Tensor product of algebras over a field; itself another algebra
In mathematics, the tensor product of two algebras over a commutative ring R is also an R-algebra. This gives the tensor product of algebras. When the ring is a field, the most common application of such products is to describe the product of algebra representations.
Let R be a commutative ring and let A and B be R-algebras. Since A and B may both be regarded as R-modules, their tensor product
![{\displaystyle A\otimes _{R}B}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4773c5f4d6541ec487737f1c3962e1aafb7fe55)
is also an R-module. The tensor product can be given the structure of a ring by defining the product on elements of the form a ⊗ b by[1]
![{\displaystyle (a_{1}\otimes b_{1})(a_{2}\otimes b_{2})=a_{1}a_{2}\otimes b_{1}b_{2))](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3886de4308e35029777a6920b87fbebed9d6090)
and then extending by linearity to all of A ⊗R B. This ring is an R-algebra, associative and unital with identity element given by 1A ⊗ 1B.[3] where 1A and 1B are the identity elements of A and B. If A and B are commutative, then the tensor product is commutative as well.
The tensor product turns the category of R-algebras into a symmetric monoidal category.[citation needed]
There are natural homomorphisms from A and B to A ⊗R B given by[4]
![{\displaystyle a\mapsto a\otimes 1_{B))](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d17374fdbe613e35bb45140b4560d24943baaa1)
![{\displaystyle b\mapsto 1_{A}\otimes b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9cf636c64b93ebd9a13daa664257c5628bcce4bf)
These maps make the tensor product the coproduct in the category of commutative R-algebras. The tensor product is not the coproduct in the category of all R-algebras. There the coproduct is given by a more general free product of algebras. Nevertheless, the tensor product of non-commutative algebras can be described by a universal property similar to that of the coproduct:
![{\displaystyle {\text{Hom))(A\otimes B,X)\cong \lbrace (f,g)\in {\text{Hom))(A,X)\times {\text{Hom))(B,X)\mid \forall a\in A,b\in B:[f(a),g(b)]=0\rbrace ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b41ec985950c28907175a6ed4509acf9d8441bab)
where [-, -] denotes the commutator.
The natural isomorphism is given by identifying a morphism
on the left hand side with the pair of morphisms
on the right hand side where
and similarly
.
The tensor product of commutative algebras is of frequent use in algebraic geometry. For affine schemes X, Y, Z with morphisms from X and Z to Y, so X = Spec(A), Y = Spec(R), and Z = Spec(B) for some commutative rings A, R, B, the fiber product scheme is the affine scheme corresponding to the tensor product of algebras:
![{\displaystyle X\times _{Y}Z=\operatorname {Spec} (A\otimes _{R}B).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b7f7dcda91bb552a0528ca3162bce64239ee34d)
More generally, the fiber product of schemes is defined by gluing together affine fiber products of this form.
- The tensor product can be used as a means of taking intersections of two subschemes in a scheme: consider the
-algebras
,
, then their tensor product is
, which describes the intersection of the algebraic curves f = 0 and g = 0 in the affine plane over C.
- More generally, if
is a commutative ring and
are ideals, then
, with a unique isomorphism sending
to
.
- Tensor products can be used as a means of changing coefficients. For example,
and
.
- Tensor products also can be used for taking products of affine schemes over a field. For example,
is isomorphic to the algebra
which corresponds to an affine surface in
if f and g are not zero.
- Given
-algebras
and
whose underlying rings are graded-commutative rings, the tensor product
becomes a graded commutative ring by defining
for homogeneous
,
,
, and
.