Relatively prime ideals

Let $𝑅$ be a commutative ring. Two ideals $𝔞, 𝔟 ⊴ 𝑅$ are said to be relatively prime iff $𝔞 + 𝔟 = ⟨ 1 ⟩$ .¹ #m/def/ring

Properties

Suppose $𝔞 1, \dots, 𝔞 𝑛$ are pairwise relatively prime. Then $𝔞 1 \dots 𝔞 𝑛 = 𝔞 1 \cap \dots \cap 𝔞 𝑛$
Suppose $𝔞 1, \dots, 𝔞 𝑛$ are each relatively prime with $𝔟$ . Then $𝔞 1 \dots 𝔞 𝑛 + 𝔟 = ⟨ 1 ⟩$ .
Suppose $𝔭, 𝔮$ are distinct nonzero prime ideals in a 1-dimensional ring. Then $𝔭 𝑠 + 𝔮 𝑡 = ⟨ 1 ⟩$ for $𝑠, 𝑡 \in ℕ$ .

Proof of 1–2

For ^P1, it suffices to show the case for $𝑛 = 2$ . For any ideals we already have $𝔞 1 𝔞 2 \subseteq 𝔞 1 \cap 𝔞 2$ . Since $𝔞 1 + 𝔞 2 = ⟨ 1 ⟩$ , it holds in particular that $𝑎 1 + 𝑎 2 = 1$ for some $𝑎 𝑖 \in 𝔞 𝑖$ . Thus for any $𝑥 \in 𝔞 1 \cap 𝔞 2$ , we have $𝑥 = 𝑥 𝑎 1 + 𝑥 𝑎 2 \in 𝔞 1 𝔞 2$ , proving ^P1.

By the hypothesis of ^P2, for each $𝑖 \in ℕ 𝑛$ there exists an $𝑏 𝑖 \in 𝔟$ such that $1 - 𝑏 𝑖 \in 𝔞 𝑖$ . Then

(1 - 𝑏 1) \dots (1 - 𝑏 𝑛) \in 𝔞 1 \dots 𝔞 𝑛

and

1 - (1 - 𝑏 1) \dots (1 - 𝑏 𝑛) \in 𝔟,

hence $1 \in 𝔞 1 \dots 𝔞 𝑛 + 𝔟$ , proving ^P2.

For ^P3 let $𝑚 = m a x {𝑠, 𝑡}$ . We show that

(𝔭 + 𝔮) 2 𝑚 \subseteq 𝔭 𝑠 + 𝔮 𝑡 .

To see this, note that every element of $(𝔭 + 𝔮) 2 𝑚$ is a sum of elements of the form $(𝑝 1 + 𝑞 1) \dots (𝑝 2 𝑚 + 𝑞 2 𝑚)$ where $𝑝 𝑖 \in 𝔭$ and $𝑞 𝑖 \in 𝔮$ . But such a term is itself a sum of terms containing either at least $𝑚$ elements of $𝔭$ or at least $𝑚$ elements of $𝔮$ , implying it is either an element of $𝔭 𝑠$ or $𝔮 𝑡$ .

Now ^P3 follows from this fact and the 1-dimensionality of $𝑅$ , since $𝔭 \neq 𝔮$ we have $𝔭 ⊊ 𝔭 + 𝔮$ , and by maximality of $𝔭$ we have $𝔭 + 𝔮 = ⟨ 1 ⟩$ .

Results

Chinese remainder theorem for rings

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

2022. Algebraic number theory course notes, §1.3.3, p. 25 ↩

Relatively prime ideals

Properties

Results

Footnotes