Maximal ideal iff prime ideal in a PID

Let $𝑅$ be a PID, and $𝐼 ◃ 𝑅$ be a nonzero ideal. Then $𝐼$ is prime iff it is maximal.¹

Proof

A maximal ideal in a commutative ring is prime. For the converse, suppose $⟨ 𝑎 ⟩ ◃ 𝑅$ is a prime ideal with $𝑎 \neq 0$ , and suppose $⟨ 𝑎 ⟩ \subseteq ⟨ 𝑏 ⟩$ for some $𝑏 \in 𝑅$ . It follows $⟨ 𝑎 ⟩ ∋ 𝑎 = 𝑏 𝑐$ for some $𝑐 \in 𝑅$ , so from primality of $⟨ 𝑎 ⟩$ we have $𝑏 \in ⟨ 𝑎 ⟩$ or $𝑐 \in ⟨ 𝑎 ⟩$ . If $𝑏 \in ⟨ 𝑎 ⟩$ it follows $⟨ 𝑏 ⟩ = ⟨ 𝑎 ⟩$ . If $𝑐 \in ⟨ 𝑎 ⟩$ it follows $𝑐 = 𝑑 𝑎$ for some $𝑑 \in 𝑅$ , so
$𝑎 = 𝑏 𝑐 = 𝑏 𝑑 𝑎$
so from cancellation $⟨ 𝑏 ⟩ ∋ 𝑏 𝑑 = 1$ , so $⟨ 𝑏 ⟩ = 𝑅$ .

tidy | en | SemBr

2009. Algebra: Chapter 0, §III.4.3, ¶4.13, pp. 151–152 ↩

Maximal ideal iff prime ideal in a PID

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Maximal ideal iff prime ideal in a PID

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