Discriminant of a number field

Let $𝐾$ be a number field of degree $𝑛$ and ${𝛼 𝑖} 𝑛 𝑖 = 1$ be an Integral basis. The discriminant $Δ 𝐾$ of $𝐾$ is given by #m/def/num/alg

Δ 𝐾 : = Δ 𝐾 : ℚ (𝛼 1, \dots, 𝛼 𝑛)

where the latter quantity is the discriminant of a separable extension and is an integer independent of the choice of integral basis.¹

Proof

Suppose ${𝛼 𝑖} 𝑛 𝑖 = 1$ and ${𝛼' 𝑖} 𝑛 𝑖 = 1$ are both integral bases for $𝐾$ . Let $𝑑 : = Δ 𝐾 : ℚ (𝛼 1, \dots, 𝛼 𝑛)$ and $𝑑' = Δ 𝐾 : ℚ (𝛼' 1, \dots, 𝛼' 𝑛)$ . We can find an appropriate change of basis matrix $𝑀 \in G L 𝑛 (ℤ)$ such that

⎡ ⎢ ⎢ ⎣ 𝛼' 1 ⋮ 𝛼' 𝑛 ⎤ ⎥ ⎥ ⎦ = 𝑀 ⎡ ⎢ ⎢ ⎣ 𝛼 1 ⋮ 𝛼 𝑛 ⎤ ⎥ ⎥ ⎦,

whence

𝑇 (𝛼' 1, \dots, 𝛼' 𝑛) = 𝑇 (𝛼 1, \dots, 𝛼 𝑛) 𝑀 𝖳, 𝑇 (𝛼 1, \dots, 𝛼 𝑛) = 𝑇 (𝛼' 1, \dots, 𝛼' 𝑛) 𝑀 - 𝖳,

Now since $d e t 𝑀$ and $d e t 𝑀 - 1$ are both integers, it follows $d e t 𝑀 = d e t 𝑀 - 1 = \pm 1$ . Thus $𝑑' = (d e t 𝑀) 2 𝑑 = 𝑑$ , as required.

Since $𝑑 \in ℚ$ is a product of algebraic integers, it follows $𝑑 \in O ℚ = ℤ$ .

For a general $ℚ$ -basis ${𝛼 𝑖} 𝑛 𝑖 = 1 \subset O 𝐾$ , we have

Δ 𝐾 : ℚ (𝛼 1, \dots, 𝛼 𝑛) = ∣ O 𝐾 ℤ 𝛼 1 + \dots + ℤ 𝛼 𝑛 ∣ 2 Δ 𝐾

where all operands are integers. We call the index on the right had side the Annoying index.

Proof

Suppose ${𝜔 𝑖} 𝑛 𝑖 = 1$ are an integral basis of $𝐾$ and write

⎡ ⎢ ⎢ ⎣ 𝛼 1 ⋮ 𝛼 𝑛 ⎤ ⎥ ⎥ ⎦ = 𝐴 ⎡ ⎢ ⎢ ⎣ 𝜔 1 ⋮ 𝜔 𝑛 ⎤ ⎥ ⎥ ⎦

for some $𝐴 \in M 𝑛, 𝑛 (ℤ)$ . Let $𝑀 = s p a n ℤ {𝛼 𝑖} 𝑛 𝑖 = 1$ . By Subgroup of a free abelian group, $| O 𝐾 / 𝑀 | = | d e t 𝐴 |$ . The fact that

𝑇 (𝛼 1, \dots, 𝛼 𝑛) = 𝑇 (𝜔 1, \dots, 𝜔 𝑛) 𝑀 𝖳

yields the desired result.

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

2022. Algebraic number theory course notes, ¶¶2.2–2.3, p. 34 ↩

Discriminant of a number field

Footnotes