Absolute norm of an ideal of $O 𝐾$

Let $O 𝐾$ be the ring of integers of a number field $𝐾$ and let $𝔞 ◃ O 𝐾$ be an ideal. Then the absolute norm $N (𝔞)$ of $𝔞$ is given by the Lagrange index ring

N (𝔞) : = | O 𝐾 / 𝔞 |,

except in the case $𝔞 = ⟨ 0 ⟩$ , where we define $N (⟨ 0 ⟩) : = 0$ .

Properties

If $𝔞 = ⟨ 𝛼 ⟩$ is a principal ideal then $N (𝔞) = | N (𝛼) |$ , where the latter is the field norm.
$N (𝔞) N (𝔟) = N (𝔞 𝔟)$ .
For any $𝑚 \in ℕ 0$ , the number of ideals $𝔞 ⊴ O 𝐾$ such that $N (𝔞) = 𝑚$ is finite.

Proof of 1–3

Let ${𝜔 𝑖} 𝑛 𝑖 = 1$ $ℤ$ -span $O 𝐾$ , whence ${𝛼 𝜔 𝑖} 𝑛 𝑖 = 1$ $ℤ$ -spans $𝔞$ . Now
$⎡ ⎢ ⎢ ⎣ 𝛼 𝜔 1 ⋮ 𝛼 𝜔 𝑛 ⎤ ⎥ ⎥ ⎦ = 𝐴 ⎡ ⎢ ⎢ ⎣ 𝜔 1 ⋮ 𝜔 𝑛 ⎤ ⎥ ⎥ ⎦$
for some $𝐴 \in G L 𝑛 (ℤ)$ , so by Subgroup of a free abelian group we have $N (𝔞) = | d e t 𝐴 |$

Now for the discriminant we have
$Δ 𝐾 : ℚ (𝛼 𝜔 1, \dots, 𝛼 𝜔 𝑛) = (d e t 𝐴) 2 Δ 𝐾 : ℚ (𝜔 1, \dots, 𝜔 𝑛) = N (𝛼) 2 Δ 𝐾 : ℚ (𝜔 1, \dots, 𝜔 𝑛)$
by basic properties of the determinant and the definition of the discriminant.

Let $𝔞, 𝔟 ⊴ O 𝐾$ . Invoking UFI, we have $𝔞 = 𝔭 𝑠 1 1 \dots 𝔭 𝑠 𝑚 𝑚$ and $𝔟 = 𝔮 𝑡 1 1 \dots 𝔮 𝑡 𝑛 𝑛$ . By the Chinese remainder theorem for rings,
$O 𝐾 𝔞 𝔟 ≅ 𝑚 ⨁ 𝑖 = 1 𝑅 𝔭 𝑠 𝑖 𝑖 \oplus 𝑛 ⨁ 𝑖 = 1 𝑅 𝔮 𝑡 𝑖 𝑖 .$
Thus it suffices to show that for any prime ideal $𝔭$ we have
$N (𝔭 𝑛) = ∣ O 𝐾 𝔭 𝑛 ∣ = N (𝔭) 𝑛 .$
Since we have the chain of ideals
$O 𝐾 ↩ 𝔭 ↩ \dots ↩ 𝔭 𝑚,$
it is enough to show that for $0 \leq 𝑘 \leq 𝑚 - 1$ we have $∣ 𝔭 𝑘 / 𝔭 𝑘 + 1 ∣ = N (𝔭)$ , since
$O 𝐾 / 𝔭 𝑘 + 1 𝔭 𝑘 / 𝔭 𝑘 + 1 ≅ ℤ O 𝐾 𝔭 𝑘 .$
by the Third isomorphism theorem. We prove the stronger result
$O 𝐾 𝔭 ≅ ℤ 𝔭 𝑘 𝔭 𝑘 + 1$
for each $𝑘$ . First, we can construct a $ℤ$ -module homomorphism
$𝜑 : O 𝐾 \to 𝔭 𝑘 / 𝔭 𝑘 + 1 𝑥 \mapsto 𝛾 𝑥 + 𝔭 𝑘 + 1$
where $𝛾 \in 𝔭 𝑘 ∖ 𝔭 𝑘 + 1$ is arbitrary. This induces a map
$˜ 𝜑 : O 𝐾 𝔭 \to 𝔭 𝑘 𝔭 𝑘 + 1$
which will turn out to be a $ℤ$ -module isomorphism.

To prove surjectivity of $𝜑$ and thus $˜ 𝜑$ , we can show that $𝐼 : = ⟨ 𝛾 ⟩ + 𝔭 𝑘 + 1 = 𝔭 𝑘$ . Since $𝔭 𝑘 ∣ ⟨ 𝛾 ⟩$ (after all we are in a Containment-division ring) we also have $𝔭 𝑘 ∣ 𝐼$ . But $𝐼 ∣ 𝔭 𝑘 + 1$ is a proper divisor, so by unique factorization $𝐼 = 𝔭 𝑘$ as required.

To prove injectivity, we can show $𝐽 : = ⟨ 𝛾 ⟩ \cap 𝔭 𝑘 + 1 = 𝛾 𝔭$ , since hence follows if $˜ 𝜑 (𝑥) = 0$ then $𝛾 𝑥 = 𝛾 𝔭$ so $𝑥 \in 𝔭$ . Since $𝛾 \in 𝔭 𝑘$ we have $𝛾 𝔭 \subseteq 𝐽$ . Conversely, let $𝑥 \in 𝐽$ and write $𝑥 = 𝛾 𝑦$ with $𝑦 \in O 𝐾$ and $𝛾 𝑦 \in 𝔭 𝑘 + 1$ , where we have the Prime order of an ideal
$o r d 𝔭 (𝛾) + o r d 𝔭 (𝑦) = o r d 𝔭 (𝛾 𝑦) \geq 𝑘 + 1 .$
Since $o r d 𝔭 (𝛾) = 𝑘$ by construction, it follows $o r d 𝔭 (𝑦) \geq 1$ whence $𝑦 \in 𝔭$ . Therefore $𝑥 = 𝛾 𝑦 \in 𝛾 𝔭$ and ^P2 is proven.

For ^P3, note that if $N (𝔞) = 𝑚$ , then $⟨ 𝑚 ⟩ ⊴ 𝔞$ . Since $O 𝐾 / ⟨ 𝑚 ⟩$ is finite by ^C1, there can only be finitely many such ideals by the Fourth isomorphism theorem, proving ^P3.

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Absolute norm of an ideal of $O 𝐾$

Properties

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Table of Contents

Absolute norm of an ideal of O𝐾

Properties

Graph View

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Absolute norm of an ideal of $O 𝐾$