# Tensor product of algebras

(Redirected from Tensor product of rings)

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.

## Definition

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

${\displaystyle A\otimes _{R}B}$

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 ab by[1][2]

${\displaystyle (a_{1}\otimes b_{1})(a_{2}\otimes b_{2})=a_{1}a_{2}\otimes b_{1}b_{2}}$

and then extending by linearity to all of AR 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]

## Further properties

There are natural homomorphisms of A and B to A ⊗RB given by[4]

${\displaystyle a\mapsto a\otimes 1_{B}}$
${\displaystyle b\mapsto 1_{A}\otimes b}$

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:

${\displaystyle Hom(A\otimes B,X)\cong \lbrace (f,g)\in Hom(A,X)\times Hom(B,X)\mid \forall a\in A,b\in B:[f(a),g(b)]=0\rbrace }$

The natural isomorphism is given by identifying a morphism ${\displaystyle \phi :A\otimes B\to X}$ on the left hand side with the pair of morphisms ${\displaystyle (f,g)}$ on the right hand side where ${\displaystyle f(a):=\phi (a\otimes 1)}$ and similarly ${\displaystyle g(b):=\phi (1\otimes b)}$.

## Applications

The tensor product of commutative algebras is of constant use in algebraic geometry. For affine schemes X, Y, Z with morphisms from X and Z to Y, so X = Spec(A), Y = Spec(B), and Z = Spec(C) for some commutative rings A,B,C, the fiber product scheme is the affine scheme corresponding to the tensor product of algebras:

${\displaystyle X\times _{Y}Z=\operatorname {Spec} (A\otimes _{B}C).}$

More generally, the fiber product of schemes is defined by gluing together affine fiber products of this form.

## Examples

• The tensor product can be used as a means of taking intersections of two subschemes in a scheme: consider the ${\displaystyle \mathbb {C} [x,y]}$-algebras ${\displaystyle \mathbb {C} [x,y]/f}$, ${\displaystyle \mathbb {C} [x,y]/g}$, then their tensor product is ${\displaystyle \mathbb {C} [x,y]/(f)\otimes _{\mathbb {C} [x,y]}\mathbb {C} [x,y]/(g)\cong \mathbb {C} [x,y]/(f,g)}$, which describes the intersection of the algebraic curves f = 0 and g = 0 in the affine plane over C.
• Tensor products can be used as a means of changing coefficients. For example, ${\displaystyle \mathbb {Z} [x,y]/(x^{3}+5x^{2}+x-1)\otimes _{\mathbb {Z} }\mathbb {Z} /5\cong \mathbb {Z} /5[x,y]/(x^{3}+x-1)}$ and ${\displaystyle \mathbb {Z} [x,y]/(f)\otimes _{\mathbb {Z} }\mathbb {C} \cong \mathbb {C} [x,y]/(f)}$.
• Tensor products also can be used for taking products of affine schemes over a field. For example, ${\displaystyle \mathbb {C} [x_{1},x_{2}]/(f(x))\otimes _{\mathbb {C} }\mathbb {C} [y_{1},y_{2}]/(g(y))}$ is isomorphic to the algebra ${\displaystyle \mathbb {C} [x_{1},x_{2},y_{1},y_{2}]/(f(x),g(y))}$ which corresponds to an affine surface in ${\displaystyle \mathbb {A} _{\mathbb {C} }^{4}}$ if f and g are not zero.