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Naoto Meguro. Amateur.
MSC2010. Primary 13C13; Secondary 13C99.
Key Words and Phrases. The tensor product, the balanced map.
The abstract. Theory of tensor products has defect.
Master Grothendieck suggested defect of the tensor product. I tried to guess it like the next.
Let A be a commutative ring. Let M and N be A-modules.
Define M⊗"AN=(⊕(m,n)∈M×N Ae(m,n))/(⊕e∈K Ae), m⊗" n=e(m,n)+(⊕e∈K Ae),
K=∪m∈M, m"∈M, n∈N, n"∈N, a∈A {e(m+m",n)-e(m,n)-e(m",n), e(m,n+n")-e(m,n)-e(m,n"), e(am,n)-e(m,an)}.
Let's think the case that A=M=N=C. t:C×C∋(z,z")→Re(zz")∈C is a C-balanced map.
Generally, v:M⊗"AN∋a1(m1⊗"n1)+…+ak(mk⊗"nk)→a1w(m1,n1)+…+akw(mk,nk) for the A-balanced map
w:M×N→L.⇒v∈HomA(M⊗"AN,L) and v(m⊗"n)=w (m,n). (L is an A-module. ⊗" is universal.)
If w=t, v(i(1⊗"1))=iv(1⊗"1)=i≠0=v(i⊗"1). So i(1⊗"1)≠i⊗"1.
Define M⊗AN=(⊕(m,n)∈M×N Ze(m,n))/(⊕e∈K Ze). Generally,
v": M⊗"AN∋a1(m1⊗"n1)+…+ak(mk⊗"nk)→w"(a1m1,n1)+…+w"(akmk,nk) for the A-balanced map
w": M×N→L".⇒ v"∈HomZ(M⊗"AN,L") and v"(m⊗"n)=w"(m,n). (L" is a Z-module.)
Assume that M⊗"AN≅M⊗AN as Z-modules by an isomorphism s for which s⊗=⊗".
When M=N=A=C, M⊗AN={z⊗1|z∈C}.
For i(1⊗"1)∈M⊗"AN,∃z((z∈C)∧(i(1⊗"1)=s(z⊗1)=z⊗"1)). When w=t, i=v(i(1⊗"1))=v(z⊗"1)=Re(z)∈R.
This is contradiction. The tensor product isn't decided uniquely. ∃P(M⊗"AN=(M⊗AN)⊕P).
Let I be an ideal of A. Put h:(A/I)⊗"AM∋a1((a"1+I)⊗"m1)+…+ak((a"k+I)⊗"mk)→a1a"1m1+…+aka"kmk+IM
When I={0} and A=M=C, h(i(1⊗"1)=h(i⊗"1)=i though i(1⊗"1)≠i⊗"1. h isn't a bijection.
(A/I)⊗"AM≅M/IM may not be formed.
Let f: M→Q be an isomorphism of A-modules. Put g: Q→0. g⊗"1N: Q⊗"AN→0.
When w": M×N∋(m,n) → f(m)⊗"n and A=M=N=C and Q=Im f ⊂C and L"=Q⊗"AN,
Ker g⊗"1N=Q⊗"AN∋ i(1⊗"1) ∉{z⊗"1 |z∈C}⊃Im v". You can define f⊗"1N=v" mechanically.
0→M⊗"AN→Q⊗"AN→0 (exact) isn't formed then though 0→0→M→Q→0 (exact).
f⊗"1N isn't decided uniquely.
Master Grothendieck suggested that the trouble is solvable. (Add the ae(m,n)-e(am,n) 's to K. You may not
need to think the balanced map.)
MSC2010. Primary 13C13; Secondary 13C99.
Key Words and Phrases. The tensor product, the balanced map.
The abstract. Theory of tensor products has defect.
1
Master Grothendieck suggested defect of the tensor product. I tried to guess it like the next.
2
Let A be a commutative ring. Let M and N be A-modules.
Define M⊗"AN=(⊕(m,n)∈M×N Ae(m,n))/(⊕e∈K Ae), m⊗" n=e(m,n)+(⊕e∈K Ae),
K=∪m∈M, m"∈M, n∈N, n"∈N, a∈A {e(m+m",n)-e(m,n)-e(m",n), e(m,n+n")-e(m,n)-e(m,n"), e(am,n)-e(m,an)}.
Let's think the case that A=M=N=C. t:C×C∋(z,z")→Re(zz")∈C is a C-balanced map.
Generally, v:M⊗"AN∋a1(m1⊗"n1)+…+ak(mk⊗"nk)→a1w(m1,n1)+…+akw(mk,nk) for the A-balanced map
w:M×N→L.⇒v∈HomA(M⊗"AN,L) and v(m⊗"n)=w (m,n). (L is an A-module. ⊗" is universal.)
If w=t, v(i(1⊗"1))=iv(1⊗"1)=i≠0=v(i⊗"1). So i(1⊗"1)≠i⊗"1.
Define M⊗AN=(⊕(m,n)∈M×N Ze(m,n))/(⊕e∈K Ze). Generally,
v": M⊗"AN∋a1(m1⊗"n1)+…+ak(mk⊗"nk)→w"(a1m1,n1)+…+w"(akmk,nk) for the A-balanced map
w": M×N→L".⇒ v"∈HomZ(M⊗"AN,L") and v"(m⊗"n)=w"(m,n). (L" is a Z-module.)
Assume that M⊗"AN≅M⊗AN as Z-modules by an isomorphism s for which s⊗=⊗".
When M=N=A=C, M⊗AN={z⊗1|z∈C}.
For i(1⊗"1)∈M⊗"AN,∃z((z∈C)∧(i(1⊗"1)=s(z⊗1)=z⊗"1)). When w=t, i=v(i(1⊗"1))=v(z⊗"1)=Re(z)∈R.
This is contradiction. The tensor product isn't decided uniquely. ∃P(M⊗"AN=(M⊗AN)⊕P).
Let I be an ideal of A. Put h:(A/I)⊗"AM∋a1((a"1+I)⊗"m1)+…+ak((a"k+I)⊗"mk)→a1a"1m1+…+aka"kmk+IM
When I={0} and A=M=C, h(i(1⊗"1)=h(i⊗"1)=i though i(1⊗"1)≠i⊗"1. h isn't a bijection.
(A/I)⊗"AM≅M/IM may not be formed.
Let f: M→Q be an isomorphism of A-modules. Put g: Q→0. g⊗"1N: Q⊗"AN→0.
When w": M×N∋(m,n) → f(m)⊗"n and A=M=N=C and Q=Im f ⊂C and L"=Q⊗"AN,
Ker g⊗"1N=Q⊗"AN∋ i(1⊗"1) ∉{z⊗"1 |z∈C}⊃Im v". You can define f⊗"1N=v" mechanically.
0→M⊗"AN→Q⊗"AN→0 (exact) isn't formed then though 0→0→M→Q→0 (exact).
f⊗"1N isn't decided uniquely.
3
Master Grothendieck suggested that the trouble is solvable. (Add the ae(m,n)-e(am,n) 's to K. You may not
need to think the balanced map.)