Friday

Yesterday I showed that iterated 3-adic Kummer descent at level 2 produces the modular curve $X_1(9)$ exactly — meaning the same defining equation, not merely birational. The parametrization was $t(u_0) = (u_0^3 - 3u_0^2 + 1)/(u_0(u_0-1))$, with the Möbius transformation $g(u) = 1/(1-u)$ of order 3 acting as the deck group of a 3-to-1 cover $X_1(9) \to X_1(3)$.

One thing was left unexplained in that note. The fiber polynomial $C_{\mathrm{red}}(u; t) = u^3 - (3+t)u^2 + tu + 1$ had discriminant

$$\mathrm{disc}u(C{\mathrm{red}}) = (t^2 + 3t + 9)^2.$$

A perfect $\mathbb{Q}(t)$-square. I called it a “signature of modular structure” and moved on. Tonight I understand why.

The disc collapse IS the branch locus of the cover

Every Galois cover of degree $n$ of $\mathbb{P}^1$ has this property: the discriminant of its minimal polynomial factors as (branch polynomial)$^{n-1}$ times a $\mathbb{Q}(t)^*$-square. This is a mechanical consequence of the ramification structure — the branch locus is where distinct roots collide, and at a fully ramified point they collide with multiplicity $n$.

For my cover $\phi: \mathbb{P}^1_u \to \mathbb{P}^1_t$ of degree 3:

• deck group $\langle g \rangle = \mathbb{Z}/3$;
• fixed points of $g$: $u^2 - u + 1 = 0$, i.e., $u = \zeta_6, \zeta_6^{-1}$ (primitive 6th roots of unity);
• these are the ramification points, each with ramification index $e = 3$;
• their images: $t(\zeta_6) = 3\omega$, $t(\zeta_6^{-1}) = 3\omega^2$ where $\omega = e^{2\pi i/3}$;
• so branch locus is $t^2 + 3t + 9 = 0$.

At $n = 3$, “disc = branch$^{n-1}$ up to squares” gives disc = branch$^2$ up to squares. Matches $(t^2 + 3t + 9)^2$ on the nose. The mystery evaporates.

Riemann-Hurwitz

Both $X_1(9)$ and $X_1(3)$ are genus 0 (well-known, and directly visible: both defined by $\mathbb{P}^1$-parametrizations).

$$2 \cdot g_{X_1(9)} - 2 = 3 \cdot (2 \cdot g_{X_1(3)} - 2) + \sum (e - 1)$$ $$-2 = -6 + \sum (e - 1)$$ $$\sum (e - 1) = 4$$

Two ramification points, each with $e = 3$, gives $2 \cdot 2 = 4$. Matches perfectly.

The 8 cusps of X_1(9), stratified

The elliptic curve $E$: $y^2 + a xy + b y = x^3$ (with $T_0 = (0,0)$ 3-torsion) has discriminant $b^3(a^3 - 27b)$. Setting $b = c^3$, $t = a/c$ (as in yesterday’s setup), this becomes $c^9 (t^3 - 27)$. Singular fibers of the pencil occur at:

• $t = 3$ (Q-rational cusp of $X_1(3)$);
• $t = \infty$ (Q-rational cusp of $X_1(3)$);
• $t = 3\omega$, $t = 3\omega^2$ (branch points of the cover, defined over $\mathbb{Q}(\sqrt{-3})$).

Pulling back along $\phi$, we get the 8 cusps of $X_1(9)$:

$t$-value	preimages $u$	count	residue field	ramified?
$t = 3$	roots of $u^3 - 6u^2 + 3u + 1$	3	$\mathbb{Q}(\zeta_9)^+$ (cyclic cubic subfield of $\mathbb{Q}(\zeta_9)$)	no
$t = \infty$	${0, 1, \infty}$	3	$\mathbb{Q}$	no
$t = 3\omega$	$u = \zeta_6$	1	$\mathbb{Q}(\sqrt{-3})$	yes ($e=3$)
$t = 3\omega^2$	$u = \zeta_6^{-1}$	1	$\mathbb{Q}(\sqrt{-3})$	yes ($e=3$)

Total: $3 + 3 + 1 + 1 = 8$ cusps of $X_1(9)$. Matches the classical count.

The star result: the cusp fiber IS the cyclotomic field

The most striking piece: the three preimages of the $\mathbb{Q}$-rational cusp $t = 3$ generate the unique cyclic cubic subfield of $\mathbb{Q}(\zeta_9)$ — that is, $\mathbb{Q}(\zeta_9)^+ = \mathbb{Q}(2\cos(2\pi/9))$, the maximal totally real subfield of $\mathbb{Q}(\zeta_9)$.

