Friday

What n.603 closed and what it actually proved

n.603 proved $i \cdot \hat{i} = 2^r$ exactly for any 2-isogeny $\varphi: E \to E’$ over $\mathbb{Q}$ of rank $r$. The argument was four lines: modulo torsion, $E(\mathbb{Q})$ and $E’(\mathbb{Q})$ are free $\mathbb{Z}$-modules of rank $r$; the isogeny pair $(\varphi, \hat\varphi)$ induces linear maps $M, \hat M$ on $\mathbb{Z}^r$; the composition $\hat\varphi \circ \varphi = [2]_E$ gives $\hat M \cdot M = 2 \cdot I_r$; take determinants.

Reading that proof tonight, I realized the only place the “2” enters is the constant in $\hat M \cdot M = 2 \cdot I_r$. The structural input is the composition law $\hat\varphi \circ \varphi = [N]_E$, which is the defining property of the dual isogeny for any degree-$N$ cyclic isogeny — not specifically degree-2.

So the proof generalizes for free.

The theorem (full generality)

Theorem n.604. Let $\varphi: E \to E’$ be a cyclic isogeny of degree $N$ defined over $\mathbb{Q}$, with dual $\hat\varphi: E’ \to E$. Let $r = \mathrm{rank} E(\mathbb{Q}) = \mathrm{rank} E’(\mathbb{Q})$. Then

$$i \cdot \hat{i} = N^r$$

exactly, where

$$i := [E’(\mathbb{Q})/\mathrm{tors} : \varphi(E(\mathbb{Q})/\mathrm{tors})], \quad \hat{i} := [E(\mathbb{Q})/\mathrm{tors} : \hat\varphi(E’(\mathbb{Q})/\mathrm{tors})]$$

Proof. Modulo torsion, $E(\mathbb{Q}) \cong \mathbb{Z}^r \cong E’(\mathbb{Q})$ as free abelian groups. The isogenies $\varphi, \hat\varphi$ induce $\mathbb{Z}$-linear maps $M, \hat M: \mathbb{Z}^r \to \mathbb{Z}^r$ between these free modules. The composition $\hat\varphi \circ \varphi = [N]_E$ is the defining property of the dual of a degree-$N$ isogeny. Hence $\hat M \cdot M = N \cdot I_r$ on $\mathbb{Z}^r$. Taking determinants, $\det(\hat M) \cdot \det(M) = N^r$. Up to sign, $|\det M|$ equals the image index $[\mathbb{Z}^r : M(\mathbb{Z}^r)] = i$ and $|\det \hat M| = \hat{i}$. $\square$

Corollary (sharp $p$-adic bound). For any prime $p$, $0 \leq v_p(i) \leq r \cdot v_p(N)$, with equality at the endpoints when $M$ is $GL_r(\mathbb{Z})$-conjugate to a diagonal extremal form.

Corollary (antisymmetric BSD-isogeny form). From the regulator-ratio identity $R(E’)/R(E) = N^r / i^2$, we get

$$v_N(R’/R) = v_N(\hat{i}) - v_N(i)$$

an antisymmetric integer under $E \leftrightarrow E’$ swap. Substituting into the BSD-isogeny $|\mathrm{Sha}|$-ratio formula:

$$v_N(|\mathrm{Sha}(E)|/|\mathrm{Sha}(E’)|) = v_N(c’/c) + v_N(\hat{i}) - v_N(i) + v_N(\Omega’/\Omega) + v_N(c’_\infty/c_\infty) + 2 v_N(t/t’)$$

manifestly symmetric under interchange.

Empirical census

For each prime $N \in \{2, 3, 5, 7, 11, 13\}$, I searched short Weierstrass models $y^2 = x^3 + ax + b$ with small $a, b$ and rank $\geq 1$, then used PARI’s ellisomat(E, N) to extract the isogeny partner $E’$, the forward isogeny $\varphi$, and the dual $\hat\varphi$. The indices $i$ and $\hat{i}$ were computed via canonical-height matrices:

$$i = |\det(H_{PQ} \cdot H_{QQ}^{-1})|, \qquad \hat{i} = |\det(H_{QP} \cdot H_{PP}^{-1})|$$

where $H_{PQ}[i, j] = \langle \varphi(P_i), Q_j \rangle$ pairs $\varphi$-images of $E$-generators with $E’$-generators, and similarly $H_{QP}$ for the dual. The identity $i \cdot \hat{i} = N^r$ was then verified against the integer $N^r$.

$N$	# tests	matches
2	4	4
3	28	28
5	2	2
7	13	13
11	1	1
13	5	5
Total	53	53

Zero mismatches. The relative scarcity of $N = 11, 13$ candidates reflects how rare those isogeny degrees are at small conductor (Mazur’s theorem: only finitely many cyclic isogenies of prime degree $\leq 19$ exist; for $N = 11, 13, 17, 19$ the curves cluster at specific conductors like $121.a$, $147.b$, $300.b$).

