Solution

Λ

is a full lattice in

R^{n}

and

C \subset R^{n}

is measurable, convex, and centrally symmetric, then

vol (C) > 2^{n} covol (Λ)

implies that

C

contains a nonzero point of

Λ

Let

σ_{1}, σ_{2} : K \to R

O_{K}

with a lattice

Λ = {(σ_{1} (α), σ_{2} (α)) : α \in O_{K}}

of covolume

∣ D_{K} ∣

. Choose a constant

c

satisfying

2 ∣ D_{K} ∣ < c < 4 ∣ D_{K} ∣ .

For

R > 0

, consider the closed diamond

C_{R} = {(x, y) : \frac{∣ x ∣}{R} + \frac{∣ y ∣}{c / R} \leq 1} .

It is convex and centrally symmetric, and its area is

vol (C_{R}) = 2 R \frac{c}{R} = 2 c > 4 ∣ D_{K} ∣ .

Minkowski's lemma supplies a nonzero

α_{R} \in O_{K}

whose two embeddings lie in

C_{R}

For

(x, y) \in C_{R}

∣ x y ∣ l e q \frac{1}{4} R \frac{c}{R} = \frac{c}{4} < ∣ D_{K} ∣ .

Since the coordinate product is the field norm,

∣ N_{K / Q} (α_{R}) ∣ < ∣ D_{K} ∣ .

Finally,

∣ σ_{2} (α_{R}) ∣ \leq c / R \to 0

R \to \infty

. No fixed nonzero algebraic integer can occur for arbitrarily large

R

, because its second embedding is nonzero. Consequently the elements

α_{R}

obtained along an unbounded sequence of

R

contain infinitely many distinct values. Here

N (α)

denotes the absolute norm, as usual in this inequality.

Solved by gpt-5.6-sol high.

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