Proof: Lemma 1: We have the natural map on simple tensorsι′:A1⊗...⊗An−1×An→(A1⊗...⊗An−1)⊗An≅A1⊗...⊗An is well defined, and together with the evidently well-defined natural mapι∗:A1×...×An→A1⊗...⊗An−1×Anwe have the relation ι=ι′∘ι∗. Proof: The lemma will follow as soon as we show ι′ is well defined. If (a1⊗...⊗an−1,an)=(a′1⊗...⊗a′n−1,a′n) then by using these two given equivalencies in the coordinates we clearly haveι′(a1⊗...⊗an−1,an)=(a1⊗...⊗an−1)⊗an=(a′1⊗...⊗a′n−1)⊗a′n=ι′(a′1⊗...⊗a′n−1,a′n) ◻Proceed by induction on n. For each an∈An defineφan:A1×...×An−1→Lφan(a1,...,an−1)=φ(a1,...,an)Note that φan is multilinearφan(a1,...,ram+r′a′m,...,an−1)=φ(a1,...,ram+r′a′m,...,an)=rφ(a1,...,am,...,an)+r′φ(a1,...,a′m,...,an)=rφan(a1,...,am,...,an−1)+r‘φan(a1,...,a′m,...,an−1)so that by induction there exists a unique homomorphism of R-modulesΦan:A1⊗...⊗An−1→Lsuch that φan=Φan∘ιΔ where ιΔ is the natural map from A1×...×An−1 to A1⊗...⊗An−1. Now defineσ′:A1⊗...⊗An−1×An→A1⊗...⊗An−1σ′(a1⊗...⊗an−1,an)=a1⊗...⊗an−1ψ:A1⊗...⊗An−1×An→Lψ=Φan∘σ′(Where an in the definition of ψ depends on the input) Prove ψ is bilinear byψ(r∑ai,1⊗...⊗ai,n−1+r′∑a′i,1⊗...⊗a′i,n−1,an)=Φan(r∑ai,1⊗...⊗ai,n−1+r′∑a′i,1⊗...⊗a′i,n−1)=rΦan(∑ai,1⊗...⊗ai,n−1)+r′Φan(∑a′i,1⊗...⊗a′i,n−1)=rψ(∑ai,1⊗...⊗ai,n−1,an)+r′ψ(∑a′i,1⊗...⊗a′i,n−1,an)as well asψ(∑ai,1⊗...⊗ai,n−1,ran+r′a′n)=Φran+r′a′n(∑ai,1⊗...⊗ai,n−1)=∑Φran+r′a′n(ai,1⊗...⊗ai,n−1)=∑(Φran+r′a′n∘ιΔ)(ai,1,...,ai,n−1)=∑φran+r′a′n(ai,1,...,ai,n−1)=∑φ(ai,1,...,ai,n−1,ran+r′a′n)=∑[rφ(ai,1,...,an)+r′φ(ai,1,...,a′n)]=∑rφ(ai,1,...,an)+∑r′φ(ai,1,...,a′n)=r∑φ(ai,1,...,an)+r′∑φ(ai,1,...,a′n)=rψ(∑ai,1⊗...⊗ai,n−1,an)+r′ψ(∑ai,1⊗...⊗ai,n−1,a′n)so that there exists a unique homomorphism of R-modulesΦ:A1⊗...⊗An→Lsuch that, in part, ψ=Φ∘ι′ on simple tensors where ι′ is described in the lemma. Now we can showΦ∘ι=Φ∘ι′∘ι∗=ψ∘ι∗=Φan∘σ′∘ι∗Since clearly σ′∘ι∗=ιΔ∘σ whereσ:A1×...×An→A1×...×An−1σ(a1,...,an)=(a1,...,an−1)we can continueΦan∘σ′∘ι∗=Φan∘ιΔ∘σ=φan∘σ=φThis completed ⇒ argument in reverse also shows ψ=Φ∘ι′⇔φ=Φ∘ι since ι∗ is surjective on simple tensors, which is seen to extend to ⇐ byψ(∑ai,1⊗...⊗ai,n−1,an)=∑ψ(ai,1⊗...⊗ai,n−1,an)=∑(Φ∘ι′)(ai,1⊗...⊗ai,n−1,an)=∑Φ(ai,1⊗...⊗an)=Φ(∑ai,1⊗...⊗an)=Φ((∑ai,1⊗...⊗ai,n−1)⊗an)=(Φ∘ι′)(∑ai,1⊗...⊗ai,n−1,an)This establishes the first part. For the converse, observe(Φ∘ι)(a1,...,ram+r′a′m,...,an)=Φ(a1⊗...⊗(ram+r′a′m)⊗...⊗an)=rΦ(a1⊗...⊗am⊗...⊗an)+r′Φ(a1⊗...⊗a′m⊗...⊗an)=r(Φ∘ι)(a1,...,am,...,an)+r′(Φ∘ι)(a1,...,a′m,...,an) ◻
Tuesday, June 11, 2013
Universal Multilinear Object (10.4 Corollary 16)
Dummit and Foote Abstract Algebra, section 10.4, corollary 16:
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Show that if R is a commutative ring, Ai is an R-module, L is an abelian group, and φ:A1×...×An→L is a multilinear map, then there exists a unique homomorphism of R-modules Φ:A1⊗...⊗An→L such that φ=Φ∘ι, 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⊗...⊗An→L then φ=Φ∘ι is a multilinear map.
