Number field

Bases for a number field

Let 𝐾 be a number field of degree 𝑛. As an 𝑛-dimensional ℚ-vector space, one may choose a basis for 𝐾.

Types

Integral basis

An integral basis {𝛼𝑖}𝑛𝑖=1 a ℤ-^basis for O𝐾 (which always exists), alg whence it is also a ℚ-basis for 𝐾.

Power basis

A power basis is a basis of the form {𝛼𝑖}𝑛−1𝑖=0 for some 𝛼 ∈𝐾, alg whose existence is guaranteed by the primitive element theorem.

Integral power basis

An integral basis which is also a power basis is called an integral power basis. alg These need not exist: A number field possessing an integral power basis is called a Monogenic field.

General properties¹

1. Suppose {𝛼𝑖}𝑛𝑖=1 ⊂O𝐾 is a (non-integral) ℚ-basis for 𝐾, and let 𝑑 =Δ𝐾:ℚ(𝛼1,…,𝛼𝑛) be the corresponding discriminant. Then {𝛼𝑖/𝑑}𝑛𝑖=1 span a ℤ-module containing O𝐾.
2. If {𝛼𝑖}𝑛𝑖=1 ⊂O𝐾 are a ℚ-bases for 𝐾 such that the discriminant Δ𝐾:ℚ(𝛼1,…,𝛼𝑛) is squarefree, then they form an Integral basis.

Proof of 1–2

Let 𝛼 ∈O𝐾. Then 𝛼 =∑𝑛𝑖=1𝑐𝑖𝛼𝑖 for 𝑐𝑖 ∈ℚ and we just need to show that 𝑑𝛼𝑖 ∈ℤ for all 𝑖 ∈ℕ𝑛. Letting {𝜎𝑖}𝑛𝑖=1 range over embeddings of 𝐾 ↪――ℚ we see

⎡⎢ ⎢⎣𝜎1(𝛼)⋮𝜎𝑛(𝛼)⎤⎥ ⎥⎦=𝑇⎡⎢ ⎢⎣𝑐1⋮𝑐𝑛⎤⎥ ⎥⎦

with 𝑇 ∈M𝑛×𝑛⁡(O――ℚ) defined as in Discriminant of a separable extension. Multiplying each side by [[Adjugate matrix|adj⁡𝑇]] and letting 𝛿 =det𝑇 we have

⎡⎢ ⎢⎣𝛽1⋮𝛽𝑛⎤⎥ ⎥⎦=𝛿⎡⎢ ⎢⎣𝑐1⋮𝑐𝑛⎤⎥ ⎥⎦

for some {𝛽𝑖}𝑛𝑖=1 ⊂O――ℚ, whence

⎡⎢ ⎢⎣𝛿𝛽1⋮𝛿𝛽𝑛⎤⎥ ⎥⎦=𝑑⎡⎢ ⎢⎣𝑐1⋮𝑐𝑛⎤⎥ ⎥⎦

so since 𝑑𝑐𝑖 =𝛿𝛽𝑖 is an algebraic integer for each 𝑖. Since 𝑑𝑐𝑖 ∈ℚ, it follows 𝑑𝑐𝑖 ∈ℤ for each 𝑖, proving ^P1 .

^P2 follows from the fact that the discriminant of algebraic integers is in an integer and ^EQ1.

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Footnotes

1. 2022. Algebraic number theory course notes, §2.1 ↩