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@goo_blog
Naoto Meguro. Amateur.
MSC 2010. Primary 13C13; Secondary 13C99.
Key Words and Phrases. tensor products, the balanced map.
The abstract. Theory of tesor products has defect.
Master Grothendieck suggested defect of the tensor products. I try 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, n∈N, m"∈M, 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)}.
If L is an A-module and w:M×N→L is an A-balanced map and
v: M⊗"AN∋a1(m1⊗"n1)+…+ak(mk⊗"nk) → a1w(m1,n1)+…+akw(mk,nk),
v∈HomA(M⊗"AN, L) and v⊗"=w. v is decided uniquely by w.
When M=N=A=L=C and w(z,z")=Re(zz") and a∉R, v(a(m⊗"n))=av(m⊗"n)=aRe(mn)∉R and
v(m"⊗"n")=Re(m"n")∈R. So a(m⊗"n)≠m"⊗"n" for a∉R and m,m"∈M and n,n"∈N.
Theorem 1. C⊗"CC≅C isn't formed.
Proof. If w(z,z")=zz", v(0⊗"0)=0 and v(1⊗"1)=1 and v(i⊗"1)=i. So 1⊗"1≠0⊗"0 and i⊗"1≠0⊗"0.
If s(∈HomA(C⊗"CC,C)) is an isomorphism, s(0⊗"0)=0 .
Put c=s(1⊗1) and c"=s(i⊗1). s is a bijection. c≠0. c"≠0.
s(i⊗"1)=c"=(c"/c)s(1⊗"1)=s((c"/c)(1⊗"1)). i⊗"1=(c"/c)(1⊗"1) for s is a bijection. When w(z,z")=Re(zz"),
0=v(i⊗"1)=v((c"/c)(1⊗"1))=(c"/c)v(1⊗"1). c"=0. This is contradiction.♦
Generally, (A/I)⊗"M≅M/IM isn't formed. (Put A=M=C and I={0}.)
If L" is a Z-module and w":M×N→L" is an A-balanced map and
v": M⊗"AN∋a1(m1⊗"n1)+…+ak(mk⊗"nk)→ w"(a1m1,n1)+…+w"(akmk,nk),
v"∈HomZ(M⊗"AN,L") and v"⊗"=w".
Put p=v" when w"=⊗" and L" =M⊗"AN. pp=p. So M⊗"AN= Im(p)⊕Ker(p).
Im(p)={m1⊗"n1+…+mk⊗"nk| mh∈M,nh∈N.}.
Put u: Im(p)→u"(m1,n1)+…+u"(mk,nk) for an A-balanced map u".
Im(p) and p are decided uniquely by A,M and N. u is decided uniquely by u". If u"=⊗",
Define M⊗AN=(⊕(m,n)∈M×N Ze(m,n))/(⊕e∈KZe), m⊗n=e(m,n)+⊕e∈K Ze
Assume that M⊗AN≅M⊗"AN as Z-modules by an isomorphism u for which u⊗=⊗". If M=N=A=C,
M⊗AN={z⊗1|z∈C}. ∃z((z∈C)∧(i(1⊗"1)=u(z⊗1)=z⊗"1. This is contradiction.
MSC 2010. Primary 13C13; Secondary 13C99.
Key Words and Phrases. tensor products, the balanced map.
The abstract. Theory of tesor products has defect.
1
Master Grothendieck suggested defect of the tensor products. I try 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, n∈N, m"∈M, 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)}.
If L is an A-module and w:M×N→L is an A-balanced map and
v: M⊗"AN∋a1(m1⊗"n1)+…+ak(mk⊗"nk) → a1w(m1,n1)+…+akw(mk,nk),
v∈HomA(M⊗"AN, L) and v⊗"=w. v is decided uniquely by w.
When M=N=A=L=C and w(z,z")=Re(zz") and a∉R, v(a(m⊗"n))=av(m⊗"n)=aRe(mn)∉R and
v(m"⊗"n")=Re(m"n")∈R. So a(m⊗"n)≠m"⊗"n" for a∉R and m,m"∈M and n,n"∈N.
Theorem 1. C⊗"CC≅C isn't formed.
Proof. If w(z,z")=zz", v(0⊗"0)=0 and v(1⊗"1)=1 and v(i⊗"1)=i. So 1⊗"1≠0⊗"0 and i⊗"1≠0⊗"0.
If s(∈HomA(C⊗"CC,C)) is an isomorphism, s(0⊗"0)=0 .
Put c=s(1⊗1) and c"=s(i⊗1). s is a bijection. c≠0. c"≠0.
s(i⊗"1)=c"=(c"/c)s(1⊗"1)=s((c"/c)(1⊗"1)). i⊗"1=(c"/c)(1⊗"1) for s is a bijection. When w(z,z")=Re(zz"),
0=v(i⊗"1)=v((c"/c)(1⊗"1))=(c"/c)v(1⊗"1). c"=0. This is contradiction.♦
Generally, (A/I)⊗"M≅M/IM isn't formed. (Put A=M=C and I={0}.)
If L" is a Z-module and w":M×N→L" is an A-balanced map and
v": M⊗"AN∋a1(m1⊗"n1)+…+ak(mk⊗"nk)→ w"(a1m1,n1)+…+w"(akmk,nk),
v"∈HomZ(M⊗"AN,L") and v"⊗"=w".
Put p=v" when w"=⊗" and L" =M⊗"AN. pp=p. So M⊗"AN= Im(p)⊕Ker(p).
Im(p)={m1⊗"n1+…+mk⊗"nk| mh∈M,nh∈N.}.
Put u: Im(p)→u"(m1,n1)+…+u"(mk,nk) for an A-balanced map u".
Im(p) and p are decided uniquely by A,M and N. u is decided uniquely by u". If u"=⊗",
Define M⊗AN=(⊕(m,n)∈M×N Ze(m,n))/(⊕e∈KZe), m⊗n=e(m,n)+⊕e∈K Ze
Assume that M⊗AN≅M⊗"AN as Z-modules by an isomorphism u for which u⊗=⊗". If M=N=A=C,
M⊗AN={z⊗1|z∈C}. ∃z((z∈C)∧(i(1⊗"1)=u(z⊗1)=z⊗"1. This is contradiction.
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