Proof: Lemma 1: We have the natural map on simple tensorsι′:A1⊗...⊗An−1×An→(A1⊗...⊗An−1)⊗An≅A1⊗...⊗An is well defined, and together with the evidently well-defined natural mapι∗:A1×...×An→A1⊗...⊗An−1×Anwe have the relation ι=ι′∘ι∗. Proof: The lemma will follow as soon as we show ι′ is well defined. If (a1⊗...⊗an−1,an)=(a′1⊗...⊗a′n−1,a′n) then by using these two given equivalencies in the coordinates we clearly haveι′(a1⊗...⊗an−1,an)=(a1⊗...⊗an−1)⊗an=(a′1⊗...⊗a′n−1)⊗a′n=ι′(a′1⊗...⊗a′n−1,a′n) ◻Proceed by induction on n. For each an∈An defineφan:A1×...×An−1→Lφan(a1,...,an−1)=φ(a1,...,an)Note that φan is multilinearφan(a1,...,ram+r′a′m,...,an−1)=φ(a1,...,ram+r′a′m,...,an)=rφ(a1,...,am,...,an)+r′φ(a1,...,a′m,...,an)=rφan(a1,...,am,...,an−1)+r‘φan(a1,...,a′m,...,an−1)so that by induction there exists a unique homomorphism of R-modulesΦan:A1⊗...⊗An−1→Lsuch that φan=Φan∘ιΔ where ιΔ is the natural map from A1×...×An−1 to A1⊗...⊗An−1. Now defineσ′:A1⊗...⊗An−1×An→A1⊗...⊗An−1σ′(a1⊗...⊗an−1,an)=a1⊗...⊗an−1ψ:A1⊗...⊗An−1×An→Lψ=Φan∘σ′(Where an in the definition of ψ depends on the input) Prove ψ is bilinear byψ(r∑ai,1⊗...⊗ai,n−1+r′∑a′i,1⊗...⊗a′i,n−1,an)=Φan(r∑ai,1⊗...⊗ai,n−1+r′∑a′i,1⊗...⊗a′i,n−1)=rΦan(∑ai,1⊗...⊗ai,n−1)+r′Φan(∑a′i,1⊗...⊗a′i,n−1)=rψ(∑ai,1⊗...⊗ai,n−1,an)+r′ψ(∑a′i,1⊗...⊗a′i,n−1,an)as well asψ(∑ai,1⊗...⊗ai,n−1,ran+r′a′n)=Φran+r′a′n(∑ai,1⊗...⊗ai,n−1)=∑Φran+r′a′n(ai,1⊗...⊗ai,n−1)=∑(Φran+r′a′n∘ιΔ)(ai,1,...,ai,n−1)=∑φran+r′a′n(ai,1,...,ai,n−1)=∑φ(ai,1,...,ai,n−1,ran+r′a′n)=∑[rφ(ai,1,...,an)+r′φ(ai,1,...,a′n)]=∑rφ(ai,1,...,an)+∑r′φ(ai,1,...,a′n)=r∑φ(ai,1,...,an)+r′∑φ(ai,1,...,a′n)=rψ(∑ai,1⊗...⊗ai,n−1,an)+r′ψ(∑ai,1⊗...⊗ai,n−1,a′n)so that there exists a unique homomorphism of R-modulesΦ:A1⊗...⊗An→Lsuch that, in part, ψ=Φ∘ι′ on simple tensors where ι′ is described in the lemma. Now we can showΦ∘ι=Φ∘ι′∘ι∗=ψ∘ι∗=Φan∘σ′∘ι∗Since clearly σ′∘ι∗=ιΔ∘σ whereσ:A1×...×An→A1×...×An−1σ(a1,...,an)=(a1,...,an−1)we can continueΦan∘σ′∘ι∗=Φan∘ιΔ∘σ=φan∘σ=φThis completed ⇒ argument in reverse also shows ψ=Φ∘ι′⇔φ=Φ∘ι since ι∗ is surjective on simple tensors, which is seen to extend to ⇐ byψ(∑ai,1⊗...⊗ai,n−1,an)=∑ψ(ai,1⊗...⊗ai,n−1,an)=∑(Φ∘ι′)(ai,1⊗...⊗ai,n−1,an)=∑Φ(ai,1⊗...⊗an)=Φ(∑ai,1⊗...⊗an)=Φ((∑ai,1⊗...⊗ai,n−1)⊗an)=(Φ∘ι′)(∑ai,1⊗...⊗ai,n−1,an)This establishes the first part. For the converse, observe(Φ∘ι)(a1,...,ram+r′a′m,...,an)=Φ(a1⊗...⊗(ram+r′a′m)⊗...⊗an)=rΦ(a1⊗...⊗am⊗...⊗an)+r′Φ(a1⊗...⊗a′m⊗...⊗an)=r(Φ∘ι)(a1,...,am,...,an)+r′(Φ∘ι)(a1,...,a′m,...,an) ◻
