Direct product module
From Maths
(Redirected from Direct product (module))
Stub grade: A
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Be sure to mention the function construction (although indexed families are really just maps...)
Contents
Definition
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:
- 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)
- 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
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]
- There exists a unique morphism[Note 3], [ilmath]\varphi:M\rightarrow\prod_{\alpha\in I}M_\alpha[/ilmath] such that:
- For any indexed family [ilmath](\varphi_\alpha:M\rightarrow M_\alpha)_{\alpha\in I} [/ilmath] of module homomorphisms
TODO: Link to diagram, this basically says it all though!
Proof of claims
Grade: B
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
The message provided is:
Do this section, or at least leave a guide, it should be routine and mirror the other instances of products
See also
- Direct sum of modules - instances of a co-product
- Characteristic property of the direct product module
Notes
- ↑ 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.
- ↑ the operation of the Abelian group that makes up a module
- ↑ Morphism - short for homomorphisms in the relevant category, in this case modules