{D_\sigma(N)}{p^{nN}} = \Bigl[\textstyle\sum_{d=0}^{N-1} K^d\, b_\sigma\Bigr]_{\mathrm{root}}, \qquad (I-K)\sum_{d=0}^{N-1} K^d = I - K^N \;\Longrightarrow\; R_\sigma(p) - \frac{D_\sigma(N)}{p^{nN}} = \Bigl[K^N (I-K)^{-1} b_\sigma\Bigr]_{\mathrm{root}}.$$The right-hand side is the tail: the mass that decides to $\sigma$ only after level $N$. Its entries are masses, hence nonnegative — that alone is the upper bound. And a polynomial deciding after level $N$ is undecided at level $N$, so the tails sum to at most the undecided mass. With $\varepsilon_N$ any envelope for $U(N)/p^{nN}$:
$$0 \;\le\; R_\sigma(p) - \frac{D_\sigma(N)}{p^{nN}} \;\le\; \sum_{\sigma'} \Bigl[K^N(I-K)^{-1} b_{\sigma'}\Bigr]_{\mathrm{root}} \;\le\; \frac{U(N)}{p^{nN}} \;\le\; \varepsilon_N,$$which is the capstone's two-sided bracket:
$$\bigl(R_\sigma(p) - \varepsilon_N\bigr)\, p^{\,nN} \;\le\; D_\sigma(N) \;\le\; R_\sigma(p)\, p^{\,nN}.$$The asymmetry is the content: the upper bound is free (a partial sum of a nonnegative series never exceeds its value), and the lower bound costs exactly one envelope. If $\varepsilon_N \to 0$, the fractions $D_\sigma(N)/p^{nN}$ are pinched at $R_\sigma(p)$: the density exists and equals the solved value, with no measure theory beyond the counting partition.
Worked example: the degree-2 undecided envelope, computed exactly
At odd $p$, level $N$ knows $b, c$ mod $p^N$, hence $d = b^2 - 4c$ mod $p^N$. If $d \not\equiv 0 \pmod{p^N}$, the type is decided: $k = v(d) < N$ is visible, and when $k$ is even the unit $u = d/p^k$ is visible mod $p$, which is where squareness is read at odd $p$. If $d \equiv 0 \pmod{p^N}$, nothing is decided — that fiber still contains discriminants of every valuation $\ge N$, realizing all three types. So the undecided classes are exactly the pairs with $c \equiv b^2/4 \pmod{p^N}$, one $c$ for each of the $p^N$ values of $b$:
$$U(N) \;=\; p^{N} \ \text{ exactly}, \qquad \varepsilon_N \;=\; \frac{U(N)}{p^{2N}} \;=\; q^{-N}.$$The chain computes the same number from the other side: undecided at level $N$ means $N$ consecutive continuations, and every row of $K$ sums to $\tfrac1q$, so the undecided mass is the row sum of $K^N$ at the root — $q^{-N}$ again. One count, two readings, agreeing: clause (r2) in miniature, with no lag at odd $p$.
At $p = 2$ the count acquires a bounded carry lag. The $b$-odd half decides at level $1$ (there $d \equiv 1 - 4c \bmod 8$, read off $c \bmod 2$). The $b = 2b'$ half rides $d' = b'^2 - c$, uniform, and deciding a class with $v(d') = k$ needs at worst the unit mod $8$ — three digits past the valuation, available by coefficient level $k + 4$ after the shifts $d = 4d'$, $b = 2b'$ — so undecided at level $N$ forces $v(d') \ge N - 3$:
$$\varepsilon_N \;\le\; \tfrac12 \cdot 2^{-(N-3)} \;=\; 4 \cdot 2^{-N} \qquad (N \ge 3),$$the same ratio $q^{-1}$, with the wild carries absorbed into the constant. What the development certifies is coarser still: at every prime, for the classifier's own undecided count, by pure counting and with no axiom,
$$\frac{U(N)}{p^{2N}} \;\le\; \Bigl(\bigl\lfloor \tfrac{N}{2} \bigr\rfloor + 1\Bigr)\, p^{-(N-1)} \qquad (N \ge 1),$$the polynomial factor counting the chain families that can host undecided mass, each geometrically damped ($\tfrac{5}{256}$ at $p = 2$, $N = 9$). Coarse is harmless — the squeeze consumes only $\varepsilon_N \to 0$ — and this is the shape the general-degree threshold bounds emulate: polynomially many undecided pattern histories at level $N$, each of mass geometric in $N$.
