Algebraic integer

Let $𝐾$ be a field with $c h a r 𝕂 = 0$ , whence $𝐾 : ℤ$ is a ring extension. An element $𝑎 \in 𝐾$ is an algebraic integer iff it is integral over $ℤ$ , #m/def/ring i.e. it is the root of some polynomial $𝑚 𝑎 (𝑥) \in ℤ [𝑥]$ . We denote the ring of algebraic integers in $𝐾$ as $O 𝐾 = O 𝐾 : ℤ$ , which is clearly an Integrally closed domain.

Properties

An algebraic number $𝑎$ is an algebraic integer iff its minimal polynomial $𝑚 𝑎 (𝑥) \in ℚ [𝑥]$ satisfies $𝑚 𝑎 (𝑥) \in ℤ [𝑥]$ .
Every algebraic number $𝛼$ is an algebraic integer divided by some integer.

Proof of 1–2

Clearly if $𝑚 𝛼 (𝑥) \in ℤ [𝑥]$ then $𝛼$ is integral over $ℤ$ . Now suppose $𝛼$ is integral over $ℤ$ , so it is the root of some monic polynomial $ℎ (𝑥) \in ℤ [𝑥]$ . Let $𝑎 1, \dots, 𝑎 𝑛$ denote the roots of $𝑚 𝑎 (𝑥)$ . Since $𝑚 𝛼 (𝑥) ∣ ℎ (𝑥)$ , it follows $ℎ (𝑎 𝑖) = 0$ for all $𝑖 \in ℕ 𝑛$ , so $𝑎 𝑖$ are each algebraic integers. Since $𝑚 𝛼 (𝑥) = \prod 𝑛 𝑖 = 1 (𝑥 - 𝑎 𝑖) 𝜇 𝑖 \in ℚ [𝑥]$ for some algebraic multiplicities $𝜇 𝑖$ has coëfficients which are the products of algebraic integers, its coëfficients are themselves algebraic integers, and thus $𝑚 𝛼 (𝑥) \in ℤ [𝑥]$ by ^P1, proving ^P1.

Let

𝑚 𝛼 (𝑥) = 𝑥 𝑛 + 𝑎 𝑛 - 1 𝑥 𝑛 - 1 + \dots + 𝑎 0 \in ℚ [𝑥]

be the minimal polynomial of $𝛼$ where $d e g 𝑓 < 𝑛$ . Then $𝑘 𝑚 𝛼 (𝑥) \in ℤ [𝑥]$ for some $𝑘 \in ℤ$ whence

0 = (𝑘 𝛼) 𝑛 + 𝑎 𝑛 - 1 𝑘 (𝑘 𝛼) 𝑛 - 1 + 𝑎 𝑛 - 2 𝑘 2 (𝑘 𝛼) 𝑛 - 2 + \dots + 𝑘 𝑛 𝑎 0 \in ℤ [𝑥]

so $𝑘 𝛼$ is an algebraic integer.

Other results

Discriminant of an algebraic integer

Special case

Ring of integers of a number field

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