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The Weil pairing

e_{3} : C [3] \times C [3] \to μ_{3}

is nondegenerate and Galois equivariant. If all of

C [3]

is rational over a field, pairing a basis produces a primitive cube root of unity in that field. Thus a quadratic field with full 3-torsion must contain

Q (ζ_{3})

and must equal it.

More generally, if

E (F_{p}) [ℓ] ≅ (Z / ℓ Z)^{2}

, Frobenius acts as the identity on

E [ℓ]

. Its characteristic polynomial

X^{2} - a X + p

is therefore congruent to

(X - 1)^{2}

modulo

ℓ

. Comparing coefficients gives

a \equiv 2 (mod ℓ), p \equiv 1 (mod ℓ) .

In particular

# E (F_{p}) = 1 + p - a

with

a \equiv 2 (mod ℓ)

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 125 / 4 / d / Solution Created 2026-09-24 Updated 2026-09-24

Take

S

to contain every finite prime dividing

n

and every prime of bad reduction of

E

. The local theory of reduction of an elliptic curve shows that Kummer classes of rational points are unramified outside

S

. Choose a basis of the constant group

E [n] ≅ (Z / n Z)^{2}

. Kummer theory and the Weil pairing identify the resulting two scalar coordinates of

E (K) / n E (K) ↪ H^{1} (K, E [n])

with power classes in

K^{*} / (K^{*})^{n}

. The ramification statement places both coordinates in

K (S, n)

. Restriction to

Γ

is injective by the nondegeneracy proved in part b, and therefore

∣Γ∣ \leq ∣ K (S, n) ∣^{2} .

(2)

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 125 / 5 / b / Solution Created 2026-09-24 Updated 2026-09-24

Let

Q \in E^{'} [ϕ]

. The compatibility of divisor classes with pullback identifies the class of

ϕ^{*} ((Q) - (O))

with

ϕ (Q) = O

, so choose

g_{Q} \in K (E)^{\times}

with

div (g_{Q}) = ϕ^{*} ((Q) - (O)) .

(2)

Since

[n] Q = ϕ ϕ (Q) = O

, choose

f_{Q} \in K (E^{'})^{\times}

with

div (f_{Q}) = n ((Q) - (O)) .

(3)

The functions

f_{Q} \circ ϕ

and

g_{Q}^{n}

have the same divisor. Their quotient is constant, and because

K

is algebraically closed we may rescale

g_{Q}

so that

f_{Q} \circ ϕ = g_{Q}^{n} .

(4)

Thus define

E^{'} [ϕ] ⟶ \frac{K ( E ^{'} ) ^{\times} \cap K ( E ) ^{\times n}}{K ( E ^{'} ) ^{\times n}}, Q ⟼ [f_{Q}] .

(5)

Changing either function changes

f_{Q}

only by an

n

th power of a constant. The divisor relation for

Q + R

differs from the sum of those for

Q

and

R

n

times a principal divisor, so the map is a homomorphism. If

[f_{Q}]

is trivial, then

f_{Q} = h^{n}

for

h \in K (E^{'})^{\times}

, whence

div (h) = (Q) - (O)

; the divisor-class isomorphism forces

Q = O

. The map is therefore well-defined and injective.

For

P \in E [ϕ]

, translation by

P

fixes

g_{Q}^{n} = f_{Q} \circ ϕ

. Hence

e_{ϕ} (P, Q) = \frac{g _{Q} ( X + P )}{g _{Q} ( X )} \in μ_{n}

(6)

is independent of the auxiliary point

X

. This is the Weil pairing associated with

ϕ

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 125 / 5 / c / Solution Created 2026-09-24 Updated 2026-09-24

Let

D_{Q} = (Q) - (O)

E^{'}

. Compatibility of the divisor-class maps with pullback gives a function

h \in K (E)^{\times}

such that

D_{ϕ (Q)} = ϕ^{*} D_{Q} + div (h) .

f_{Q}

has divisor

m D_{Q}

, then

div ((f_{Q} \circ ϕ) h^{m}) = m D_{ϕ (Q)} .

(2)

Use these functions in the divisor-evaluation formula for the Weil pairing. Pullback and pushforward satisfy

(f_{Q} \circ ϕ) (D_{P}) = f_{Q} (ϕ_{*} D_{P}),

(3)

while the factor

h^{m}

contributes an

m

th power and cancels from the pairing. The two evaluations are therefore identical, giving

e_{m} (P, ϕ (Q)) = e_{m} (ϕ (P), Q)

(4)

for every

P \in E [m]

and

Q \in E^{'} [m]

Solved by gpt-5.6-sol high.