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Is there a difference between saying that the direct sum of $|S|$ copies of a ring $R$ is the set of all functions $f: S \to R$ such that they are zero except in finitely many places and saying that $\oplus_{s \in S} R$ is the set of all tuples $(r_s, r_{s^\prime}, r_{s^{\prime \prime}}, \dots )$ such that only finitely many $r_s$ are non-zero?

It seems to be the same and the second seems more intuitive but the first one is used on Wikipedia for some reason.

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Yes, it is the same set-theoretically! –  Andrea Apr 21 '12 at 19:59
    
@Andrea Thank you! –  Rudy the Reindeer Apr 21 '12 at 20:15
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@ClarkKent: That's the point: there are different constructions of the direct sum of modules, depending on what you think a "direct product of sets" is. If you think direct products of sets are sets of tuples, then direct sums are sets of tuples; if you think direct products of sets are sets of functions, then direct sums of modules are sets of functions. The point is that both constructions satisfy the universal property of the direct sum, and so they are (functionally) isomorphic. –  Arturo Magidin Apr 21 '12 at 23:20
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Of course, note that the way you are writing the tuples implies that your set $S$ is not only countable, but already well-ordered. As such, the notation is somewhat misleading. –  Arturo Magidin Apr 22 '12 at 1:57
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@ClarkKent: Yes, $(r_s)_{s\in S}$, which gives the tuple as a family, would be a reasonable notation. Of course, then you are coming closer to the identification of a "tuple"/family with a function, since a family is just a function with domain the index set. –  Arturo Magidin Apr 22 '12 at 20:03

2 Answers 2

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The "tuples" characterization is the same thing unless you think there's no such thing as a tuple $(\ldots,\bullet, \bullet,\bullet,\ldots)$ when the number of components is uncountable. I.e. the "tuples" way of saying it seems to assume the number of copies you're summing is at most countably infinite.

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To avoid this implication I deliberately used $(r_s , r_{s^\prime}, \dots )$ rather than $(r_1, r_2, \dots )$. Maybe this still assumes that $S$ is countable? Is there a way to write a tuple for an uncountable $S$? –  Rudy the Reindeer Apr 21 '12 at 20:35
    
So the other answer is wrong because this seems to be an essential difference between the two? –  Rudy the Reindeer Apr 21 '12 at 20:36
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Well, this is all really a matter of how you construe the notation. But the "functions" characterization doesn't seem to have the same potential for confusion about the meaning. Maybe the "tuples" characterization is more intuitive if you're accustomed to it. –  Michael Hardy Apr 22 '12 at 0:12

There is no essential difference. You may consider the way of expressing a direct product $$\prod_{\alpha\in I} A.$$ We may define it as the set of functions from $I\rightarrow A$ because to every $\alpha\in I$ it uniquely associates an element in $A$. The direct sum construction you quoted is a special case of this.

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It's not really a special case, because the direct sum is a proper subobject of the direct product in this context! –  Zhen Lin Apr 21 '12 at 20:15

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