tensor product
Definition (Standard). Let be a commutative ring, and let be -modules. There exists an -module , called the
tensor product of and over , together with a canonical
bilinear homomorphism
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distinguished, up to isomorphism, by the following universal
property.
Every bilinear -module homomorphism
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lifts to a unique -module homomorphism
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such that
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for all Diagramatically:
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The tensor product can be constructed by taking the free
-module generated by all formal symbols
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and quotienting by the obvious bilinear relations:
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Note. In order to make the base ring clear,
the tensor product is sometimes written as .
Basic properties. Let be a commutative ring and be -modules, then, as modules, we have the following isomorphisms:
- 1.
,
- 2.
,
- 3.
- 4.
Definition (Categorical). Using the language of categories, all
of the above can be expressed quite simply by stating that for all
-modules , the functor is left-adjoint to the
functor .
Comments
Operators on tensor product spaces
I have a question: what is the meaining of
$A \otimes A$ defined on $\widehat{L^2}(\mathbb{R}^2)$, which is just the space of symmetrix $L^2$ functions on $\mathbb{R}^2$ and $A$ is defined as: $A=-\frac{d^2}{dx^2} + x^2 + 1$, i.e. a densly defined linear operator?
Would it help that for Hermite functions $e_n$ we have
$A e_n = (2n_2)e_n$, that is Hermite functions are a basis of $L^2(\mathbb{R})?
A \otimes_R B
Perhaps the entry should mention the $\otimes_R$ syntax?
Re: A \otimes_R B
The entry is currently world-editable, so I've added this.
Tensor Product
I have a quick question about the tensor product of two R-modules A,B.
If, say, B is also an R'-module (R,R' commutative rings with identity) then an R'-module structure can be defined on A*B (Spanier, Algebraic Topology). My problem is it doesn't say how this is defined and i haven't seen this in any other book. If i define it in the most natural way, say if x belongs to A*B then x is a finite sum of elements of the form a(i)*b(i) then let xr'=a(i)*(b(i)r'); but i can't proove that this is well defined in general since the elements of the form a*b do not form a basis for A*B in general.
Re: Tensor Product
I don't see what is not well-defined there, that is the natural method.
See Hungerford's Algebra or Rotman's Advanaced Modern Algebra, or probably any book at that level.
Given that we tensor over R and B is a left R' module, then given r\in R',
(a*b)r:= a*(br)
the only relation to check is the tensore relations: given s\in R
then (as*b)r = (a*sb)r: check:
(as*b)r=(as)*(br)=a*s(br)=a*(sb)r=(a*sb)r
also the sum:
((a+c)*b)r=(a+c)*(br)=a*(br)+c*(br)=(a*b)r+(c*b)r
...so what is missing?
Now it is not the case that this has to be a free R'-module so a basis need not follow. For example, Z_4 * Z_6 = Z_2 and although Z_2 is a Z_6 (and a Z_4) module, it isn't a free Z_4 or Z_6 module (here R=Z and take your pick about R'=Z_6 or Z_4.)
Re: Tensor Product
from the "definition" (a*b)r'=a*(br') and the fact that the set of elements of the from a*b generates A*B as an R-module, the distributivity and associativity relations for multiplication with elements from R', namely
(x1+x2)r'=x1r'+x2r', x(r1'+r2')=xr1'+xr2', (xr1')r2'=x(r1'r2')
where x1,x2 are in A*B, r1',r2' are in R',
are easily checked so indeed this would make A*B an R'-module; but what i meant by well defined is, if say an element x from A*B can be written as a sum of elements of the form a*b in two different ways (possible since A*B is not a free R-module hence doesn't have a basis), e.g.
x=Sum over i a1(i)*b1(i)=Sum over i a2(i)*b2(i)
then it has to be shown that for any r' from R'
xr'=Sum over i a1(i)*(b1(i)r')=Sum over i a2(i)*(b2(i)r'),
otherwise from the prescription (a*b)r'=a*(br') the quantity xr' is ambiguous.
Re: Tensor Product
never mind i've got it, i'm pretty sure R' signifies a factor ring of R in the book, thus B is a R' module means there is an ideal K in R s.t. R'=R/K and K is included in the annihilator of B; then A*B is also an R' module with the definition x(r+K)=xr.
Re: Tensor Product
I don't see a problem here either. You don't need special properties on R' at all. This is universal in a way. If A is a right R-module, B is right R-module (R need not be commutative or unital) and B is also a left S-module (no condition on S), the A tensor B over R is a left S-module. No assumption of free, finitely generated, commutative, nothing. The same goes for Hom(A,B). It is crucial for establishing
homology theory in module categories. but the proof and definition is as I described.
You may want to look into Hungerford or Rotman, both have descent accounts of this.
Re: Tensor Product
I meant B is a left R-module, sorry.