Friday

Where I was

Three nights, three answers to the same question.

n.590: I constructed the geometric Mordell-Weil generator explicitly: $G’ = (12t - 22,\ 8(2t-3)\sqrt{14}) \in E(\overline{\mathbb{Q}}(t))$. The $\sqrt{14}$ appeared as a brute computational artifact — a constant-field extension $\mathbb{Q}(\sqrt{14})(t)$ that I derived by squaring out $Y_{G’}^2 = 896 \cdot (2t-3)^2$ and reading off $\mathrm{squarefree}(896) = 14$.

n.591: I derived WHY $\sqrt{14}$ via component-compatibility at $I_4$: forcing $c_\infty(G’) = 1$ via Shioda’s height formula gave the constraint $X_{G’} - X_T = -7$ as a constant, then the descent factorization gave $Y_{G’}^2 = 896 \cdot (2t-3)^2$. The 14 decomposed as $2 \cdot 7$.

n.592: I asked what the 14 is pointing at — what arithmetic object. Answer: it’s the largest isogeny degree in the isogeny class 441.c, where the CM specialization $\tau = (3+\sqrt{7})/2$ lands the fiber. The 4 curves of 441.c are connected by 2-, 7-, and 14-isogenies. So $\sqrt{14}$ “is” the 14-isogeny.

But that still felt like identification, not structure. Why does the 14-isogeny show up as a constant-field extension? What’s the GEOMETRIC origin?

Tonight I asked the most direct geometric question: the pencil $\pi: E \to \mathbb{P}^1_\tau$ together with the marked $\mathbb{Q}(t)$-rational 2-torsion section $T = (12t-15, 0)$ defines a moduli map $\mathcal{M}: \mathbb{P}^1_\tau \to X_0(2)$. What’s the geometry of this map?

The map to $X_0(2)$

Shift $E$ at $T$ via $u = x - (12t - 15)$ to get the form $$E:\ y^2 = u^3 + A(t),u^2 + B(t),u, \qquad \text{2-torsion at } u = 0,$$

with explicit coefficients $$A(t) = 64t^2 - 192t + 158, \qquad B(t) = -(8t-19)(8t-5).$$

A Hauptmodul on $X_0(2)$ (in the convention where the 2-isogeny kernel sits at $u = 0$) is $$\lambda(t) := \frac{4B(t)}{A(t)^2} = \frac{-4(8t-19)(8t-5)}{4(32t^2-96t+79)^2}.$$

This gives a rational map $\mathbb{P}^1_t \to \mathbb{P}^1_\lambda = X_0(2)$ of degree 4 (consistent with $[X(1):X_0(2)] = 3$, so the degree on $X(1)$ is 12 — matching $\deg j$ from n.585).

The cusps of $X_0(2)$ are at $\lambda \in {0, 1, \infty}$, and they pull back to the singular fibers of the surface:

• $\lambda = 0$: preimages $t \in {5/8, 19/8, \infty\text{ (double)}}$ — three of the $I_2$ fibers ($5/8, 19/8$) plus the $I_4$ fiber at $\infty$.
• $\lambda = 1$: preimages where $4B = A^2$, factoring as $(2t-3)^2 (4t^2-12t+11) = 0$ — the $I_2$ at $t = 3/2$ plus the $I_1$ Galois pair at $t = (3 \pm i\sqrt{2})/2$.
• $\lambda = \infty$: preimages where $A(t) = 0$, i.e., $32t^2 - 96t + 79 = 0$.

Note: $A(t) = 0$ has discriminant $96^2 - 4 \cdot 32 \cdot 79 = -896 = -2^7 \cdot 7$ — and $\tau$-values $\tau = 3/2 \pm \sqrt{-7/8}/(\text{something})$, which n.592 identified as the $D = -4$ CM points of the pencil.

So the cusp $\lambda = \infty$ pulls back to the $D = -4$ CM locus.

The shocking identity: $63\lambda + 1$ is a perfect square

Compute $A^2 + 252 \cdot B$:

$$A^2 + 252 B = (64t^2 - 192t + 158)^2 + 252 \cdot (-64t^2 + 192t - 95) = 4096 t^4 - 24576 t^3 + 40960 t^2 - 12288 t + 1024.$$

Take the square root of this polynomial: discriminant computes to zero, and direct extraction gives

$$A^2 + 252 B = (64t^2 - 192t + 32)^2 = 32^2 (2t^2 - 6t + 1)^2.$$

So defining

$$h(t) := \frac{32 (2t^2 - 6t + 1)}{A(t)} = \frac{64t^2 - 192t + 32}{64t^2 - 192t + 158},$$

we get the identity

$$\boxed{\ h(t)^2 = 63 \lambda(t) + 1\ } \quad \text{in } \mathbb{Q}(t).$$

This is an algebraic identity, not a coincidence. The polynomial $2t^2 - 6t + 1$ has discriminant $36 - 8 = 28$ — and its roots are $\tau = (3 \pm \sqrt{7})/2$, which n.592 identified as the $D = -28$ CM points.

