MHB Direct Products and Sums of Modules - Notation

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The discussion revolves around understanding the notation used by John Dauns in "Modules and Rings," specifically in section 1-2 regarding products and direct sums of modules. Participants clarify that the notation for the product, represented as x = (x_1, x_2, x_3) for I = {1,2,3}, indicates an ordered triple where the order matters. The distinction between the direct product (Π M_i) and the direct sum (⊕ M_i) is emphasized, noting that the direct sum consists of elements with only a finite number of non-zero components, while the direct product can include elements with infinitely many non-zero components. For finite index sets, both the direct product and direct sum yield the same result, which simplifies the understanding of their relationship. Overall, the discussion seeks clarity on the notation and the implications of these mathematical constructs.
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I am reading John Dauns book "Modules and Rings". I am having problems understanding the notation in section 1-2 (see attachment)

My issue is understanding the notation on Section 1-2, subsection 1-2.1 (see attachment).

Dauns is dealing with the product \Pi \{ M_i | i \in I \} \equiv \Pi M_i and states in (ii) - see attachment

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Alternatively, the product can be viewed as consisting of all strings or sets

x = \{ x_i | i \in I \} \equiv (x_i)_{i \in I} \equiv (x_i) \equiv ( \_ \_ \_ \ , x_i , \_ \_ \_ ) , x_i \in M_i; i-th

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I am not sure of the meaning of the above set of equivalences. Can someone briefly elaborate ... preferably with a simple example

If we take the case of I = {1,2,3} and consider the product M_1 \times M_2 \times M_3 then does Dauns notation mean

x = (x_1, x_2, x_3) where order in the triple matters (mind you if it does what are we to make of the statement x = \{ x_i | i \in I \}Can someone confirm that x = (x_1, x_2, x_3) is a correct interpretation of Dauns notation?

===============================================================================================

Dauns then goes on to define the direct sum as follows:

The direct sum \oplus \{ M_i | i \in I \} \equiv \oplus M_i is defined as the submodule \oplus M_i \subseteq \Pi M_i consisting of those elements x = (x_i) \in \Pi M_i having at most a finite number of non-zero coordinates or components. Sometimes \oplus M_i , \Pi M_i are called the external direct sum and the external direct product respectively.================================================================================================

Can someone point out the difference between \oplus M_i , \Pi M_i in the case of the example involving M_1, M_2, M_3 - I cannot really see the difference! For example, what elements exactly are in \Pi M_i that are not in \oplus M_i

I would be grateful if someone can clarify these issues.

Peter

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Peter said:
If we take the case of I = {1,2,3} and consider the product M_1 \times M_2 \times M_3 then does Dauns notation mean

x = (x_1, x_2, x_3) where order in the triple matters (mind you if it does what are we to make of the statement x = \{ x_i | i \in I \}
Yes, if $x\in M_1 \times M_2 \times M_3$, then $x$ is an ordered triple $(x_1,x_2,x_3)$ such that $x_i\in M_i$ for $i=1,2,3$. The notation $x = \{ x_i\mid i \in I \}$ is perhaps unfortunate. It would be valid if $x_i$ carried the information of the module they came from, e.g., $x = \{ (x_i,i)\mid i \in I \}$. Then it is possible to order such set according to the second component of its elements.

Alternatively, $x$ may be viewed as a function from $\{1,2,3\}$ to $M_1\cup M_2\cup M_3$ with the restriction that $x(i)$ is always in the correct module $M_i$. Such construction is called dependent product in programming (type theory).

Peter said:
Can someone point out the difference between \oplus M_i , \Pi M_i in the case of the example involving M_1, M_2, M_3 - I cannot really see the difference! For example, what elements exactly are in \Pi M_i that are not in \oplus M_i
For finite products and sums (i.e., when the index set is finite), direct product and direct sum are exactly the same. See Wikipedia.
 
Thanks Evgeny ... A most helpful post ...

Peter
 

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