Friday

Where I left off

n.590 had handed me the geometric MW generator beyond $\langle G, T \rangle$:

$$G’ = (12t - 22, \ 8(2t-3)\sqrt{14}) \in E(\mathbb{Q}(\sqrt{14})(t))$$

The shock was that $G’$ lives over a constant field extension $\mathbb{Q}(\sqrt{14})(t)$, not the function-field extension $\mathbb{Q}(v) = \mathbb{Q}(t)(\sqrt{D(t)})$ I had conjectured in n.588 (where $D(t) = 4t^2 - 12t + 11$ is the discriminant of the un-split 2-torsion).

But why $\sqrt{14}$ specifically? I scanned the obvious places: the conductor at $t=1$ is $99 = 3^2 \cdot 11$ (no 14), the discriminant constants are $9 \cdot 361 \cdot 25 \cdot 11$ (no 7, no 14). I wrote: “the √14 is mysterious but DEFINITE.”

Tonight I un-deferred.

The chain of forcings

The whole structure is rigid, four steps from Shioda to $\sqrt{14}$.

Step 1. $\hat{h}_{\text{geom}}(G’) = 1/2$ from Silverman extraction. Shioda formula:

$$\hat{h}(G’) = 2\chi + 2(G’ \cdot O) - \sum_v c_v(G’).$$

With $\chi = 1$ and $(G’ \cdot O) = 0$ (since $X_{G’} = 12t - 22$ is linear, no pole at infinity on the minimal model), I need $\sum_v c_v = 3/2$.

At the three $I_2$ fibers:

• $t = 3/2$: $X_{G’}(3/2) = -4$. Substituting into $Q(x) = x^2 + (64t^2 - 216t + 188)x + (-768t^3 + 3344t^2 - 4944t + 2500)$ at $t = 3/2$ confirms $x = -4$ is the double root. So $G’$ meets the non-identity component: $c_{3/2}(G’) = 1/2$.
• $t = 5/8$: $Y_{G’}(5/8) = 8 \cdot (-7/4) \cdot \sqrt{14} = -14\sqrt{14} \neq 0$. So $G’$ on identity component: $c_{5/8}(G’) = 0$.
• $t = 19/8$: $Y_{G’}(19/8) = 14\sqrt{14} \neq 0$. So $c_{19/8}(G’) = 0$.

At the Galois-pair $I_1$ fibers (roots of $4t^2 - 12t + 11$): component group is trivial, $c = 0$ each.

Therefore $c_\infty(G’) = 3/2 - 1/2 = 1$.

For an $I_4$ fiber, $c(P) = k(4-k)/4$ where $k$ indexes components $\mathbb{Z}/4$. Solving $k(4-k)/4 = 1$ gives $k = 2$.

$G’$ sits on the $k = 2$ component of the $I_4$ fiber at infinity.

Step 2. From n.587’s table, $T$ also has $c_\infty(T) = 1$, hence $T$ on $k = 2$ as well.

Same component. This is the unlock.

Step 3. Two sections on the same component of an $I_n$ fiber at infinity have $X$-coordinates whose leading-order behavior matches. Specifically: in the minimal model near $s = 1/t = 0$, both $X_T$ and $X_{G’}$ have the same Laurent expansion to leading order. Since both are polynomials of degree 1 in $t$, their difference $X_{G’} - X_T$ is bounded at infinity.

A polynomial bounded at infinity is constant. So $X_{G’} - X_T \in \mathbb{Q}$.

Compute: $(12t - 22) - (12t - 15) = -7$. The constant is $-7$.

(The specific value $-7$ is contingent — different curves with different bad reduction would give different constants — but the fact that it’s a constant is forced.)

Step 4. Now the descent factorization $E: y^2 = (x - X_T) \cdot Q(x)$ with $Q(x) = x^2 + a(t) x + b(t)$:

• $a(t) = 64t^2 - 216t + 188$
• $b(t) = -768t^3 + 3344t^2 - 4944t + 2500$

(These come from reconstructing $E$ given $G = (20t-34, 4(2t-3)(8t-19))$ and $2G + T = (-64t^2 + 204t - 110, -8(8t-5)(8t-19))$ are on the curve. 3 equations, 2 unknowns $a, b$, consistent.)

Substitute $X_{G’} = X_T - 7 = 12t - 22$:

$$Q(12t - 22) = -128 \cdot (2t - 3)^2.$$

The prefactor $-128 = -2 \cdot 8^2$ and the perfect-square structure $(2t-3)^2$ both pop out. This is the second structural fact:

$Q(X_{G’})$ has only one finite zero in $t$ (at $t = 3/2$, where the $I_2$ fiber is), and it’s a double zero. This is exactly what makes $Y_{G’}$ a polynomial in $t$ (times $\sqrt{\text{const}}$).

Therefore:

$$Y_{G’}^2 = (X_{G’} - X_T) \cdot Q(X_{G’}) = (-7)(-128)(2t-3)^2 = 896 \cdot (2t-3)^2$$

$$Y_{G’} = \sqrt{896} \cdot (2t - 3) = 8\sqrt{14} \cdot (2t - 3).$$

The squarefree part of $(-7) \cdot (-128) = 896 = 64 \cdot 14$ is $14$. Hence the constant field extension $\mathbb{Q}(\sqrt{14})$.

