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Tuesday, June 11, 2013

Universal Multilinear Object (10.4 Corollary 16)

Dummit and Foote Abstract Algebra, section 10.4, corollary 16:

MathJax TeX Test Page Show that if R is a commutative ring, Ai is an R-module, L is an abelian group, and φ:A1×...×AnL is a multilinear map, then there exists a unique homomorphism of R-modules Φ:A1...AnL such that φ=Φι, where ι is the natural map from A1×...×An to An...An. Conversely, show that if there is a homomorphism of R-modules Φ:A1...AnL then φ=Φι is a multilinear map.

Proof: Lemma 1: We have the natural map on simple tensorsι:A1...An1×An(A1...An1)AnA1...An is well defined, and together with the evidently well-defined natural mapι:A1×...×AnA1...An1×Anwe have the relation ι=ιι. Proof: The lemma will follow as soon as we show ι is well defined. If (a1...an1,an)=(a1...an1,an) then by using these two given equivalencies in the coordinates we clearly haveι(a1...an1,an)=(a1...an1)an=(a1...an1)an=ι(a1...an1,an) Proceed by induction on n. For each anAn defineφan:A1×...×An1Lφan(a1,...,an1)=φ(a1,...,an)Note that φan is multilinearφan(a1,...,ram+ram,...,an1)=φ(a1,...,ram+ram,...,an)=rφ(a1,...,am,...,an)+rφ(a1,...,am,...,an)=rφan(a1,...,am,...,an1)+rφan(a1,...,am,...,an1)so that by induction there exists a unique homomorphism of R-modulesΦan:A1...An1Lsuch that φan=ΦanιΔ where ιΔ is the natural map from A1×...×An1 to A1...An1. Now defineσ:A1...An1×AnA1...An1σ(a1...an1,an)=a1...an1ψ:A1...An1×AnLψ=Φanσ(Where an in the definition of ψ depends on the input) Prove ψ is bilinear byψ(rai,1...ai,n1+rai,1...ai,n1,an)=Φan(rai,1...ai,n1+rai,1...ai,n1)=rΦan(ai,1...ai,n1)+rΦan(ai,1...ai,n1)=rψ(ai,1...ai,n1,an)+rψ(ai,1...ai,n1,an)as well asψ(ai,1...ai,n1,ran+ran)=Φran+ran(ai,1...ai,n1)=Φran+ran(ai,1...ai,n1)=(Φran+ranιΔ)(ai,1,...,ai,n1)=φran+ran(ai,1,...,ai,n1)=φ(ai,1,...,ai,n1,ran+ran)=[rφ(ai,1,...,an)+rφ(ai,1,...,an)]=rφ(ai,1,...,an)+rφ(ai,1,...,an)=rφ(ai,1,...,an)+rφ(ai,1,...,an)=rψ(ai,1...ai,n1,an)+rψ(ai,1...ai,n1,an)so that there exists a unique homomorphism of R-modulesΦ:A1...AnLsuch that, in part, ψ=Φι on simple tensors where ι is described in the lemma. Now we can showΦι=Φιι=ψι=ΦanσιSince clearly σι=ιΔσ whereσ:A1×...×AnA1×...×An1σ(a1,...,an)=(a1,...,an1)we can continueΦanσι=ΦanιΔσ=φanσ=φThis completed argument in reverse also shows ψ=Φιφ=Φι since ι is surjective on simple tensors, which is seen to extend to byψ(ai,1...ai,n1,an)=ψ(ai,1...ai,n1,an)=(Φι)(ai,1...ai,n1,an)=Φ(ai,1...an)=Φ(ai,1...an)=Φ((ai,1...ai,n1)an)=(Φι)(ai,1...ai,n1,an)This establishes the first part. For the converse, observe(Φι)(a1,...,ram+ram,...,an)=Φ(a1...(ram+ram)...an)=rΦ(a1...am...an)+rΦ(a1...am...an)=r(Φι)(a1,...,am,...,an)+r(Φι)(a1,...,am,...,an) 

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