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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.

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    $\begingroup$ Do you see that in $\mathbb{Z}$ for example, $a$ and $b$ are coprime iff $(a) + (b) = \mathbb{Z}$? $\endgroup$
    – Eric Auld
    Sep 9, 2015 at 2:04
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    $\begingroup$ In $\mathbb Z$, we have $(a)+(b)=(gcd(a,b))$. $\endgroup$
    – Rachmaninoff
    Sep 9, 2015 at 3:01
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    $\begingroup$ See Bézout's Identity $\endgroup$
    – Viktor Vaughn
    Sep 9, 2015 at 3:43
  • $\begingroup$ To Eric Auld and Rachmaninoff: What does it mean to have $a$ and $b$ in parenthesis, as in $(a)$ and $(b)$? I like your ideas, would love to adopt it if I could understand the parenthesis notation. Thanks again to you and SpamIAM. $\endgroup$
    – A.Magnus
    Sep 9, 2015 at 11:19

2 Answers 2

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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.

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  • $\begingroup$ Thank you, you have earned my up-vote and green check-mark. Thanks again for responding to me. $\endgroup$
    – A.Magnus
    Sep 10, 2015 at 14:46
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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.

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