tensor product

Summary. The tensor product is a formal bilinear multiplicationof two modules or vector spaces. In essence, it permits us to replacebilinear maps from two such objects by an equivalent linear map fromthe tensor product of the two objects. The origin of this operationlies in classic differential geometry and physics, which had need ofmultiply indexed geometric objects such as the first and secondfundamental forms, and the stress tensor — see Tensor Product (Classical) (http://planetmath.org/TensorProductClassical).

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

$$\otimes :A\times B\to A\otimes B,$$

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

$$\varphi :A\times B\to C,$$

lifts to a unique $R$-module homomorphism

$$\tilde{\varphi }:A\otimes B\to C,$$

such that

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

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

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

$$a\otimes b,a\in A,b\in B,$$

and quotienting by the obvious bilinear relations:

$(a_1+a_2)\otimes b$$=a_1\otimes b+a_2\otimes b,$$a_1,a_2\in A,b\in B$

$a\otimes (b_1+b_2)$$=a\otimes b_1+a\otimes b_2,$$a\in A,b_1,b_2\in B$

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

Note.

Basic . 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, allof the above can be expressed quite simply by stating that for all$R$-modules $M$, the functor $(-)\otimes M$ is left-adjoint to thefunctor $\mathrm{Hom}(M,-)$.