Number field

Discriminant of a number field

Let 𝐾 be a number field of degree 𝑛 and {𝛼𝑖}𝑛𝑖=1 be an Integral basis. The discriminant Δ𝐾 of 𝐾 is given by alg

Δ𝐾:=Δ𝐾:ℚ(𝛼1,…,𝛼𝑛)

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 {𝛼𝑖}𝑛𝑖=1 and {𝛼′𝑖}𝑛𝑖=1 are both integral bases for 𝐾. Let 𝑑 :=Δ𝐾:ℚ(𝛼1,…,𝛼𝑛) and 𝑑′ =Δ𝐾:ℚ(𝛼′1,…,𝛼′𝑛). We can find an appropriate change of basis matrix 𝑀 ∈GL𝑛⁡(ℤ) such that

⎡⎢ ⎢⎣𝛼′1⋮𝛼′𝑛⎤⎥ ⎥⎦=𝑀⎡⎢ ⎢⎣𝛼1⋮𝛼𝑛⎤⎥ ⎥⎦,

whence

𝑇(𝛼′1,…,𝛼′𝑛)=𝑇(𝛼1,…,𝛼𝑛)𝑀𝖳,𝑇(𝛼1,…,𝛼𝑛)=𝑇(𝛼′1,…,𝛼′𝑛)𝑀−𝖳,

Now since det𝑀 and det𝑀−1 are both integers, it follows det𝑀 =det𝑀−1 = ±1. Thus 𝑑′ =(det𝑀)2𝑑 =𝑑, as required.

Since 𝑑 ∈ℚ is a product of algebraic integers, it follows 𝑑 ∈Oℚ =ℤ.

For a general ℚ-basis {𝛼𝑖}𝑛𝑖=1 ⊂O𝐾, we have

Δ𝐾:ℚ(𝛼1,…,𝛼𝑛)=∣O𝐾ℤ𝛼1+⋯+ℤ𝛼𝑛∣2Δ𝐾

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

Proof

Suppose {𝜔𝑖}𝑛𝑖=1 are an integral basis of 𝐾 and write

⎡⎢ ⎢⎣𝛼1⋮𝛼𝑛⎤⎥ ⎥⎦=𝐴⎡⎢ ⎢⎣𝜔1⋮𝜔𝑛⎤⎥ ⎥⎦

for some 𝐴 ∈M𝑛,𝑛⁡(ℤ). Let 𝑀 =spanℤ⁡{𝛼𝑖}𝑛𝑖=1. By Subgroup of a free abelian group, |O𝐾/𝑀| =|det𝐴|. The fact that

𝑇(𝛼1,…,𝛼𝑛)=𝑇(𝜔1,…,𝜔𝑛)𝑀𝖳

yields the desired result.

See also Discriminant of an algebraic integer.

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Footnotes

1. 2022. Algebraic number theory course notes, ¶¶2.2–2.3, p. 34 ↩