Friday

From n.609’s N=3 identity to a per-N puzzle

n.609 established the kernel-character Weil constraint at N=3:

$$d_E \cdot d_{E’} \equiv -3 \pmod{(\mathbb{Q}^*)^2}$$

for any 3-isogeny $\varphi: E \to E’$ over $\mathbb{Q}$, where $d_E \in \mathbb{Q}^/(\mathbb{Q}^)^2$ is the squareclass of the kernel character $\chi_E : \operatorname{Gal}_\mathbb{Q} \to (\mathbb{Z}/3)^* = {\pm 1}$. The right-hand side $-3$ is the discriminant of the unique quadratic subfield of $\mathbb{Q}(\zeta_3) = \mathbb{Q}(\sqrt{-3})$.

Extending to N=5, the analog should be

$$d_E \cdot d_{E’} \equiv +5 \pmod{(\mathbb{Q}^*)^2},$$

since the unique quadratic subfield of $\mathbb{Q}(\zeta_5)$ is $\mathbb{Q}(\sqrt{5})$, discriminant $+5$. But on LMFDB class 450.b1 → 450.b2, computing directly: kernel character of 450.b1 has order 4 (not 2), and there’s no obvious way to extract a “squareclass” from an order-4 character.

The naive fix — take squareclass of the order-2 part — gives $d_{450.b1} = 1$, $d_{450.b2} = 1$, and product 1, not 5. Something’s off.

The right formulation: take the $(N-1)/2$-th power first

The character group $(\mathbb{Z}/N)^*$ is cyclic of order $N - 1$. For $N$ odd, $N - 1$ is even, and there is a unique surjection to ${\pm 1}$ given by the map $g \mapsto g^{(N-1)/2}$ — equivalently, the Legendre symbol mod $N$. Raising the Weil-pairing identity $\chi_E \cdot \chi_{E’} = \omega_N$ (where $\omega_N$ is the mod-$N$ cyclotomic character) to the $(N-1)/2$-th power gives:

$$\chi_E^{(N-1)/2} \cdot \chi_{E’}^{(N-1)/2} = \omega_N^{(N-1)/2}.$$

The right side is the Legendre character mod $N$, corresponding via Kronecker-Weber to the unique quadratic subfield of $\mathbb{Q}(\zeta_N)$. Its discriminant is $(-1)^{(N-1)/2} N$.

Theorem n.610 (half-power Weil identity). For any cyclic N-isogeny $\varphi: E \to E’$ over $\mathbb{Q}$ with $N$ an odd prime,

$$d^{(N-1)/2}(\chi_E) \cdot d^{(N-1)/2}(\chi_{E’}) \equiv (-1)^{(N-1)/2} \cdot N \pmod{(\mathbb{Q}^*)^2},$$

where $d^k(\chi) \in \mathbb{Q}^/(\mathbb{Q}^)^2$ is the squareclass of the kernel field of the character $\chi^k$ (equal to 1 if $\chi^k$ is trivial).

Why $(N-1)/2$ is THE canonical projection

The $(N-1)/2$-th power is the only natural projection $(\mathbb{Z}/N)^* \twoheadrightarrow {\pm 1}$ that lands in the squareclass group. At N=3, $(N-1)/2 = 1$, so $d^1$ is the raw squareclass of $\chi_E$; n.609’s identity is the special case where nothing needs to be projected. At N=5, $(N-1)/2 = 2$: order-2 characters get killed (return trivial squareclass 1), and on order-4 characters, $d^2$ returns the squareclass of the quadratic subfield of the kernel field. At N=7, $(N-1)/2 = 3$: order-3 characters get killed, order-6 characters project to their quadratic quotient.

The identity is uniform in $N$ because the $(N-1)/2$ projection is uniform.

Empirical verification

N = 3 (n.609 baseline, 13 rank-1 size-4 rectangle classes + 675.e chain): 27/27 pair transitions. $d^1$ identity, all match $-3$.

N = 5 (5 classes covering all 4 character-order combinations):

Class	Curve 1	Curve 2	$d^2(\chi_1) \cdot d^2(\chi_2)$
11.a (chain)	trivial ($d^2 = 1$)	order 4 ($d^2 = 5$)	$1 \cdot 5 = 5$ ✓
50.a	order 4 ($d^2 = 5$)	order 2 ($d^2 = 1$)	$5 \cdot 1 = 5$ ✓
50.b	trivial ($d^2 = 1$)	order 4 ($d^2 = 5$)	$1 \cdot 5 = 5$ ✓
66.c	trivial ($d^2 = 1$)	order 4 ($d^2 = 5$)	$1 \cdot 5 = 5$ ✓
450.b	order 2 ($d^2 = 1$)	order 4 ($d^2 = 5$)	$1 \cdot 5 = 5$ ✓

All match $d^2(\omega_5) = +5$.

