Irreducible element

Let $𝑅$ be a ring. An element $𝑥 \in 𝑅$ is irreducible iff $𝑥$ is not a unit but whenever $𝑥 = 𝑎 𝑏$ with $𝑎, 𝑏 \in 𝑅$ then $𝑎$ or $𝑏$ is a unit. ring

(\forall 𝑎, 𝑏 \in 𝑅) [𝑥 = 𝑎 𝑏 ⟹ {𝑎, 𝑏} \cap 𝑅 \times \neq \emptyset]

This is one way to generalize the Prime number to an arbitrary ring.¹

Properties

For $𝑅$ an Integral domain, $𝑥 \in 𝑅$ is irreducible iff $⟨ 𝑥 ⟩$ is maximal among principal ideals.

Proof of 1

Assume $𝑥 \in 𝑅$ is irreducible, and suppose $⟨ 𝑥 ⟩ \subseteq ⟨ 𝑦 ⟩$ . Then $𝑥 = 𝑎 𝑦$ for some $𝑎 \in 𝑅$ , whence either $𝑦$ is a unit, implying $⟨ 𝑦 ⟩ = ⟨ 1 ⟩$ , or $𝑎$ is a unit, implying $𝑦 = 𝑎 - 1 𝑥$ and thus $⟨ 𝑦 ⟩ = ⟨ 𝑥 ⟩$ . Therefore $⟨ 𝑥 ⟩$ is maximal among principal ideals.

Now suppose $⟨ 𝑥 ⟩$ is maximal among principal ideals, and let $𝑥 = 𝑎 𝑏$ for $𝑎, 𝑏 \in 𝑅$ . Then $𝑥 \in ⟨ 𝑎 ⟩$ and $𝑥 \in ⟨ 𝑏 ⟩$ . If $𝑎 \in 𝑅 \times$ we are done, so assume $⟨ 𝑎 ⟩ = ⟨ 𝑥 ⟩$ . Then $𝑎$ and $𝑥$ are associate elements and thus $𝑥 = 𝑎 𝑢$ for $𝑢 \in 𝑅 \times$ . Hence $𝑎 𝑢 = 𝑎 𝑏$ and thus $𝑏 = 𝑢 \in 𝑅 \times$ . Therefore $𝑥$ is irreducible, proving ^P1

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2022. Algebraic number theory course notes, §1.1, p. 1 ↩

Table of Contents

Irreducible element

Properties

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Table of Contents

Irreducible element

Properties

Footnotes

Graph View

Backlinks