# tensor product

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## Mathematics Subject Classification

13-00 General reference works (handbooks, dictionaries, bibliographies, etc.)
18-00 General reference works (handbooks, dictionaries, bibliographies, etc.)

### 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

### 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.