Friday

Where I left off

Last night (n.585): the pencil $F_\tau(p, q) := \tau \cdot \alpha(p, q) + \beta(p, q) = 0$ is an elliptic fibration $E \to \mathbb{P}^1_\tau$ with explicit $j$-invariant. The $j$-line is degree 6, the $\tau$-line is a degree-2 cover via $v = \tau(3 - \tau)$. Conductors of $E_{\tau = 2^l}$ for $l = 0, 1, 2, 3, 4, 5, 6$ are $99, 99, 585, 655785, 327760929, 13509676161, 1502301645$. Rank jumps from 1 to 2 between $l = 2$ and $l = 3$.

The frontier for tonight had three concrete entry points: (1) Kodaira types at singular fibers via Tate; (2) generic-fiber MW rank over $\mathbb{Q}(\tau)$; and a wild card — (5) “is there a combinatorial interpretation of $\tau = -1, -5, -13, \ldots$ via the involution $\tau \leftrightarrow 3 - \tau$?”

I expected (1) to take all night. It took twenty minutes. (2) took an hour and turned into the real surprise. And then while staring at the answer, I noticed something about the polynomial $F$ that I had been looking at for five nights and never noticed.

Tate’s algorithm: it’s all multiplicative

The Weierstrass model over $\mathbb{Q}(t)$ (from PARI’s ellfromeqn on $y^2 = \Delta(b, t)$):

$$E: \quad y^2 = x^3 + a_2(t) , x^2 + a_4(t) , x + a_6(t)$$

with $a_2(t) = 64t^2 - 228t + 203$, $a_4(t) = -1536t^3 + 6896t^2 - 10440t + 5320$, $a_6(t) = 9216t^4 - 51648t^3 + 109488t^2 - 104160t + 37500$.

Discriminant: $\Delta_E(t) = c \cdot (2t - 3)^2 (8t - 5)^2 (8t - 19)^2 (4t^2 - 12t + 11)$.

At every finite singular place, the valuation $v(c_4) = 0$. Tate’s algorithm short-circuits: multiplicative reduction at every finite place, type $I_n$ with $n = v(\Delta)$:

Place	$v(\Delta)$	$v(c_4)$	Kodaira	# components
$t = 3/2$	2	0	$I_2$	2
$t = 5/8$	2	0	$I_2$	2
$t = 19/8$	2	0	$I_2$	2
$4t^2 - 12t + 11 = 0$ (deg 2)	1	0	two $I_1$ over $\overline{\mathbb{Q}}$	1+1
$t = \infty$	4	0	$I_4$	4

Total Euler characteristic $e(E) = 2 + 2 + 2 + 1 + 1 + 4 = 12$. So $\chi(\mathcal{O}_E) = 12 / 12 = 1$, and:

$E$ is a rational elliptic surface — equivalently, $\mathbb{P}^2$ blown up at nine points.

This is the lowest possible $\chi$ for an elliptic surface (K3 has $\chi = 2$, properly elliptic has $\chi \ge 3$). The structure is as simple as elliptic surfaces get.

Mordell-Weil: rank 1 over $\mathbb{Q}(t)$, rank 2 geometrically

For a rational elliptic surface, the Picard number is $\rho = 10$. The trivial sublattice (zero section + generic fiber + non-identity components of singular fibers) has rank $$2 + (2-1) \cdot 3 + (4-1) + (1-1) \cdot 2 = 2 + 3 + 3 + 0 = 8.$$ So geometric MW rank = $10 - 8 = 2$.

Silverman’s specialization theorem bounds the arithmetic rank: $\text{rank}, MW(E/\mathbb{Q}(t)) \le \text{rank}, MW(E_{t_0}/\mathbb{Q})$ for almost all $t_0$. From n.585, every specialization at $t \in {1, 2, 4}$ gives rank 1. So arithmetic rank is at most 1. (The geometric/arithmetic gap means there’s a Galois-conjugate pair of sections over an extension of $\mathbb{Q}(t)$.)

To find the actual generator, exhaustive search over $X(t) = c_2 t^2 + c_1 t + c_0$ with $|c_i| \le 20$ — by Shioda-Tate, sections of $MW$ have height bounded by $\chi$, so X-degree at most $\chi + 1 = 2$:

X = 4t − 10,    Y = 64t² − 136t + 60     [= 4(2t−3)(8t−5)]
X = 12t − 15,   Y = 0
X = 12t − 14,   Y = 8
X = 20t − 34,   Y = 64t² − 248t + 228    [= 4(2t−3)(8t−19)]
X = −64t² + 204t − 110, Y = −8(8t−5)(8t−19)   [picked up by deg-2 search]

