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I am reading p. 20 of Introduction to Commutative Algebra by Atiyah and Macdonald. Given a module decomposition

$$ A=\mathfrak{a}_1\oplus\cdots\oplus\mathfrak{a}_n $$ of ring $A$ as direct sum of ideals $\mathfrak{a}_i$ What does module decomposition mean in this line?

  1. Internal direct sum
  2. External direct sum
  3. $\mathfrak{a}_i \cap \mathfrak{a}_j=0$ for $i\neq j$
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  • $\begingroup$ As far as I know, the third thing is not a thing that is used anywhere. $\endgroup$
    – rschwieb
    Oct 15, 2020 at 13:03

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This means internal direct sum. By definition, $\mathfrak{a}_i\cap \mathfrak{a}_j=0$ for $i\ne j$ and $\mathfrak{a}_1+\cdots+\mathfrak{a}_n=A$.

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  • $\begingroup$ He didn't define internal direct sum but he uses it , and it confuses me for hours. $\endgroup$
    – Lord Shadow
    Oct 15, 2020 at 3:44
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    $\begingroup$ Well there isn't really a big difference between internal and external direct sum. I wrote an answer math.stackexchange.com/questions/3651553/… which explains in part what I mean, if you're interested. That is, it shows how the two notions are closely related. $\endgroup$
    – Alekos Robotis
    Oct 15, 2020 at 3:48
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    $\begingroup$ @LordShadow The contextual cue is that the things you are summing are subsets of the thing they sum to. On the other hand if you just take two random $\mathbb F$ vector spaces $V,W$, $V\oplus W$ could not initially be called "internal' because $V,W$ are not common subsets of a set at the beginning. However, they are (isomorphic to) common subsets of their external direct product, and you can say that the external product of two things is itself an internal product of two subsets, each of which is a copy of $V,W$. That's basically what's described at the link Alekos gave. $\endgroup$
    – rschwieb
    Oct 15, 2020 at 13:07

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