Coprime Ideals in Commutative Rings

Question

I am working on a project on generalization of the Chinese Remainder Theorem in commutative rings, which inevitably have to go through the definition of coprimality in commutative rings. I came across page 65 of this paper here which says:

Definition 2.12: The ideals $\mathscr I$ and $\mathscr J$ of $R$ a commutative ring are relatively prime if $\mathscr I + \mathscr J = R.$

Honestly, I don't feel comfortable present it as it is on my paper without any narration explaining the rationale behind it. It looks so intuitive but I am lost on how to elaborate it, I have been searching left and right for rationales but could not find any. Any links or pointers would be very much appreciated.

Thank you for your time and help.

Answer 1

This is essentially a generalisation of the notion of coprimality in the integers. If two integers are coprime, then their gcd is 1 (and by Bézout's identity gcd can be expressed as a linear combination of those two numbers), which is the generator of the entire ring $Z$. Hence the sum of their ideals is $Z$. So while I can't say what to expect in a ring without a multiplicative identity, i.e 1, but in the rings with 1 this seems to be the best way to define coprimality.

Answer 2

By definition, let R be a ring. Two ideals I, J of R are called coprime /comaximal / relatively prime if I+J = R. Suppose that I and J are coprime.

For any r,t ∈ R, there is an s ∈ R such that s+I = r+I and s+J = t+J.

Proof: Since R = I + J any element of R can be written as i + j for some i ∈ I and j ∈ J.

In particular, r = i + j, t = i1 + j1 for some i, i1 ∈ I and j, j1 ∈ J. Let s = j + i1. Since s − r = i1 − i ∈ I, we have s + I = r + I. Similarly, since s − t = j − j1 ∈ J, we have s + J = t + J.