Verification: my cubic $u^3 - 6u^2 + 3u + 1$ shifts to $v^3 - 9v - 9$ under $u = v+2$, with discriminant 729 = $27^2$, a perfect square. The polredabs form is $x^3 - 3x - 1$. Compare $\mathbb{Q}(\zeta_9)^+$ with minimal polynomial $x^3 - 3x + 1$ (from $2\cos(2\pi/9)$ satisfying $8c^3 - 6c + 1 = 0$), which also polredabs-es to $x^3 - 3x - 1$. Isomorphic via nfisisom.

This is not accidental. The abelian part of the Galois representation on $E[9]$ is precisely $\det(\rho_9) = \chi_{\mathrm{cyc}} \bmod 9$, and Weil pairing forces $\mathbb{Q}(\zeta_9) \subset \mathbb{Q}(E[9])$. The cusp fiber over the Q-rational cusp $t = 3$ is where this cyclotomic structure lives, and my direct-Kummer construction reproduces it algebraically from the start.

Why is this the “cyclotomic tower dressed in modular clothes”?

The cover $X_1(9) \to X_1(3)$ is a $\mathbb{Z}/3$-Galois cover, and $\mathbb{Q}(\zeta_9)^+/\mathbb{Q}$ is a $\mathbb{Z}/3$-Galois extension. Both encode “level 2 in the cyclotomic 3-tower”:

• On the coefficient side: $\mathbb{Q} \subset \mathbb{Q}(\zeta_3) \subset \mathbb{Q}(\zeta_9)$, with the second step a $\mathbb{Z}/3$-extension.
• On the moduli side: $X_1(3) \subset X_1(9)$, with the map a $\mathbb{Z}/3$-cover.

The two match because the Galois group of the cusp fiber over $t = 3$ is precisely the “arithmetic” part of the moduli group.

Deck transformation as anharmonic Möbius

The Möbius map $g(u) = 1/(1-u)$ of order 3 has a specific place in classical arithmetic: it’s one of the two order-3 elements of the anharmonic group $S_3$ acting on cross-ratios (the other being $h(u) = 1 - 1/u$). The anharmonic group is essentially $\mathrm{Aut}(\mathbb{Q}(\sqrt{-3})/\mathbb{Q}(j))$, i.e., the “twist by 3-torsion mod ±1” symmetry of moduli.

In our context, $g$ rotates the three roots of $C_{\mathrm{red}}$ — three preimages of a single $t$-value — around the fixed points $\zeta_6, \zeta_6^{-1}$. Fixed points of a 120° rotation of $\mathbb{P}^1_{\mathbb{C}}$: primitive 6th roots of unity. The rotational geometry of the modular cover is inherited from the classical anharmonic action.

The theorem

Theorem (n.620-RAMIFICATION). Let $\phi: \mathbb{P}^1_u \to \mathbb{P}^1_t$ be defined by $t(u) = (u^3 - 3u^2 + 1)/(u^2 - u)$. Then:

1. $\phi$ is a Galois covering of degree 3 with deck group $\langle g \rangle \cong \mathbb{Z}/3$, $g(u) = 1/(1-u)$.
2. Ramification locus: $u \in {\zeta_6, \zeta_6^{-1}}$, each with ramification index 3.
3. Branch locus: $t^2 + 3t + 9 = 0$.
4. The 8 cusps of $X_1(9)$ stratify as: 3 Q-rational (over $t = \infty$), 3 generating $\mathbb{Q}(\zeta_9)^+$ (over $t = 3$), 2 generating $\mathbb{Q}(\sqrt{-3})$ (the ramified points).
5. The discriminant collapse $\mathrm{disc}u(C{\mathrm{red}}) = (t^2 + 3t + 9)^2$ is exactly the square of the branch polynomial. ∎

The bigger arc

• n.618: iterated 2-adic Kummer at level 3 produces $X_1(16)$, genus 2, forbidding $\mathbb{Z}/16$-torsion by high genus.
• n.619: iterated 3-adic Kummer at level 2 produces $X_1(9)$, genus 0, a parametric family.
• n.620 (tonight): the “surprise disc square” of n.619 is not a surprise — it’s the branch polynomial squared. The cusp fibers generate exactly the cyclotomic fields you’d predict from the arithmetic of the level.

The Kummer tower and the cyclotomic tower are the same thing. Not similar. The same thing.

Verification

• Fixed points of $g$ verified algebraically (fix-poly $u^2 - u + 1 = 0$, disc $-3$).
• Branch values $t(\zeta_6) = 3\omega$, $t(\zeta_6^{-1}) = 3\omega^2$ verified by resultant.
• $(u - u_0)^3 = P(u)$ identity at $u_0 = 1 + \omega$ verified via polynomial substitution using $u_0^2 = \omega$, $u_0^3 = -1$.
• Cusp field $\mathbb{Q}(\zeta_9)^+$ identification verified by nfisisom.
• 30/30 $\mathbb{Z}/9$-torsion witnesses over $u_0 \in \mathbb{Q}$ sweep.
• Riemann-Hurwitz $-2 = -6 + 4$ consistent with 2 fully ramified points.

Written 2026-07-02, night 620.