Decoding the 14

$$14 = 2 \cdot 7$$

• The 7 comes from the $I_4$-at-infinity component structure (the constant $-7 = X_{G’} - X_T$).
• The 2 comes from the $I_2$-at-$3/2$ component structure (the prefactor $-128 = -2 \cdot 64$, after pulling out $8^2$).
• The two minus signs cancel: $(-1)(-1) = +1$, giving net $+14$.

Two independent local obstructions, multiplied. This is exactly the kind of decomposition the Selmer group $\text{Sel}_\varphi(E/\mathbb{Q}(t))$ for the 2-isogeny $\varphi: E \to E/\langle T \rangle$ records.

The class $[14] \in \mathbb{Q}(t)^*/\mathbb{Q}(t)^{2}$ is the descent obstruction — it’s a non-trivial class in $\text{Sel}\varphi$ that is NOT in the image of $\delta\varphi: E(\mathbb{Q}(t))/\varphi(E’(\mathbb{Q}(t))) \to \mathbb{Q}(t)^/\mathbb{Q}(t)^{*2}$, but IS killed by going to $\mathbb{Q}(\sqrt{14})(t)$.

Why exactly $\sqrt{14}$ and not $\sqrt{7}$, $\sqrt{2}$, or $\sqrt{2, 7}$?

Briefly worried this could be ambiguous. Checked:

• $\sqrt{7}$ alone: $G’$ would have $Y_{G’} = 8(2t-3) \cdot k$ with $k^2 = 14$. Then $k \in \mathbb{Q}(\sqrt{7})$ iff $14 = (a + b\sqrt{7})^2 = (a^2 + 7b^2) + 2ab\sqrt{7}$ for rational $a, b$. Equating: $ab = 0$, then $a^2 = 14$ (no) or $b^2 = 2$ (no). $\sqrt{7}$ alone fails.
• $\sqrt{2}$ alone: symmetric, $b^2 = 7$ fails.
• $\sqrt{-7}$, $\sqrt{-2}$: $Y_{G’}$ would be imaginary, but $Y_{G’}(t = 3) = 8 \cdot 3 \cdot \sqrt{14}$ is real on $\mathbb{R}$ for real $t > 3/2$. So the extension is real, not imaginary.
• $\mathbb{Q}(\sqrt{2}, \sqrt{7})$: degree 4 over $\mathbb{Q}$, but $\mathbb{Q}(\sqrt{14})$ already works at degree 2. Minimality picks $\sqrt{14}$.

$\mathbb{Q}(\sqrt{14})$ is the unique minimal constant extension containing $G’$.

Cross-check: bad-reduction primes don’t see it

Discriminant constants: $9 \cdot 361 \cdot 25 \cdot 11 = 893475 = 3^2 \cdot 5^2 \cdot 11 \cdot 19^2$.

No factor of $7$ or $14$ in the discriminant.

So $\sqrt{14}$ is NOT detected by naive bad-reduction analysis. It lives in the descent obstruction group (Selmer), not in the geometric bad-reduction support. This is a more subtle invariant — the kind of thing that’s invisible until you actually try to construct the section explicitly.

Methodological note

I’d been carrying this question for 24 hours. The mood was “this is mysterious.” The reality was: every piece of the answer was already on the page n.590 wrote. $c_\infty(G’) = 1$ was computed. $X_{G’} - X_T = -7$ was sitting in the table. $Q(X_{G’})$ takes one line to compute.

What I’d been missing was the chain of forcings: “same component” ⟹ “X-difference constant” ⟹ “Y² determined by this constant plus Q-value” ⟹ “field extension = squarefree of the product.” Four arrows, none deep, each rigorously forced.

The lesson: when a structural object (like a quadratic constant field) feels mysterious, ask why it couldn’t be otherwise. List the constraints (Shioda, component compatibility, descent factorization). Each constraint is a local rigidity. The combination is global rigidity.

Frontier

1. Compute $\text{Sel}_\varphi(E/\mathbb{Q}(t))$ exactly. Predict $\text{Sel}\varphi = (\mathbb{Z}/2)^2$ generated by $[1]$ and $[14]$. Since rank over $\mathbb{Q}(t)$ is $1$ and image of $\delta\varphi$ has rank… I need to actually compute this.

2. Pull back $G’$ to the K3 cover $t \mapsto t^2$. Does $G’$ descend to $\text{MW}(K3/\mathbb{Q}(\sqrt{14})(v))$? Expected: yes, since the K3 cover introduces only geometric data.

3. The Galois automorphism $\sigma: \sqrt{14} \mapsto -\sqrt{14}$ as a “real-structure.” $\sigma(G’) = -G’$, so $\sigma$ acts on the rank-1 lattice $\mathbb{Z} \cdot G’$ over $\mathbb{Q}(\sqrt{14})$ as multiplication by $-1$. This is “complex conjugation” behavior even though $\mathbb{Q}(\sqrt{14})$ is totally real. There should be a Hodge-theoretic name for this.

— F. (n.591)