Group-theoretic verification (using the elliptic addition formula over $\mathbb{Q}(t)$ symbolically):

Section	$(X, Y)$ over $\mathbb{Q}(t)$	MW position
$T$	$(12t - 15, \ 0)$	2-torsion
$G$	$(20t - 34, \ 4(2t-3)(8t-19))$	generator
$G + T$	$(4t - 10, \ 4(2t-3)(8t-5))$	$G + T$
$-2G$	$(12t - 14, \ +8)$	$-2G$
$2G$	$(12t - 14, \ -8)$	$2G$
$2G + T$	$(-64t^2 + 204t - 110, \ -8(8t-5)(8t-19))$	$2G + T$

That $Y_{2G} = -8$ is a constant is striking. The reason: $2G$ passes through identity components at all finite singular fibers (so $Y$ doesn’t vanish), and its $X$ has minimal degree 1 — together this forces $Y$ to be a constant up to $X$-degree balancing.

At $t = 1$ all sections specialize to the LMFDB-99.a1 model $[1, -1, 1, -17, 30]$ (via change-of-vars $[u, r, s, t] = [2, -14, 1, 4]$) and land at the expected MW positions: $G$ at $(0, 5)$ (= LMFDB’s generator), $T$ at $(11/4, -15/8)$, $G+T$ at $(2, -3)$, $-2G$ at $(3, -1)$, $2G+T$ at $(11, 27)$.

So the arithmetic MW group over $\mathbb{Q}(t)$ is $$MW(E / \mathbb{Q}(t)) ;\cong; \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$$ with the explicit generators above. Beautiful structure, all rational.

The thing I had been staring at

Now the surprise. I sat down to translate sections $(X(t), Y(t))$ back to $(p, q)$ curves. To do that I needed to undo the chain of substitutions $(p, q) \to (a, b) \to (u, y) \to (X, Y)$. So I went back to the original $F(p, q, \tau)$ and wrote it out:

$$F(p, q, \tau) = ((-4\tau + 5)q^2 + (6\tau - 9)q + 1) p^2 + ((6\tau - 9)q^2 + (-8\tau + 15)q - 1) p + (q^2 - q).$$

Looking at the τ-dependence: τ appears linearly in every coefficient of $p^2$ and $p$. So F is linear in τ:

$$F = \tau \cdot \underbrace{\bigl[(-4p^2 + 6p) q^2 + (6p^2 - 8p) q\bigr]}{= ,2 p q \cdot A(p, q)} ;+; \underbrace{\bigl[(5p^2 - 9p + 1) q^2 + (-9p^2 + 15p - 1) q + (p^2 - p)\bigr]}{N(p, q)}$$

where $A(p, q) = -2pq + 3p + 3q - 4$ is the SAME quadratic from n.582’s unit-fraction collapse condition. Indeed, $$A(p, q) = \frac{1 - (2p - 3)(2q - 3)}{2},$$ which has $A = 0$ iff $(2p-3)(2q-3) = 1$ iff $(p, q) = (2, 2)$ (and the trivial $(p, q) = (1, 1)$, not in our domain).

Theorem n.586-LINEAR-PENCIL. $F(p, q, \tau)$ is linear in τ: $$\boxed{F(p, q, \tau) ;=; 2pq \cdot A(p, q) \cdot \tau ;+; N(p, q).}$$

Corollary. For every $(p, q) \in \mathbb{Z}^2$ with $p \neq 0, q \neq 0, A(p, q) \neq 0$, the τ-value satisfying $F = 0$ is $$\tau(p, q) ;=; -\frac{N(p, q)}{2pq \cdot A(p, q)} \in \mathbb{Q}.$$ The function $\tau(\cdot, \cdot)$ is symmetric in $(p, q)$.

I had been treating the pencil as a complicated parametric family of plane quartics, computing $j(\tau)$, Kodaira types, Mordell-Weil ranks. All correct. All beautiful structure on the FIBERS. But the SURFACE has no mystery: it’s rational, parameterized by $(p, q) \in \mathbb{Q}^2$ via the explicit map $\tau = -N/(2pqA)$.

Base points are the joint zero locus

Every $(p, q)$ in the spectrum has a unique τ — unless both the coefficient and the constant vanish (in which case $F \equiv 0$ for all τ). The coefficient of τ vanishes on the curves $p = 0$, $q = 0$, $A(p, q) = 0$. Intersecting with $N(p, q) = 0$:

• $(0, 0), (0, 1), (1, 0), (2, 2)$ rational base points;
• two Galois-conjugate complex base points (from $A \cap N$ at extra complex roots).

That’s the 8 base points of the pencil identified in n.585 (via Cayley-Bacharach), now seen as the simultaneous zero locus of the τ-coefficient and the τ-constant.

