Direct product module

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Be sure to mention the function construction (although indexed families are really just maps...)


Let [ilmath](R,+,*,0)[/ilmath] be a ring (with out without unity), and let [ilmath](M_\alpha)_{\alpha\in I} [/ilmath] be an indexed family of left [ilmath]R[/ilmath]-modules. We can construct a new module, denoted [ilmath]\prod_{\alpha\in I}M_\alpha[/ilmath] that is a categorical product of the members of the family [ilmath](M_\alpha)_{\alpha\in I} [/ilmath][1]:

  • [ilmath]\prod_{\alpha\in I}M_\alpha[/ilmath] is the underlying set of the module (we define [ilmath]M:=\prod_{\alpha\in I}M_\alpha[/ilmath] for convenience). This is a standard Cartesian product[Note 1]. The operations are:
    1. Addition: [Note 2] [ilmath]+:M\times M\rightarrow M[/ilmath] by [ilmath]+:((x_\alpha)_{\alpha\in I},(y_\alpha)_{\alpha\in I})\mapsto(x_\alpha+y_\alpha)_{\alpha\in I} [/ilmath] (standard componentwise operation)
    2. Multiplication/Action: [ilmath]\times:R\times M\rightarrow M[/ilmath] given by [ilmath]\times:(r,(x_\alpha)_{\alpha\in I})\mapsto(rx_\alpha)_{\alpha\in I} [/ilmath], again standard componentwise definition.
Claim 1: this is indeed an [ilmath]R[/ilmath]-module.[1]

With this definition we also get canonical projections, for each [ilmath]\beta\in I[/ilmath][1]:

  • [ilmath]\pi_\beta:M\rightarrow M_\beta[/ilmath] given by [ilmath]\pi:(x_\alpha)_{\alpha\in I}\mapsto x_\beta[/ilmath]
Claim 2: the canonical projections are module homomorphisms[1]
Claim 3: the [ilmath]R[/ilmath]-module [ilmath]M[/ilmath] is the unique [ilmath]R[/ilmath]-module such that all the projections are module homomorphisms.[1]

Characteristic property of the direct product module

[ilmath]\begin{xy} \xymatrix{ & & \prod_{\alpha\in I}M_\alpha \ar[dd] \\ & & \\ M \ar[uurr]^\varphi \ar[rr]+<-0.9ex,0.15ex>|(.875){\hole} & & X_b \save (15,13)+"3,3"*+{\ldots}="udots"; (8.125,6.5)+"3,3"*+{X_a}="x1"; (-8.125,-6.5)+"3,3"*+{X_c}="x3"; (-15,-13)+"3,3"*+{\ldots}="ldots"; \ar@{->} "x1"; "1,3"; \ar@{->}_(0.65){\pi_c,\ \pi_b,\ \pi_a} "x3"; "1,3"; \ar@{->}|(.873){\hole} "x1"+<-0.9ex,0.15ex>; "3,1"; \ar@{->}_{\varphi_c,\ \varphi_b,\ \varphi_a} "x3"+<-0.9ex,0.3ex>; "3,1"; \restore } \end{xy}[/ilmath]

TODO: Description

Let [ilmath](R,*,+,0)[/ilmath] be a ring (with or without unity) and let [ilmath](M_\alpha)_{\alpha\in I} [/ilmath] be an arbitrary indexed family of [ilmath]R[/ilmath]-modules. Let [ilmath]\prod_{\alpha\in I}M_\alpha[/ilmath] be their direct product, as usual. Then[1]:
  • For any [ilmath]R[/ilmath]-module, [ilmath]M[/ilmath] and
    • For any indexed family [ilmath](\varphi_\alpha:M\rightarrow M_\alpha)_{\alpha\in I} [/ilmath] of module homomorphisms
      • There exists a unique morphism[Note 3], [ilmath]\varphi:M\rightarrow\prod_{\alpha\in I}M_\alpha[/ilmath] such that:
        • [ilmath]\forall\alpha\in I[\pi_\alpha\circ\varphi=\varphi_\alpha][/ilmath]

TODO: Link to diagram, this basically says it all though!

Proof of claims

Grade: B
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Do this section, or at least leave a guide, it should be routine and mirror the other instances of products

See also


  1. An alternate construction is that [ilmath]\prod_{\alpha\in I}M_\alpha[/ilmath] consists of mappings, [ilmath]f:I\rightarrow\bigcup_{\alpha\in I}M_\alpha[/ilmath] where [ilmath]\forall\alpha\in I[f(\alpha)\in M_\alpha][/ilmath] this is just extra work as you should already be familiar with considering tuples as mappings.
  2. the operation of the Abelian group that makes up a module
  3. Morphism - short for homomorphisms in the relevant category, in this case modules


  1. 1.0 1.1 1.2 1.3 1.4 1.5 Abstract Algebra - Pierre Antoine Grillet