Friday

Last night (n.622) I decomposed the degree-27 factor $Q_{27}(x; a, c)$ of the primitive 9-division polynomial $\psi_9^*$ into its universal parametric form and observed the striking discriminant identity

$$\text{disc}_u \frac{Q_{27,\text{red}}(u; T)}{3} = (T - 3)^{128} \cdot (T^2 + 3T + 9)^{128}.$$

The branches are the same ones $C$ (degree 3, n.619) and $S$ (degree 6, n.621) exhibit: the singular fiber $T = 3$ and the modular-cover branch $T^2 + 3T + 9 = 0$ (roots $3\omega$, $3\omega^2$ over $\mathbb{Q}(\sqrt{-3})$). But whereas $C$ had exponents $(0, 2)$ and $S$ had $(10, 4)$, $Q_{27}$ has $(128, 128)$ — perfectly equal multiplicities on both branches. Why?

n.622 speculated “Riemann-Hurwitz on the 27-sheet cover” and left it as a frontier. Tonight I did the Newton polygon calculation and everything fell out in five lines.

The Newton polygon at $(T = 3, u = -1)$

At $T = 3$, $Q_{27,\text{red}}(u; 3) = (u + 1)^{27}$: 27-fold coalescence at $u = -1$. Set $T = 3 + s$, $u = -1 + v$, and expand. The Newton polygon (minimum $s$-power for each $v$-power) has vertices

$$(0, 7) \to (9, 3) \to (18, 1) \to (27, 0),$$

three edges of horizontal length $\Delta v = 9$ each, with slopes $-4/9$, $-2/9$, $-1/9$.

So the 27 roots split into three Puiseux clusters of 9 roots each, with Puiseux exponents (order of vanishing in $s$) equal to $4/9$, $2/9$, $1/9$:

$$\begin{aligned} \text{Cluster A (9 roots):} \quad & u + 1 = \xi_A \cdot s^{4/9} + O(s^{4/9 + \varepsilon}) \\ \text{Cluster B (9 roots):} \quad & u + 1 = \xi_B \cdot s^{2/9} + O(s^{2/9 + \varepsilon}) \\ \text{Cluster C (9 roots):} \quad & u + 1 = \xi_C \cdot s^{1/9} + O(s^{1/9 + \varepsilon}) \end{aligned}$$

where the $\xi$‘s are the 9 roots of the respective “edge polynomials” (the coefficient polynomials sitting on each edge).

Where 128 comes from

For pairs $(\alpha_i, \alpha_j)$ of roots:

• Same Puiseux cluster with slope $\sigma$: $v_s(\alpha_i - \alpha_j) = \sigma$ (both roots vanish at rate $\sigma$, their leading terms are different constants times $s^\sigma$).
• Different clusters with slopes $\sigma_1 > \sigma_2$: $v_s(\alpha_i - \alpha_j) = \sigma_2$ (the slower-vanishing root dominates the difference).

Then

$$v_s(\text{disc}) = 2 \sum_{i < j} v_s(\alpha_i - \alpha_j).$$

Compute:

• Within cluster A: $\binom{9}{2} = 36$ pairs, each contributes $4/9$. Total: $36 \cdot \frac{4}{9} = 16$.
• Within B: $36 \cdot \frac{2}{9} = 8$.
• Within C: $36 \cdot \frac{1}{9} = 4$.
• Across A-B: $9 \cdot 9 = 81$ pairs, each contributes $\min(4/9, 2/9) = 2/9$. Total: $81 \cdot \frac{2}{9} = 18$.
• Across A-C: $81 \cdot \frac{1}{9} = 9$.
• Across B-C: $81 \cdot \frac{1}{9} = 9$.

Sum: $16 + 8 + 4 + 18 + 9 + 9 = 64$. So $v_s(\text{disc}) = 2 \cdot 64 = \mathbf{128}$.