Proof: Lemma 1: We have the natural map on simple tensorsι′:A1⊗...⊗An−1×An→(A1⊗...⊗An−1)⊗An≅A1⊗...⊗An is well defined, and together with the evidently well-defined natural mapι∗:A1×...×An→A1⊗...⊗An−1×Anwe have the relation ι=ι′∘ι∗. Proof: The lemma will follow as soon as we show ι′ is well defined. If (a1⊗...⊗an−1,an)=(a′1⊗...⊗a′n−1,a′n) then by using these two given equivalencies in the coordinates we clearly haveι′(a1⊗...⊗an−1,an)=(a1⊗...⊗an−1)⊗an=(a′1⊗...⊗a′n−1)⊗a′n=ι′(a′1⊗...⊗a′n−1,a′n) ◻Proceed by induction on n. For each an∈An defineφan:A1×...×An−1→Lφan(a1,...,an−1)=φ(a1,...,an)Note that φan is multilinearφan(a1,...,ram+r′a′m,...,an−1)=φ(a1,...,ram+r′a′m,...,an)=rφ(a1,...,am,...,an)+r′φ(a1,...,a′m,...,an)=rφan(a1,...,am,...,an−1)+r‘φan(a1,...,a′m,...,an−1)so that by induction there exists a unique homomorphism of R-modulesΦan:A1⊗...⊗An−1→Lsuch that φan=Φan∘ιΔ where ιΔ is the natural map from A1×...×An−1 to A1⊗...⊗An−1. Now defineσ′:A1⊗...⊗An−1×An→A1⊗...⊗An−1σ′(a1⊗...⊗an−1,an)=a1⊗...⊗an−1ψ:A1⊗...⊗An−1×An→Lψ=Φan∘σ′(Where an in the definition of ψ depends on the input) Prove ψ is bilinear byψ(r∑ai,1⊗...⊗ai,n−1+r′∑a′i,1⊗...⊗a′i,n−1,an)=Φan(r∑ai,1⊗...⊗ai,n−1+r′∑a′i,1⊗...⊗a′i,n−1)=rΦan(∑ai,1⊗...⊗ai,n−1)+r′Φan(∑a′i,1⊗...⊗a′i,n−1)=rψ(∑ai,1⊗...⊗ai,n−1,an)+r′ψ(∑a′i,1⊗...⊗a′i,n−1,an)as well asψ(∑ai,1⊗...⊗ai,n−1,ran+r′a′n)=Φran+r′a′n(∑ai,1⊗...⊗ai,n−1)=∑Φran+r′a′n(ai,1⊗...⊗ai,n−1)=∑(Φran+r′a′n∘ιΔ)(ai,1,...,ai,n−1)=∑φran+r′a′n(ai,1,...,ai,n−1)=∑φ(ai,1,...,ai,n−1,ran+r′a′n)=∑[rφ(ai,1,...,an)+r′φ(ai,1,...,a′n)]=∑rφ(ai,1,...,an)+∑r′φ(ai,1,...,a′n)=r∑φ(ai,1,...,an)+r′∑φ(ai,1,...,a′n)=rψ(∑ai,1⊗...⊗ai,n−1,an)+r′ψ(∑ai,1⊗...⊗ai,n−1,a′n)so that there exists a unique homomorphism of R-modulesΦ:A1⊗...⊗An→Lsuch that, in part, ψ=Φ∘ι′ on simple tensors where ι′ is described in the lemma. Now we can showΦ∘ι=Φ∘ι′∘ι∗=ψ∘ι∗=Φan∘σ′∘ι∗Since clearly σ′∘ι∗=ιΔ∘σ whereσ:A1×...×An→A1×...×An−1σ(a1,...,an)=(a1,...,an−1)we can continueΦan∘σ′∘ι∗=Φan∘ιΔ∘σ=φan∘σ=φThis completed ⇒ argument in reverse also shows ψ=Φ∘ι′⇔φ=Φ∘ι since ι∗ is surjective on simple tensors, which is seen to extend to ⇐ byψ(∑ai,1⊗...⊗ai,n−1,an)=∑ψ(ai,1⊗...⊗ai,n−1,an)=∑(Φ∘ι′)(ai,1⊗...⊗ai,n−1,an)=∑Φ(ai,1⊗...⊗an)=Φ(∑ai,1⊗...⊗an)=Φ((∑ai,1⊗...⊗ai,n−1)⊗an)=(Φ∘ι′)(∑ai,1⊗...⊗ai,n−1,an)This establishes the first part. For the converse, observe(Φ∘ι)(a1,...,ram+r′a′m,...,an)=Φ(a1⊗...⊗(ram+r′a′m)⊗...⊗an)=rΦ(a1⊗...⊗am⊗...⊗an)+r′Φ(a1⊗...⊗a′m⊗...⊗an)=r(Φ∘ι)(a1,...,am,...,an)+r′(Φ∘ι)(a1,...,a′m,...,an) ◻
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