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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-29/1/ii/solution

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Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 29 / 1 / ii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

The given relation implies

π (1 - ϕ_{1}) = (1 + ϕ_{2}) π .

Take isogeny degrees and cancel the positive integer

de g π

. Multiplicativity of isogeny degree gives

de g (1 - ϕ_{1}) = de g (1 + ϕ_{2}) .

As above,

de g (1 - ϕ_{i}) = # E_{i} (F_{q})

and

de g ϕ_{i} = q

. The degree parallelogram law on

End (E_{2})

now yields

de g (1 + ϕ_{2}) + de g (1 - ϕ_{2}) = 2 de g 1 + 2 de g ϕ_{2} = 2 (q + 1) .

Substitution proves

# E_{1} (F_{q}) + # E_{2} (F_{q}) = 2 (q + 1) .

Equivalently, their Frobenius traces are negatives of one another. The isogeny degree argument works over the algebraic closure, so it does not wrongly assume that

π

is defined over

F_{q}

; the stated quadratic-extension hypothesis is compatible with the anticommutation relation.

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