# Tensor product of algebras

(Redirected from Tensor product of R-algebras)

In mathematics, the tensor product of two R-algebras is also an R-algebra. This gives us a tensor product of algebras. The special case R = Z gives us a tensor product of rings, since rings may be regarded as Z-algebras.[1]

## 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, we may form their tensor product

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

which is also an R-module. We can give the tensor product the structure of an algebra by defining the product on elements of the form ab by[2][3]

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

and then extending by linearity to all of AR B. This product is R-bilinear, associative, and unital with an identity element given by 1A ⊗ 1B,[4] where 1A and 1B are the identities of A and B. If A and B are both commutative then the tensor product is commutative as well.

The tensor product turns the category of all R-algebras into a symmetric monoidal category.[citation needed]

## Further properties

There are natural homomorphisms of A and B to A ⊗RB given by[5]

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

These maps make the tensor product a 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 an 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 morphism ${\displaystyle (f,g)}$ 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 algebras is of constant use in algebraic geometry: working in the opposite category to that of commutative R-algebras, it provides pullbacks of affine schemes, otherwise known as fiber products.

## 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 ${\displaystyle \mathbb {C} [x,y]/(f)\otimes _{\mathbb {C} [x,y]}\mathbb {C} [x,y]/(g)\cong \mathbb {C} [x,y]/(f,g)}$.
• 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 point. 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 ${\displaystyle \mathbb {A} _{\mathbb {C} }^{4}}$.