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 ${𝛼 𝑖} 𝑛 𝑖 = 1$ a $ℤ$ -basis for $O 𝐾$ (which always exists), #m/def/num/alg whence it is also a $ℚ$ -basis for $𝐾$ .

Power basis

A power basis is a basis of the form ${𝛼 𝑖} 𝑛 - 1 𝑖 = 0$ for some $𝛼 \in 𝐾$ , #m/def/num/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. #m/def/num/alg These need not exist: A number field possessing an integral power basis is called a Monogenic field.

General properties¹

Suppose ${𝛼 𝑖} 𝑛 𝑖 = 1 \subset O 𝐾$ is a (non-integral) $ℚ$ -basis for $𝐾$ , and let $𝑑 = Δ 𝐾 : ℚ (𝛼 1, \dots, 𝛼 𝑛)$ be the corresponding discriminant. Then ${𝛼 𝑖 / 𝑑} 𝑛 𝑖 = 1$ span a $ℤ$ -module containing $O 𝐾$ .
If ${𝛼 𝑖} 𝑛 𝑖 = 1 \subset O 𝐾$ are a $ℚ$ -bases for $𝐾$ such that the discriminant $Δ 𝐾 : ℚ (𝛼 1, \dots, 𝛼 𝑛)$ is squarefree, then they form an Integral basis.

Proof of 1–2

Let $𝛼 \in O 𝐾$ . Then $𝛼 = \sum 𝑛 𝑖 = 1 𝑐 𝑖 𝛼 𝑖$ for $𝑐 𝑖 \in ℚ$ and we just need to show that $𝑑 𝛼 𝑖 \in ℤ$ for all $𝑖 \in ℕ 𝑛$ . Letting ${𝜎 𝑖} 𝑛 𝑖 = 1$ range over embeddings of $𝐾 ↪ ―― ℚ$ we see

⎡ ⎢ ⎢ ⎣ 𝜎 1 (𝛼) ⋮ 𝜎 𝑛 (𝛼) ⎤ ⎥ ⎥ ⎦ = 𝑇 ⎡ ⎢ ⎢ ⎣ 𝑐 1 ⋮ 𝑐 𝑛 ⎤ ⎥ ⎥ ⎦

with $𝑇 \in M 𝑛 \times 𝑛 (O ―― ℚ)$ defined as in Discriminant of a separable extension. Multiplying each side by $a d j 𝑇$ and letting $𝛿 = d e t 𝑇$ we have

⎡ ⎢ ⎢ ⎣ 𝛽 1 ⋮ 𝛽 𝑛 ⎤ ⎥ ⎥ ⎦ = 𝛿 ⎡ ⎢ ⎢ ⎣ 𝑐 1 ⋮ 𝑐 𝑛 ⎤ ⎥ ⎥ ⎦

for some ${𝛽 𝑖} 𝑛 𝑖 = 1 \subset O ―― ℚ$ , whence

⎡ ⎢ ⎢ ⎣ 𝛿 𝛽 1 ⋮ 𝛿 𝛽 𝑛 ⎤ ⎥ ⎥ ⎦ = 𝑑 ⎡ ⎢ ⎢ ⎣ 𝑐 1 ⋮ 𝑐 𝑛 ⎤ ⎥ ⎥ ⎦

so since $𝑑 𝑐 𝑖 = 𝛿 𝛽 𝑖$ is an algebraic integer for each $𝑖$ . Since $𝑑 𝑐 𝑖 \in ℚ$ , it follows $𝑑 𝑐 𝑖 \in ℤ$ 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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2022. Algebraic number theory course notes, §2.1 ↩