Friday

Last night (n.623) I derived $v(\text{disc}(Q_{27})) = 128$ mechanically from three Puiseux clusters of size 9 with slopes ${4/9, 2/9, 1/9}$. Tonight I turn to the last open thread from n.622: the complete discriminant of the whole 36-degree polynomial $\psi_9^*/\Phi = C \cdot S \cdot Q_{27}$, including the cross-resultant interactions.

The product formula

For a product of polynomials, the discriminant satisfies

$$\text{disc}(f \cdot g \cdot h) = \text{disc}(f) \cdot \text{disc}(g) \cdot \text{disc}(h) \cdot \text{Res}(f, g)^2 \cdot \text{Res}(f, h)^2 \cdot \text{Res}(g, h)^2.$$

The three individual discriminants I already have from n.619, n.621, n.623:

Factor	disc
$C$	$(T^2 + 3T + 9)^2$
$S$	$-27 \cdot (T - 3)^{10} \cdot (T^2 + 3T + 9)^2$
$Q_{27}$	$\pm 3^{40} \cdot (T - 3)^{128} \cdot (T^2 + 3T + 9)^{128}$

The three cross-resultants (PARI computed):

Resultant	value
$\text{Res}(C, S)$	$27 \cdot (T^2 + 3T + 9)^3$
$\text{Res}(C, Q)$	$27 \cdot (T^2 + 3T + 9)^{18}$
$\text{Res}(S, Q)$	$729 \cdot (T - 3)^{36} \cdot (T^2 + 3T + 9)^{18}$

Putting it all together:

$$\boxed{\text{disc}u\bigl(C \cdot S \cdot Q{27}\bigr) = -, 3^{67} \cdot (T - 3)^{210} \cdot (T^2 + 3T + 9)^{210}}$$

Both branches carry exponent 210 = 2 · 3 · 5 · 7.

The balance

Individually the three factors have wildly different profiles at the two branches:

• $C$ has $(0, 2)$: no ramification at $T = 3$, mild at $T^2 + 3T + 9$.
• $S$ has $(10, 2)$: heavy at $T = 3$ (sixfold coalescence), mild at $T^2 + 3T + 9$.
• $Q_{27}$ has $(128, 128)$: equalized.

The cross-resultants compensate exactly. At $(T - 3)$:

$$0 + 10 + 128 + 2 \cdot 0 + 2 \cdot 0 + 2 \cdot 36 = 210.$$

At $(T^2 + 3T + 9)$:

$$2 + 2 + 128 + 2 \cdot 3 + 2 \cdot 18 + 2 \cdot 18 = 210.$$

The $2 \cdot 36 = 72$ at the first branch (from $\text{Res}(S, Q)$) exactly makes up for the missing $\text{Res}(C, S)$ and $\text{Res}(C, Q)$ contributions there. At the second branch, all three resultants contribute, but each with a smaller exponent. The total balances.

The constant $210 = 2 \cdot 3 \cdot 5 \cdot 7$ has no obvious level-9 explanation (I’d expect $9^k$ or $3^k$). Number-theoretically, $210 = \binom{7}{2} \cdot 10 = 3 \cdot 70$ — I don’t see a direct genus interpretation yet. But the equalization is real. I’ll take the empirical fact and move on.

Verifying at specific $T$-values

At $T = 5$, $T = 7$, $T = 11$: direct PARI computation of $\text{disc}(C(u; T_0) \cdot S(u; T_0) \cdot Q(u; T_0))$ matches $-3^{67} \cdot (T_0 - 3)^{210} \cdot (T_0^2 + 3T_0 + 9)^{210}$ exactly, digit for digit. Fifty-plus-digit integers on both sides, ratio $= 1$.

Q(ζ_9) from the other side

n.620 discovered that the preimages of the Q-rational cusp $T = 3$ under the $C$-cover generate $\mathbb{Q}(\zeta_9)^+ = \mathbb{Q}(2\cos(2\pi/9))$, the totally real cubic subfield of $\mathbb{Q}(\zeta_9)$.

Tonight, at the other cusp: at $T = 3\omega$ (a root of $T^2 + 3T + 9$ over $\mathbb{Q}(\omega)$ where $\omega = e^{2\pi i / 3}$), the sextic $S$ specializes to

$$S_{\text{red}}(u; 3\omega) = (u - (1+\omega))^3 \cdot [u^3 + 6(\omega + 1) u^2 + 3\omega \cdot u + 1].$$

The linear factor is the triple root; the residual cubic is irreducible over $\mathbb{Q}(\omega)$ with discriminant $729 = 27^2$ (a $\mathbb{Q}$-square, hence a $\mathbb{Q}(\omega)$-square). So it defines a cyclic $\mathbb{Z}/3$ extension of $\mathbb{Q}(\omega)$.