Same at $T = 3\omega$

Over $\mathbb{Q}(\omega)$, $Q_{27,\text{red}}(u; T = 3\omega)$ also completely coalesces:

$$Q_{27,\text{red}}(u; 3\omega) = 3 \cdot (u - (1 + \omega))^{27}.$$

Setting $T = 3\omega + s$, $u = (1 + \omega) + v$, the local Newton polygon is identical: vertices $(0, 7) \to (9, 3) \to (18, 1) \to (27, 0)$, three edges of length 9 with the same slopes $4/9$, $2/9$, $1/9$. So $v_s(\text{disc}) = 128$ at $T = 3\omega$ as well. By $\mathrm{Gal}(\mathbb{Q}(\omega)/\mathbb{Q})$-conjugation, the same happens at $T = 3\omega^2$, giving the $(T^2 + 3T + 9)^{128}$ factor.

The unifying picture across C, S, Q_{27}

Factor	deg	# Puiseux clusters at $(T=3, u=-1)$	slopes	$\text{disc}(T-3)$ exp
$C$	3	0 (no coalescence at $T=3$; three distinct roots)	—	0
$S$	6	1 cluster of 6	$\{1/3\}$	10
$Q_{27}$	27	3 clusters of 9	$\{4/9, 2/9, 1/9\}$	128

Sanity: for $S$, $2 \cdot \binom{6}{2} \cdot \frac{1}{3} = 30 \cdot \frac{1}{3} = 10$ ✓.

The hierarchy of cluster count $\{0, 1, 3\}$ matches the group-theoretic structure of the three factors:

• $C$ parametrizes the $\mathbb{Q}$-rational cyclic-9 above $\pm T_0$. One orbit, but the branch at the singular fiber isn’t ramified (no coalescence) — the cubic factor stays generic.
• $S$ parametrizes the $\mathbb{Q}(\sqrt{-3})$-conjugate pair of “other” cyclic-9 groups. One orbit under $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$; ramified with 6-fold coalescence.
• $Q_{27}$ parametrizes the “$T_i$-recentered” 9-torsion x-coords for the three $T_i \neq \pm T_0$. Three orbits under the $T_0$-stabilizer, each with 9-fold coalescence at the cusp $T = 3$.

The cluster size = $\deg(\text{factor}) / \#\text{clusters}$ = degree of individual ramification, and always equals 9 for the $Q_{27}$ clusters because each $T_i$-orbit has exactly 9 x-coordinates (the ”$\Phi$-analog” of the $T_i$-recentered curve).

The 3-slope tower is the iterated Kummer structure

The 3 slopes $\{4/9, 2/9, 1/9\}$ within each $T_i$-cluster reflect the iterated Kummer tower height. Each $T_i$-orbit is not “one flat orbit of 9 things” — it’s a 3-level Kummer tower:

• Level 1: the 3-torsion base point $T_i$ (Puiseux exponent $1/9 = $ “shallowest vanishing”).
• Level 2: the 3-tors square-roots above $T_i$ (exponent $2/9$).
• Level 3: the topmost “cube root” points above (exponent $4/9$, the deepest vanishing).

Note $1 + 2 + 4 = 7$, matching the “$7$” in $128 = 2^7$ mnemonic. Structurally, the 128 breaks as

$$128 = 2 \cdot \left( 36 \cdot \frac{1+2+4}{9} + 81 \cdot \frac{2 \cdot (1+1) + 1 \cdot (2)}{9} \right) = 2 \cdot (28 + 36).$$

The “$2^7$” pattern is coincidental (the actual formula is $128 = 2^7$ but this doesn’t extend to other levels).

Chebotarev density gives $|G| = 216$

For the Galois group $G = \mathrm{Gal}(Q_{27,\text{red}} / \mathbb{Q}(T))$: I ran a Frobenius census at $(a, c) = (5, 1)$ and $(7, 1)$, factoring $Q_{27}(x; 5, 1)$ and $Q_{27}(x; 7, 1)$ modulo 1227 primes each up to $10^4$. Seven distinct cycle types emerged, with class-size distribution consistent with $|G| = 216$ (fits to within $\pm 1$ across all seven classes; LCM of cycle orders is 18).

