Northcott Multilinear Algebra Universal Property Proof Northcott Multilinear Algebra poses a pro



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Northcott Multilinear Algebra Universal Property Proof
Northcott Multilinear Algebra poses a problem. Consider R-modules M1,,Mp, M and N. Consider multilinear mapping
Northcott calls the universal problem as the problem to find M and multilinear mapping ϕ:M1××MpM such that there is exactly one R-module homomorphism h:MN such that hϕ=ψ.
If λ and λ exist I understand why the equalities at the end of the sentence follow, based on the satisfaction of the universal problem. I can't see however why homomorphisms λ and λ should exist.
I did more group theory many years ago and this is my first serious foray into "modules" so I wouldn't be surprised if there is something obvious I'm missing.
my thoughts: Clearly M and M′ are both homomorphic to N through h and h′, I'm not sure if this says anything about a relationship between M and M′ though.
If h′ were injective I could say something like λ(m)=h1(h(m)) but I don't know if there is any guarantee that h′ is injective..
Likewise, if ϕ were injective I could define λ(m)=ϕ(ϕ1(m)) but again I don't know why this would be the case...
I've tried replacing M and N with more familiar vector spaces and R-module homomorphisms by multilinear maps for better intuition but no luck.. I do know that if M and M′ are vector spaces with the same dimension then there is an isomorphism between them. I guess more generally if M and M′ have different dimensions (say dim(M)>dim(M)) then there is a homomorphism from M into a subspace of M′ and another homormophism from M′ onto M. Maybe this carries over to modules and is in the right direction for what I need...?

Answer & Explanation

Clark Carson

Clark Carson

Beginner2022-02-16Added 17 answers

The existence of λ and λ are furnished by the maps ϕ \phi via the universal property; that is, we are taking N=M, M respectively. To spell this out further, the map
is bilinear by assumption, and so the universal property of the pair (M,ϕ) says that there is exactly one morphism λ:MM such that λϕ=ϕ. You obtain the morphism λ:MM analogously by switching the roles of M and M′.
This is a typical situation that arises in universal problems: the data of any two solutions contains information which allows you to relate the two solutions through the universal property. In this case, the crucial piece of information is this bilinear map ϕ:M1××MpM which allows you to relate any two solutions.
Rachel Frazier

Rachel Frazier

Beginner2022-02-17Added 14 answers

In my post I misunderstood the universal problem. I thought the universal problem supplied a (M,ϕ) given a particular map ψ. But, more powerfully, a solution to the universal problem (for multilinear maps from M1××Mp) is a pair (M,ϕ) with M an R-module and ϕ:M1××MpM multilinear, such that for ANY multilinear ψ:M1××MpN there exists a unique R-module homomorphism h:MN such that hϕ=ψ.
In this case we have the following. Suppose we have multilinear ψ:M1××MpN and both (M,ϕ) and (M,ϕ) solve the universal problem. ϕ is a multilinear map from M1××Mp into M' so, because (M,ϕ) solves the univeral problem from M1××Mp,ϕ can be uniquely decomposed as
With λ:MM a R-module homomorphism. Likewise for ϕ so that
with λ:MM also an R-module homomorphism.

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