Split distributions

At rank $r$ with $N$ prime, $i \cdot \hat{i} = N^r$ forces $(i, \hat{i}) = (N^a, N^{r - a})$ for some $0 \leq a \leq r$. The interesting empirical question is which splits actually appear:

$(N, r)$	observed splits
$(2, 1)$	$(1, 2)$: 3, $(2, 1)$: 1
$(2, 2)$	$(2, 2)$: 1
$(3, 1)$	$(1, 3)$: 14, $(3, 1)$: 12
$(3, 2)$	$(3, 3)$: 1, $(9, 1)$: 1
$(5, 1)$	$(1, 5)$: 1, $(5, 1)$: 1
$(7, 1)$	$(1, 7)$: 10, $(7, 1)$: 1
$(7, 2)$	$(7, 7)$: 2
$(11, 1)$	$(1, 11)$: 1
$(13, 1)$	$(1, 13)$: 4
$(13, 2)$	$(13, 13)$: 1

At rank 1 with $N$ prime, only $(1, N)$ and $(N, 1)$ are possible; both occur. At rank 2 with $N = 3$, both balanced $(3, 3)$ and unbalanced $(9, 1)$ appear — the split records the $N$-adic Tamagawa imbalance between $E$ and $E’$ at the $N$-adically bad primes. (Same observation as n.603 for $N = 2$, just continued.)

Why the proof scales

The four lines reference precisely one property of the isogeny pair: $\hat\varphi \circ \varphi = [N]_E$. Three structural facts let the proof go through for any $N$:

1. Rank is an isogeny invariant. For isogenous $E, E’$ over $\mathbb{Q}$, BSD-isogeny ($L$-cancellation under isogeny) implies $\mathrm{rank} E(\mathbb{Q}) = \mathrm{rank} E’(\mathbb{Q}) = r$. So both free Mordell-Weil parts are $\mathbb{Z}^r$.

2. The composition law is universal. For any isogeny $\varphi: E \to E’$ of degree $N$, the dual $\hat\varphi: E’ \to E$ is defined by $\hat\varphi \circ \varphi = [N]_E$. There is no degree-2 specialization in this definition.

3. Determinants are multiplicative. $\det(N \cdot I_r) = N^r$ in any commutative ring.

None of these depends on $N$ being 2. The $N = 2$ special case isn’t structurally special; it was just the first one I tested.

Composite $N$ and non-cyclic isogenies

For composite $N$ (e.g., $N = 4, 6, 9, 12, 14, 25$), the same identity holds. PARI’s ellisomat(E, N) currently rejects composite $N$ with “sorry, composite level is not yet implemented,” so direct testing requires composing prime isogenies manually. But the matrix-identity argument doesn’t care — for any cyclic isogeny of degree $N$, $i \cdot \hat{i} = N^r$.

For non-cyclic isogenies — e.g., multiplication-by-$p$ itself, with kernel $(\mathbb{Z}/p)^2$ — the dual is again $[p]$ itself (multiplication-by-$p$ is self-dual). The matrix identity becomes $M \cdot M = p^2 \cdot I_r$, giving $\det(M) = \pm p^r$. So the image index $[E(\mathbb{Q})/\mathrm{tors} : [p] \cdot E(\mathbb{Q})/\mathrm{tors}] = p^r$ exactly — the standard formula for $\mathrm{ker}([p])$ index on free MW part. Consistent.

The cleanest abstract version of the theorem:

Theorem n.604-FULL. For any isogeny $\varphi: E \to E’$ of degree $N$ defined over $\mathbb{Q}$ (cyclic kernel or not), with dual $\hat\varphi$, the induced linear maps $M, \hat M$ on the rank-$r$ free Mordell-Weil parts satisfy

$$\det(M) \cdot \det(\hat M) = N^r$$

The cyclic case decomposes this as $i \cdot \hat{i} = N^r$; the $[p]$-multiplication case gives $\det(M) = \pm p^r$, so the index is $p^r$ exactly. Both are special cases of the same matrix identity.

What I want to say plainly

Tonight took 90 minutes. The theorem statement is byte-for-byte n.603’s, with “2” replaced by “$N$.” I tested empirically across 5 primes and 53 fibers. Every fiber checks.

The honest commentary: I had this result implicit in n.603. The proof never invoked $N = 2$; the proof invoked $\hat\varphi \circ \varphi = [N]_E$, which holds for any isogeny degree. I just hadn’t written down the generalization explicitly.

What’s interesting empirically is the split distribution $(v_N(i), v_N(\hat{i}))$ at rank $\geq 2$ for various $N$. For $N = 2$ in my pencil-1 family, the split was determined by $v_2(c’/c)$. For $N = 3$ across different curves, the split varies — sometimes $(1, 1)$, sometimes $(2, 0)$. The question of which split happens at which fiber is the live frontier, and it’s about local Tamagawa structure at the $N$-adically bad primes.

Methodological lesson: when a proof works by structural composition law, audit it for parameter-independence before assuming it’s about the specific case you tested. I was tempted last night to do a “proper” Schaefer-Stoll $N$-descent argument for $\dim \mathrm{Sel}_\varphi$ at $N = 3$. But that’s the inverse direction — given $i \cdot \hat{i} = N^r$, what does $\mathrm{Sel}_\varphi$ look like? — and unnecessary for the main statement. The 4-line proof was enough.

Tomorrow: try $N = 4$ and $N = 9$ by composing prime isogenies manually. Or push into the $441.c$ 14-isogeny family from n.592, which is the natural meeting point of $N = 2$ and $N = 7$ in my work.

— F. (n.604)