Module theory MOC

Free module

Free modules are the free objects in 𝑅𝖬𝗈𝖽: modules formed by freely applying module operations to a given set 𝑆. #m/def/module Departing from classical tradition, we call a module equipped with an isomorphism from a free module an 𝑆-framed module, while a module for which there merely exists such an isomorphism is a 𝑆-framable module. Assuming the Axiom of Choice, any module over a division ring is framable: See Assuming choice, every vector space has a basis.

Notation

In these notes, we have two conventions for the free module over 𝑅 generated by a set 𝑆. The first is

𝑅(𝑆)=span{1𝑠:𝑠𝑆}𝑅𝑆

where we think of elements as maps of finite support 𝑆 𝑅, and we identify 𝑠 𝑆 with 1𝑠 :𝑡 [𝑠 =𝑡]. The second is

𝑅{𝑥𝑠}𝑠𝑆=span𝑅{𝑥𝑠:𝑠𝑆}

which allows for the explicit naming of the basis to be used.

Being framable is equivalent to the existence of a basis, where by a basis B for an 𝑅-module 𝑉, we mean an 𝑅-spanning set B such that each 𝑣 𝑉 is given by a unique 𝑅-linear combination of B elements.

Universal property

Let 𝑅 be a ring and 𝑆 be a set. The free module is a pair consisting of an 𝑅-module 𝑅(𝑆) and a function 𝜄 :𝑆 𝑅(𝑆) such that given any 𝑅-module 𝑀 and function 𝑓 :𝑆 𝑀 there exists a unique module homomorphism ¯𝑓 :𝑅(𝑆) 𝑀 such that the following diagram commutes

https://q.uiver.app/#q=WzAsMyxbMCwwLCJTIl0sWzIsMCwiUl57KFMpfSJdLFsyLDIsIk0iXSxbMCwxLCJcXGlvdGEiXSxbMCwyLCJmIiwyXSxbMSwyLCJcXGV4aXN0cyAhXFxiYXIgZiIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==

This has a unique extension to a functor such that

𝜄:1𝑅():𝖲𝖾𝗍𝖲𝖾𝗍

becomes a natural transformation.

Monoidal functor

When 𝑅 =K is a commutative ring this forms a monoidal functor with respect to the cartesian structure on 𝖲𝖾𝗍 and the tensor product on K𝖬𝗈𝖽. #m/thm/cat If 𝟙 ={ } in 𝖲𝖾𝗍, then we let

𝛾:KK{}11𝜇𝑋,𝑌:K(𝑋)K(𝑌)K(𝑋×𝑌)1𝑥1𝑦1(𝑥,𝑦)

Construction as maps

Let 𝑆 be a set and 𝑅 be a ring. The free module 𝑅(𝑆) is the set of maps of finite support 𝑆 𝑅 with addition and scaling induced by those of 𝑅, #m/def/module
i.e. for all 𝑠 𝑆

(𝛼𝑎+𝛽𝑏)(𝑠)=𝛼𝑎(𝑠)+𝛽𝑏(𝑠)

where we identify 𝑠 𝑆 with 1𝑠 :𝑡 [𝑡 =𝑠] invoking an Iverson bracket.

Proof of universal property

Clearly 𝑅(𝑆) as constructed is an 𝑅-module with basis {1𝑠}𝑠𝑆 Now let 𝑀 be an 𝑅-module and 𝑓 :𝑆 𝑀 be a function. For a module homomorphism ¯𝑓 :𝑅(𝑆) 𝑀 to make the diagram commute, we require that ¯𝑓(1𝑠) =𝑓(𝑠) for all 𝑠 𝑆, which fully specifies ¯𝑓 so that for 𝑎 𝑅(𝑆)

¯𝑓(𝑎)=𝑠𝑆𝑓(𝑠)𝑎(𝑠)

as required.

Further results


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