# Tensor product of algebras

{{ safesubst:#invoke:Unsubst||$N=Refimprove |date=__DATE__ |$B= {{#invoke:Message box|ambox}} }} 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.

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

$A\otimes _{R}B$ which is also an R-module. We can give the tensor product the structure of an algebra by defining

$(a_{1}\otimes b_{1})(a_{2}\otimes b_{2})=a_{1}a_{2}\otimes b_{1}b_{2}$ and then extending by linearity to all of A ⊗RB. This product is easily seen to be R-bilinear, associative, and unital with an identity element given by 1A ⊗ 1B, where 1A and 1B are the identities of A and B. If A and B are both commutative then the tensor product is as well.

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

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

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

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