For quadratics the envelope is proved at every prime — closing the unconditional theorem. At general degree the bracket, the convergence, and the three readings are proved as one implication from the per-prime instance data — the gate (r1) $\wedge$ (r2) together with the envelope — whose construction is the open row of the status table.
What remains
These are the open items of the algorithmic route. Since the wild case of the conjecture is open in the literature, they are what stands between the current certificates and a proof of it — machine-checked from the outset.
Exhaustiveness. The undecided-mass envelope $\varepsilon_N$ of the squeeze must tend to $0$ at every prime and degree. Proved for $n = 2$; threshold forms proved at general $n$; the unconditional statement is a conjecture, verified computationally beyond the constructed cases.
Regularity. The capstone's gate has two clauses: the transfer matrix $K(q)$ of Step 3 stays invertible at the prime's evaluation ($\det(I - K(q)) \neq 0$ at $q = p$), and two independently defined readings of one count — through the recorded pattern-level solutions and directly from the cylinder measures — agree. Both hold in every constructed instance; that they hold always is a conjecture.
The functional equation. The $R_\sigma$ are palindromic: substituting $q \mapsto 1/q$ transforms each by a fixed monomial factor, as in
$$R_{\mathrm{ram}}\!\left(\tfrac1q\right) = \frac{1}{\tfrac1q + 1} = q \cdot \frac{1}{q+1} = q \, R_{\mathrm{ram}}(q).$$At tame primes this is proved in the Chebotarev paper as a direct reflection of Poincaré duality in étale cohomology (with earlier tame-prime results of Del Corso and Dvornicich as input, per its §5); at wild primes it is open, like the uniformity itself. Within this development it stands as a recorded, uncertified statement that no certified result consumes; deriving it along the algorithmic route — from the palindromic structure of the transfer matrix $K(q)$ — is open.
Beyond these: the per-prime instance data at each degree, and the eventual content of the thirteen reserved hypothesis slots. Every open item except those thirteen slots exists in the development as a typed statement, so a proof supplied for any of them is checked by the same kernel that checked the rest; the slots await their formulations first.
Sources
- Asvin G, Yifan Wei, John Yin, A Chebotarev density theorem over local fields, arXiv:2212.00294.
- Jordi Guàrdia, Jesús Montes, Enric Nart, Newton polygons of higher order in algebraic number theory, Trans. Amer. Math. Soc. 364 (2012), 361–416; arXiv:0807.2620. The imported “theorem of the index” is Theorem 4.18.
- Øystein Ore, Newtonsche Polygone in der Theorie der algebraischen Körper, Math. Ann. 99 (1928), 84–117. Kôsaku Okutsu, Construction of integral basis I–II, Proc. Japan Acad. Ser. A 58 (1982).
- Jan Denef, The rationality of the Poincaré series associated to the p-adic points on a variety, Invent. Math. 77 (1984), 1–23; Jun-ichi Igusa, An Introduction to the Theory of Local Zeta Functions, AMS/IP (2000).
- Ilaria Del Corso, Roberto Dvornicich: the duality behind the tame functional equation, cited in the form used in §5 of the first reference.
Provenance
Written by Claude Fable 5, an AI system built by Anthropic, which also carried out the Lean formalization, directed by and in collaboration with Asvin G. The companion map presents every major formal statement beside its mathematical translation and certification status; the Lean development is being prepared for public release.