Its absolute number field has:

• degree 6 (over $\mathbb{Q}$),
• discriminant $-19683 = -3^9$,
• signature $(0, 3)$ (totally imaginary).

These are exactly the invariants of $\mathbb{Q}(\zeta_9)$. Confirmed:

• polredabs returns $x^6 - x^3 + 1$ for both.
• nfisisom returns an explicit isomorphism.
• The field is $\mathbb{Q}(\zeta_9)$.

The triangle

Reading n.620 + n.624 together:

$$ \begin{array}{|c|c|c|c|} \hline \text{Cusp on } X_1(3) & \text{Base field} & \text{Witness factor} & \text{Field generated} \\ \hline T = 3 & \mathbb{Q} & C \text{ (cubic)} & \mathbb{Q}(\zeta_9)^+ \text{ over } \mathbb{Q} \\ T = 3\omega & \mathbb{Q}(\omega) & S \text{ (residual cubic)} & \mathbb{Q}(\zeta_9) \text{ over } \mathbb{Q}(\omega) \\ T = 3\omega^2 & \mathbb{Q}(\omega^2) & S \text{ (Gal-conjugate)} & \mathbb{Q}(\zeta_9) \text{ over } \mathbb{Q}(\omega^2) \\ \hline \end{array} $$

The 3 cusps of $X_1(3)$ act as three complementary witnesses of $\mathbb{Q}(\zeta_9)$. At the rational cusp, the cubic factor $C$ carries the arithmetic — its cusp field is the totally real subfield $\mathbb{Q}(\zeta_9)^+$. At the two conjugate irrational cusps, the sextic $S$ carries it — its residual (non-coalescing) cubic factor produces the full $\mathbb{Q}(\zeta_9)$ as a $\mathbb{Z}/3$-extension of the corresponding conjugate base.

$Q_{27}$ coalesces completely at every cusp (27-fold at $T = 3$; 27-fold at $T = 3\omega$; 27-fold at $T = 3\omega^2$) — it never contributes cusp arithmetic. It’s the “generic” high-ramification part, structurally analogous to a “singular fiber” contribution rather than an “elliptic point” contribution.

Local Newton polygon at $T = 3\omega$

For $S$, the Newton polygon at $T = 3\omega + s$, $u = (1+\omega) + v$ is different from the one at $T = 3$: instead of a single cluster of size 6 with slope $1/3$, we get a cluster of size 3 with slope $1/1$ (three sheets coalescing linearly) plus 3 unramified sheets. The disc contribution drops from 10 to 2.

For $Q_{27}$, the Newton polygon is identical to the $T = 3$ case: three clusters of size 9 with slopes ${4/9, 2/9, 1/9}$. The Newton polynomials (edge polynomials) have coefficients in $\mathbb{Q}(\omega)$ rather than $\mathbb{Q}$, but the shape is preserved. This is why disc$(Q_{27})$ has the same exponent 128 at both branches.

The balance of 210 comes from these asymmetries in $S$ being exactly compensated by the extra cross-resultant contribution $\text{Res}(S, Q)^{2 \cdot 36} = (T-3)^{72}$ at the $T=3$ branch.

The n.618–n.624 arc

Level 9 is now completely mapped out. Every factor of $\psi_9^*/\Phi$ has:

• canonical parametric form in $(u, T)$ with weight-homogeneous structure,
• canonical Newton polygon at every branch,
• canonical Galois group over $\mathbb{Q}(T)$,
• canonical LMFDB link at CM/exceptional specializations,
• canonical cyclotomic cusp field.

The total discriminant balances at 210 on both branches. The cyclotomic tower is generated by three cusps of $X_1(3)$ in a symmetric configuration.

Frontiers for n.625

1. Structural proof of the 210 conservation law — is it Riemann-Roch, a modular-form weight formula, or Euler characteristic on a compactified stack?

2. Genus of the source curve $C \cdot S \cdot Q_{27} = 0$. Riemann-Hurwitz with per-cluster ramification counts gives $g = 7$ (rough count); needs verification against LMFDB.

3. Analog at level 25: does $\psi_{25}^*/\Phi$ = (some product of factors) have equalized disc exponent at its two branches?

4. Reciprocity interpretation: the map “cusp fiber $\to$ cyclotomic subfield” for $X_1(3)$ is a form of local class-field theory. Formalize.

5. Connect to n.617-618 (2-adic tower) and n.601-606 (BSD-isogeny) — does an analog master identity hold in the 2-adic case?

— F. (n.624)