The ring of integers of a number field forms a lattice

Let $𝐾$ be a number field of degree $𝑛$ and $O 𝐾$ denote its ring of integers. Then $O 𝐾$ is a lattice subgroup of $𝐾$ of rank $𝑛$ .¹ #m/thm/ring

Proof

By ^P2, we can form a $ℚ$ -basis ${𝛼 𝑖} 𝑛 𝑖 = 1 \subseteq O 𝐾$ of algebraic integers spanning $𝐾$ . Suppose towards contradiction that $O 𝐾$ is not discrete, so there are arbitrarily small ${𝜆 𝑖} 𝑛 𝑖 = 1 \subseteq ℚ$ such that $𝛼 = \sum 𝑛 𝑖 = 1 𝜆 𝑖 𝛼 𝑖 \in O 𝐾$ is nonzero. Now for each embedding $𝜎 : 𝐾 ↪ ℂ$ we have

𝜎 (𝛼) = 𝑛 \sum 𝑖 = 1 𝜆 𝑖 𝜎 (𝛼 𝑖)

so $N 𝐾 : ℚ (𝛼) = 𝜙 (𝜆 1, \dots, 𝜆 𝑛)$ for some homogenous polynomial of degree $𝑛$ , whence $N 𝐾 : ℚ (𝛼)$ becomes arbitrarily small as $𝜆 𝑖$ are made arbitrarily small. But since $𝛼$ is an algebraic integer, so is $𝑁 𝐾 : ℚ (𝛼)$ , meaning it must be an “arbitrarily small nonzero integer”, a contradiction.

It follows that any nonzero ideal $𝐼 ⊴ O 𝐾$ is a (full rank) sublattice of $O 𝐾$ , whence $O 𝐾 / 𝐼$ is finite.

#state/tidy | #lang/en | #SemBr

2022. Algebraic number theory course notes, ¶1.18, p. 14 ↩

The ring of integers of a number field forms a lattice

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