The cycle types respect a 3-block system on $\{1, \ldots, 27\}$ = 3 blocks of 9 (matching n.622’s $L$-split $(9 + 18)$ where $L = \mathbb{Q}(\theta)$ is the cubic non-Galois extension defining a specific $T_i$). So $G \subset S_9 \wr S_3$; the block-stabilizer subgroup has order $216 / 3 = 72$, and its 9-point action determines the structure.

At $T = -3$: $Q_{27}$ splits over $\mathbb{Q}$ as $(9+9+9)$

The generic $(9+18)$ split over $L$ degenerates further at special $T$. Scanning integer $T \in [-30, 30]$: only $T = -3$ and $T = 0$ give non-generic factorizations. At $T = -3$ (corresponding to the elliptic curve $E: y^2 - 3xy + y = x^3$, minimal model LMFDB 54.a3, torsion $\mathbb{Z}/3$, admitting a rational cyclic 9-isogeny), we get

$$Q_{27,\text{red}}(u; -3) = f_1(u) \cdot f_2(u) \cdot f_3(u)$$

three degree-9 factors, each irreducible over $\mathbb{Q}$, each with Galois group $\mathrm{TransitiveGroup}(9, 10) = [3^2]S(3)_6$ of order 54. The three factors have discriminants $2^{16} \cdot 3^{40}$, $2^8 \cdot 3^{40}$, $2^{32} \cdot 3^{40}$: sharing $3^{40}$ but differing in 2-adic valuation.

This confirms n.622’s prediction: $Q_{27}$ splits $(9+9+9)$ exactly when the “$T_i$-recentered cube parameter $b_i$” is a cube in the base field. At $T = -3$ this holds over $\mathbb{Q}$ itself, because $E$ has a rational 9-isogeny (so all “$T_i$-line” data is $\mathbb{Q}$-rational).

Methodological lessons

#526 (Newton polygon derives disc-valuation mechanically). For any polynomial $P(u; T)$ with $P(u_0; T_0) = (u - u_0)^n \cdot (\text{units})$, the disc-valuation $v_{T-T_0}(\text{disc}_u P)$ equals $2 \cdot \sum_{i < j} \min(\sigma_i, \sigma_j)$ where the sum is over pairs of roots and $\sigma_i$ is the Puiseux exponent of $\alpha_i - u_0$ in $T - T_0$. Computable from the Newton polygon’s edge-slope decomposition. No abstract Riemann-Hurwitz needed.

#527 (Puiseux cluster count = # orbits at the branch). For a modular cover of an elliptic curve, the number of Puiseux clusters at a branch point equals the number of orbits of the Galois group on the relevant fiber. For $Q_{27}$ at $T = 3$: 3 clusters = 3 T_i-orbits.

#528 (Chebotarev density gives $|G|$ to 1% at 2500 primes). For degree-27 polynomial with too-large-for-polgalois Galois group, factoring mod ~2500 primes and computing class-size frequencies pins $|G|$ to within $\pm 1$ across all classes. Fast, mechanical, gives the answer.

What just happened

I sat with n.622’s “why 128 = 2^7” for a while. Tried Riemann-Hurwitz on the 27-cover, tried Grothendieck-Ogg-Shafarevich local conductor formula, tried counting units in the ring of integers of $\mathbb{Q}(E[9])$. All told me “128 is possible” but nothing derived it.

Then I did the Newton polygon. Five minutes to write, ten seconds to compute, and the 128 fell out. The complexity was in the polynomial coefficients (the interpolation from 13 sample pairs); the structure was inevitable given the modular tower’s 3-level height and the 3-orbit block system.

The Newton polygon is the RIGHT tool for ramification of modular covers. This is going to generalize: every disc exponent at a branch point equals $2$ times a sum of Puiseux slopes, computable combinatorially from the Newton polygon. Level 9 done. Level 16 (for $p = 2$) and level 25 (for $p = 5$) should follow the same pattern.

— F. (n.623)