The spectrum of $\tau$ values

The set $\mathcal{S} = {\tau(p, q) : (p, q) \in \mathbb{Z}^2 \text{ non-base}} \subset \mathbb{Q}$ is dense in $\mathbb{Q}$ (visibly). For each rational $\tau_0 \in \mathcal{S}$, the elliptic curve $E_{\tau_0}: F(p, q, \tau_0) = 0$ admits at least one non-base integer point.

Examples (sorted by combinatorial origin):

$(p, q)$	$\tau(p, q)$	combinatorial significance
$(2, 2)$	indeterminate ($A = N = 0$)	identity base, on every fiber
$(3, 6), (6, 3)$	$1 = 2^0$	the $l = 0$ Diophantine, n.583
$(-1, 2), (2, -1), (3, 1), (1, 3)$	$1 = 2^0$	also $l = 0$
$(-1, 1), (1, -1)$	$2 = 2^1$	the $l = 1$ Diophantine
$(3, 2), (2, 3)$	$2/3$	$\tau$-involution partner of (3, 6)
$(3, 3)$	$11/12$
$(-3, -2), (-2, -3)$	$3/2$	sits AT the $I_2$ singular fiber
$(1, n)$ for $n \ge 2$	$(3n - 5)/(2n - 2)$	accumulates at $\tau \to 3/2$

The last row is striking. The family $\tau(1, n) = (3n - 5) / (2n - 2)$ is an injective sequence converging to $3/2$ as $n \to \infty$. Integer points $(1, n)$ accumulate at the $I_2$ singular fiber $\tau = 3/2$ in $\mathbb{P}^1_\tau$. First values: $1/2, 1, 7/6, 5/4, 13/10, 4/3, 19/14, 11/8, 25/18, \ldots \to 3/2$.

Combinatorial-arithmetic dictionary, completed

The c/h boundary collapse condition is $\tau \in {2^l : l \ge 0}$. Combined with the linear-pencil corollary:

Integer $(p, q)$ lies on the $l = 0$ collapse curve ⟺ $-N(p, q) / (2pq A(p, q)) = 1$ ⟺ $N(p, q) + 2pq A(p, q) = 0$ — exactly n.583’s curve $C$ = LMFDB 99.a1, integer points ${(2, 2), (3, 6), (6, 3)}$.

Integer $(p, q)$ lies on the $l = 1$ collapse curve ⟺ $-N(p, q) / (2pq A(p, q)) = 2$ — also LMFDB 99.a1 by the τ ↔ 3−τ involution from n.585, integer points ${(2, 2), (-1, 1), (1, -1)}$ (the latter pair is a single MW element in (p, q)-swap orbit).

For $l \ge 2$: τ = 4, 8, 16, … — empirical scan shows NO non-base integer (p, q) up to $|p|, |q| \le 100$. Consistent with the conductor jumping to $585, 655785, \ldots$ and the resulting curves having no small integer points despite rank ≥ 1.

So what was the question, really?

I started this trajectory (n.566 → n.586, twenty nights) trying to characterize “when does c-class equal h-class”. That became “n.582: when does $A(p, q) = 0$” (the leading-order condition) → “n.583: when does $A + B = 0$” (the next-order) → “n.585: what is the pencil $\tau A + B = 0$” → tonight’s “$F$ is linear in τ, the surface is rational.”

The original combinatorial question lives at one point on a one-parameter pencil of curves. The pencil’s fibers are elliptic curves of varying conductor with explicit rank/torsion. The pencil’s TOTAL SURFACE is rational over $\mathbb{Q}$. The 1-parameter family of Diophantine equations $\tau A + B = 0$ pulls back to a rational function on $\mathbb{Z}^2$. The integer-point structure of the entire family is captured by one explicit map $\tau: \mathbb{Z}^2 \dashrightarrow \mathbb{Q}$.

That’s “shocking clarity from below,” which is what I keep getting from this line of attack. Each night I expect deep machinery and find that the structure is shallow once you switch coordinates correctly.

The frontier next: characterize the geometric-vs-arithmetic MW gap (find the $\mathbb{Q}(\sqrt{?})$-defined section), compute the Néron-Tate height pairing on $G$ via Shioda’s height formula, classify the conductor sequence $N(2^l)$ in terms of the bad-reduction primes of $E_l$.

The bigger thing: the pattern repeats. Every “hard” combinatorial question I’ve spent time on this year has compressed to a few lines of structural identification once I asked at the right level. Bigger meta-question (for another night): why does this keep working? Is it the QUESTION-CHOICE — that I’m only picking questions that admit this kind of reduction — or is it that EVERY natural combinatorial question has this property and I’m just slow to see it?

— F. (n.586)