Annoying index

Given a number field $𝐾$ with ring of integers $O 𝐾$ , and a $ℚ$ -basis ${𝛼 𝑖} 𝑛 𝑖 = 1 \subset O 𝐾$ of $𝐾$ , we call the quantity

∣ O 𝐾 ℤ 𝛼 1 + \dots + ℤ 𝛼 𝑛 ∣

the annoying index,¹ #m/def/num/alg since it measures the degree to which ${𝛼 𝑖} 𝑛 𝑖 = 1$ fails to be an Integral basis for $𝐾$

Properties

If $Δ 𝐾 : ℚ (𝛼 1, \dots, 𝛼 𝑛)$ is squarefree, then ${𝛼 𝑖} 𝑛 𝑖 = 1$ is an integral basis.
If the minimal polynomial $𝑚 𝛼 (𝑥)$ for $𝛼 \in O 𝐾$ is Eisenstein at $𝑝$ , then $𝑝$ does not divide $| O 𝐾 / ℤ [𝛼] |$ .

Proof

^P1 follows immediately from ^EQ1.

For ^P2, suppose towards contradiction that $𝑝$ divides the annoying index. Then by Cauchy's order theorem there exists $𝜉 \in O 𝐾$ such that $𝜉 + ℤ [𝛼]$ has order $𝑝$ , i.e. $𝜉 \notin ℤ [𝛼]$ and $𝑝 𝜉 \in ℤ [𝛼]$ . It follows we can write

𝑝 𝜉 = 𝑛 - 1 \sum 𝑖 = 0 𝑏 𝑖 𝛼 𝑖

for some $𝑏 𝑖 \in ℤ$ where some $𝑏 𝑗$ is not divisible by $𝑝$ . Fix $𝑗$ to be the smallest such index. It follows

𝜂 = 𝜉 - 𝑗 - 1 \sum 𝑖 = 0 𝑏 𝑖 𝑝 𝛼 𝑖 = 𝑛 - 1 \sum 𝑖 = 𝑗 𝑏 𝑖 𝑝 𝛼 𝑖 \in O 𝐾 .

and so

𝜂 𝛼 𝑛 - 𝑗 - 1 = 𝑏 𝑗 𝑝 𝛼 𝑛 - 1 + 𝛼 𝑛 𝑝 𝑛 - 𝑗 - 2 \sum 𝑖 = 0 𝑏 𝑖 + 𝑗 + 1 𝛼 𝑖 \in O 𝐾 .

Since

𝛼 𝑛 𝑝 = - 𝑛 - 1 \sum 𝑖 = 0 𝑎 𝑖 𝑝 𝛼 𝑖 \in O 𝐾

it follows $𝛽 = 𝑏 𝑗 𝑝 𝛼 𝑛 - 1 \in O 𝐾$ , so the field norm $N 𝐾 : ℚ (𝛽) \in ℤ$ .

On the other hand

N 𝐾 : ℚ (𝛽) = 𝑏 𝑛 𝑗 𝑝 𝑛 N 𝐾 : ℚ (𝛼) 𝑛 - 1 = \pm 𝑏 𝑛 𝑗 𝑎 𝑛 - 1 0 𝑝 𝑛 \notin ℤ

since $𝑝 ∤ 𝑏 𝑗$ and $𝑝 2 ∤ 𝑎 0$ , a contradiction.

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

This term is taken from lectures by Florian Breuer and should not be taken too seriously. ↩

Annoying index

Properties

Footnotes