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Hometensor product

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# tensor product

Summary. The tensor product is a formal bilinear multiplication of two modules or vector spaces. In essence, it permits us to replace bilinear maps from two such objects by an equivalent linear map from the tensor product of the two objects. The origin of this operation lies in classic differential geometry and physics, which had need of multiply indexed geometric objects such as the first and second fundamental forms, and the stress tensor — see Tensor Product (Classical).

Definition (Standard). Let $R$ be a commutative ring, and let $A,B$ be $R$-modules. There exists an $R$-module $A\otimes B$, called the
*tensor product* of $A$ and $B$ over $R$, together with a canonical
bilinear homomorphism

$\otimes:A\times B\rightarrow A\otimes B,$ |

distinguished, up to isomorphism, by the following universal property. Every bilinear $R$-module homomorphism

$\phi:A\times B\rightarrow C,$ |

lifts to a unique $R$-module homomorphism

$\tilde{\phi}:A\otimes B\rightarrow C,$ |

such that

$\phi(a,b)=\tilde{\phi}(a\otimes b)$ |

for all $a\in A,\;b\in B.$ Diagramatically:

$\xymatrix{\ar[dr]^{(}.55)\phi\ar[r]^{\otimes}A\times B&A\otimes B\ar@{-->}[d]^% {(}.4){\exists!\,\tilde{\phi}}\\ &C}$ |

The tensor product $A\otimes B$ can be constructed by taking the free $R$-module generated by all formal symbols

$a\otimes b,\quad a\in A,\;b\in B,$ |

and quotienting by the obvious bilinear relations:

$\displaystyle(a_{1}+a_{2})\otimes b$ | $\displaystyle=a_{1}\otimes b+a_{2}\otimes b,$ | $\displaystyle a_{1},a_{2}\in A,\;b\in B$ | |||

$\displaystyle a\otimes(b_{1}+b_{2})$ | $\displaystyle=a\otimes b_{1}+a\otimes b_{2},$ | $\displaystyle a\in A,\;b_{1},b_{2}\in B$ | |||

$\displaystyle r(a\otimes b)$ | $\displaystyle=(ra)\otimes b=a\otimes(rb)$ | $\displaystyle a\in A,\;b\in B,\;r\in R$ |

Note. In order to make the base ring $R$ clear, the tensor product $A\otimes B$ is sometimes written as $A\otimes_{R}B$.

Basic properties. Let $R$ be a commutative ring and $L,M,N$ be $R$-modules, then, as modules, we have the following isomorphisms:

1. $R\otimes M\cong M$,

2. $M\otimes N\cong N\otimes M$,

3. $(L\otimes M)\otimes N\cong L\otimes(M\otimes N)$

4. $(L\oplus M)\otimes N\cong(L\otimes N)\oplus(M\otimes N)$

Definition (Categorical). Using the language of categories, all of the above can be expressed quite simply by stating that for all $R$-modules $M$, the functor $(-)\otimes M$ is left-adjoint to the functor $\mathrm{Hom}(M,-)$.

## Mathematics Subject Classification

13-00*no label found*18-00

*no label found*

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## 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.