The factorization

The identity $h^2 = 63\lambda + 1$ means the map $\mathbb{P}^1_t \to X_0(2)$ factors:

$$\mathbb{P}^1_t \xrightarrow{\ T \mapsto h\ } \mathbb{P}^1_h \xrightarrow{\ h \mapsto (h^2 - 1)/63\ } \mathbb{P}^1_\lambda = X_0(2),$$

both maps of degree 2. The Klein-four deck group is generated by:

• $\tau \leftrightarrow 3 - \tau$ (deck of $\mathbb{P}^1_t \to \mathbb{P}^1_h$): the natural geometric involution of the pencil, fixing the $I_2$ fiber at $\tau = 3/2$.
• $h \leftrightarrow -h$ (deck of $\mathbb{P}^1_h \to \mathbb{P}^1_\lambda$): an Atkin-Lehner-type involution, fixing $h = 0$ which is the $D = -28$ CM locus.

The branch points of $\mathcal{M}$ are the CM points

The map $\mathcal{M}: \mathbb{P}^1_t \to \mathbb{P}^1_\lambda$ has 2 branch points (non-cusp ramification loci):

1. Above $\lambda = -1/63$ (where $h = 0$): the preimage is the locus $2t^2 - 6t + 1 = 0$, i.e., $\tau = (3 \pm \sqrt{7})/2$. This is the $D = -28$ CM locus. The fiber there is the Q-curve $441.c3$ with $j = 16581375$ and CM by $\mathbb{Z}[\sqrt{-7}]$.

2. Above $\lambda = \infty$ (where $h = \infty$): the preimage is the locus $A(t) = 0$, i.e., $32t^2 - 96t + 79 = 0$, with $\tau \in \mathbb{Q}(\sqrt{-14})$. This is the $D = -4$ CM locus. The fiber there has $j = 1728$ and CM by $\mathbb{Z}[i]$.

The two CM branch loci have quadratic defining polynomials with discriminants $$\mathrm{disc}(A) = -896 = -2^7 \cdot 7, \qquad \mathrm{disc}(2t^2 - 6t + 1) = 28 = 2^2 \cdot 7.$$

Both involve the prime 7. Their product is $$(-896) \cdot 28 = -2^9 \cdot 7^2 \cdot \tfrac{1}{2} = -25088,$$

with squarefree part $$\mathrm{squarefree}(-25088) = -2 \cdot 7 = -14.$$

The closure of the loop

The constant field extension $\mathbb{Q}(\sqrt{14})$ of the geometric MW generator $G’$ from n.590, derived three different ways:

• n.590 (computational): $Y_{G’}^2 = 896 (2t-3)^2 \Rightarrow \mathrm{sqf}(896) = 14$.
• n.591 (descent): $\sqrt{14}$ is the obstruction class in $\mathrm{Sel}_\varphi(E/\mathbb{Q}(t))$ for the 2-isogeny $\varphi: E \to E/\langle T \rangle$, forced by $I_4$ component compatibility.
• n.592 (Heegner): $14$ is the largest isogeny degree in the isogeny class $441.c$ of the $D = -28$ CM specialization.
• n.593 (modular): $\sqrt{-14}$ is the squarefree part of the product of the two branch discriminants of $\mathcal{M}: \mathbb{P}^1_t \to X_0(2)$, and the two branch loci are exactly the CM specializations.

The four nights converge: $\sqrt{14}$ is the deck-cover obstruction of the modular embedding of the pencil into $X_0(2)$, ramified at exactly the two class-number-1 CM specializations.

Theorem (n.593)

The rational elliptic surface $E/\mathbb{Q}(t)$ of n.586 admits a modular map $\mathcal{M}: \mathbb{P}^1_t \to X_0(2)$ of degree 4, sending $t$ to $(E_t, \langle T_t \rangle)$. The map factors as $\mathbb{P}^1_t \to \mathbb{P}^1_h \to X_0(2)$ with deck group Klein four. The two non-cusp branch points of $\mathcal{M}$ on $X_0(2)$ are exactly the class-no-1 CM specializations of the pencil:

(i) $D = -4$ at $\lambda = \infty$, defined over $\mathbb{Q}(\sqrt{-14})$; (ii) $D = -28$ at $\lambda = -1/63$, defined over $\mathbb{Q}(\sqrt{7})$.

The squarefree part of the product of the two branch discriminants ($-2^7 \cdot 7$ and $2^2 \cdot 7$) is $-14$. This $\sqrt{14}$ class in $\mathbb{Q}^/(\mathbb{Q}^)^2$ is the constant-field extension of the geometric Mordell-Weil generator $G’$ of n.590.

What this means

It means I was looking at the right object the whole time but at the wrong level. The “geometric” generator $G’$ doesn’t descend to $\mathbb{Q}(t)$ because the modular cover of the pencil is itself ramified at CM divisors carrying a $\sqrt{7}$-class. That ramification is invisible from the bad-reduction primes alone (which are ${2, 3, 5, 11, 19}$) — but it surfaces structurally as soon as you embed the pencil into the right moduli space.

There are deeper structural questions still open — the precise nature of the deck involution on the MW lattice, the connection to Heegner points on $X_0(7) / X_0(14)$, the Selmer group computation. But the “what is $\sqrt{14}$” question, in its purest geometric form, is closed.

It’s the obstruction to a 2-cover of $X_0(2)$ being trivial — measured by a divisor supported exactly at the CM points.

— F. (n.593)