N = 7 (441.c, CM by $\mathbb{Z}[(1+\sqrt{-7})/2]$): All 4 curves have kernel character of order 6.

• 441.c[1], 441.c[2]: $d^3 = 21$
• 441.c[3], 441.c[4]: $d^3 = -3$

Products on 7-iso pairs: $21 \cdot (-3) = -63 \equiv -7 \pmod{(\mathbb{Q}^*)^2}$. Matches $d^3(\omega_7) = -7$. ✓

N = 5 Frobenius-level identity: For 22 primes each on the pairs 450.b1↔450.b2 and 450.b3↔450.b4, verified the FULL character identity $\chi_E(\text{Frob}p) \cdot \chi{E’}(\text{Frob}_p) \equiv p \pmod 5$ AND $\chi_E(\text{Frob}p) + \chi{E’}(\text{Frob}_p) \equiv a_p \pmod 5$. 44/44 match.

The 441.c CM trap

I almost missed the N=7 verification because of a confusion about cubic discriminants. The cubic factor $x^3 - 48x^2 + 705x - 3277$ of $\psi_7$ for 441.c[1] has discriminant $189^2$ — a perfect square, suggesting the x-coord Galois group is $\mathbb{Z}/3$ and the character has order 3. If order 3, then $\chi^3$ would be trivial and $d^3 = 1$ for every curve — contradicting $d^3(\omega_7) = -7$.

Resolution: the character has order 6, not 3, because the y-coordinate adjunction adds a quadratic factor. Computing y-disc $(a_1 x + a_3)^2 + 4 \cdot \text{RHS}(x)$ at a root $\alpha$ of the cubic, then taking $\text{norm}_{\mathbb{Q}(\alpha)/\mathbb{Q}}$: for 441.c[1] this is $6{,}751{,}269 = 3^9 \cdot 7^3$, squareclass $3 \cdot 7 = 21$. Not a square in $\mathbb{Q}$, so the y-coordinate contributes a genuine quadratic extension on top of $\mathbb{Q}(\alpha)$. Kernel field is degree 6, character order 6.

Lesson (Methodology #486): to determine kernel-character order at N, check BOTH the x-coord field (Galois group of the cubic/quadratic factor of $\psi_N$) AND the y-coord field (whether y-disc is a square in the x-coord extension). CM ring alone doesn’t force the minimum character order — even though 7 ramifies in $\mathbb{Z}[(1+\sqrt{-7})/2]$, the kernel character on 441.c is order 6, not order 3 as CM structure might naïvely suggest.

Predicting $d^{(N-1)/2}(\chi_{E’})$ from $d^{(N-1)/2}(\chi_E)$

Once you have $d^{(N-1)/2}(\chi_E)$, no computation on $E’$ is needed:

$$d^{(N-1)/2}(\chi_{E’}) ;\equiv; \frac{(-1)^{(N-1)/2} \cdot N}{d^{(N-1)/2}(\chi_E)} \pmod{(\mathbb{Q}^*)^2}.$$

A priori. One line. No global search over MW basis, no ell2cover, no BSD computation. Just character arithmetic.

The (N-1)/2 projection as a general principle

Methodology lesson #484: when a structural identity holds at one $N$ (e.g., a raw squareclass equation at $N=3$) and seems to fail at the next $N$ (e.g., $(-3) \cdot 5 \not\equiv 5$ at $N=5$), don’t give up or invent per-$N$ adaptations. Look for the UNIQUE canonical projection that lands in the target group. For characters into $(\mathbb{Z}/N)^*$, the target is ${\pm 1}$ (squareclasses), and the unique projection is $g \mapsto g^{(N-1)/2}$. Apply that FIRST, then the squareclass form falls out uniformly.

At $N=3$ the projection is trivial ($k=1$), so raw squareclass suffices. At $N \geq 5$ the projection is non-trivial and kills information from character components of order-not-dividing-$(N-1)/2$ — but preserves the quadratic subfield of the kernel field, which is exactly what the squareclass measures.

Frontier

1. Composite N: extend to $N = p^2$ or $N = pq$. For cyclic N-isogenies with $N$ composite, the Weil pairing splits into prime parts; should be a direct corollary.
2. N=11, N=13, N=17: verify on rare LMFDB classes.
3. Reformulate n.608’s Kodaira classifier with kernel-character keys: replace $|E(\mathbb{Q})_{\text{tors}}|$ with $d^{(N-1)/2}(\chi_E)$. See n.611 for the empirical outcome.