kaliitcri

2022-02-15

Northcott Multilinear Algebra Universal Property Proof

Northcott Multilinear Algebra poses a problem. Consider R-modules$M}_{1},\dots ,{M}_{p$ , M and N. Consider multilinear mapping

$\psi :{M}_{1}\times \cdots \times {M}_{p}\to N$

Northcott calls the universal problem as the problem to find M and multilinear mapping$\varphi :{M}_{1}\times \cdots \times {M}_{p}\to M$ such that there is exactly one R-module homomorphism $h:M\to N$ such that $h\circ \varphi =\psi$ .

If$\lambda \text{}\text{and}\text{}{\lambda}^{\prime}$ 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 $\lambda \text{}\text{and}\text{}{\lambda}^{\prime}$ 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$\lambda \left(m\right)={h}^{{}^{\prime}-1}\left(h\left(m\right)\right)$ but I don't know if there is any guarantee that h′ is injective..

Likewise, if$\varphi$ were injective I could define $\lambda \left(m\right)={\varphi}^{\prime}\left({\varphi}^{-1}\left(m\right)\right)$ 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$\mathrm{dim}\left({M}^{\prime}\right)>\mathrm{dim}\left(M\right))$ 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...?

Northcott Multilinear Algebra poses a problem. Consider R-modules

Northcott calls the universal problem as the problem to find M and multilinear mapping

If

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

Likewise, if

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

Clark Carson

Beginner2022-02-16Added 17 answers

The existence of $\lambda \text{}\text{and}\text{}{\lambda}^{\prime}$ are furnished by the maps $\varphi}^{\prime}\text{}\text{phi$ via the universal property; that is, we are taking $N={M}^{\prime}$ , M respectively. To spell this out further, the map

$\varphi}^{\prime}:{M}_{1}\times \cdots \times {M}_{p}\to {M}^{\prime$

is bilinear by assumption, and so the universal property of the pair$(M,\varphi )$ says that there is exactly one morphism $\lambda :M\to {M}^{\prime}$ such that $\lambda \circ \varphi ={\varphi}^{\prime}$ . You obtain the morphism ${\lambda}^{\prime}:{M}^{\prime}\to M$ 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$\varphi :{M}_{1}\times \cdots \times {M}_{p}\to M$ which allows you to relate any two solutions.

is bilinear by assumption, and so the universal property of the pair

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

Rachel Frazier

Beginner2022-02-17Added 14 answers

In my post I misunderstood the universal problem. I thought the universal problem supplied a $(M,\varphi )$ given a particular map $\psi$ . But, more powerfully, a solution to the universal problem (for multilinear maps from $M}_{1}\times \cdots \times {M}_{p$ ) is a pair $(M,\varphi )$ with M an R-module and $\varphi :{M}_{1}\times \cdots \times {M}_{p}\to M$ multilinear, such that for ANY multilinear $\psi :{M}_{1}\times \cdots \times {M}_{p}\to N$ there exists a unique R-module homomorphism $h:M\to N$ such that $h\circ \varphi =\psi$ .

In this case we have the following. Suppose we have multilinear$\psi :{M}_{1}\times \cdots \times {M}_{p}\to N$ and both $(M,\varphi )\text{}\text{and}\text{}({M}^{\prime},{\varphi}^{\prime})$ solve the universal problem. $\varphi}^{\prime$ is a multilinear map from $M}_{1}\times \cdots \times {M}_{p$ into M' so, because $(M,\varphi )$ solves the univeral problem from $M}_{1}\times \cdots \times {M}_{p},{\varphi}^{\prime$ can be uniquely decomposed as

$\lambda \circ \varphi ={\varphi}^{\prime}$

With$\lambda :M\to {M}^{\prime}$ a R-module homomorphism. Likewise for $\varphi$ so that

${\lambda}^{\prime}\circ {\varphi}^{\prime}=\varphi$

with${\lambda}^{\prime}:{M}^{\prime}\to M$ also an R-module homomorphism.

In this case we have the following. Suppose we have multilinear

With

with

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