Let $A$ and $B$ be algebras over a commutative ring $R$ . As modules, we can form the tensor product $A\otimes B$ . The resulting structure is again an $R$ -module. Since $A$ and $B$ have the additional structure of being algebras, we would like this structure being preserved in $A\otimes B$ as well. This can indeed be done, if we define a ``multiplication'' $\cdot: (A\otimes B)\times (A\otimes B)\to A\otimes B$ by: $$(a\otimes b)\cdot (c\otimes d):=ac\otimes bd, \qquad a,c\in A\mbox{ and }b,d\in B$$ and requiring that $\cdot$ distributes over $+$ , that is, $$x\cdot (y+z):=x\cdot y+x\cdot z,\qquad(y+z)\cdot x:=y\cdot x+z\cdot x, \qquad x,y,z\in A\otimes B.$$ With this and the fact that $\cdot$ is associative, $(A\otimes B, +, \cdot)$ becomes a ring. To turn $A\otimes B$ into an $R$ -algebra, we need to verify that $\cdot$ is bilinear, that is: $$r(x\cdot y)=(rx)\cdot y = x\cdot (ry)\qquad x,y\in A\otimes B\mbox{ and }r\in R.$$ Because of the distributivity of $\cdot$ over $+$ , it is enough to verify the case when $x=a\otimes b$ and $y=c\otimes d$ . Then $r((a\otimes b)\cdot (c\otimes d))=r(ac\otimes bd)= (rac)\otimes bd = (ra\otimes b)\cdot (c\otimes d)=(r(a\otimes b))\cdot (c\otimes d)$ . This shows that $r(x\cdot y)=(rx)\cdot y$ . Since $(rac)\otimes bd = ac\otimes (rbd)$ , we also have $r(x\cdot y)=x\cdot (ry)$ . Therefore, $A\otimes B$ is an algebra over $R$ .

Here are some basic properties of the tensor product of algebras:

• if $A$ and $B$ are both unital, so is $A\otimes B$ , with $1\otimes 1$ as the multiplicative identity
• if $A$ and $B$ are both commutative, so is $A\otimes B$
• $A\otimes B\cong B\otimes A$
• $(A\otimes B)\otimes C\cong A\otimes (B\otimes C)$
• $(A\oplus B)\otimes C\cong (A\otimes C) \oplus (B\otimes C)$
• $R\otimes A\cong A$
• Assume both $A$ and $B$ are unital algebras, the canonical injections $A\to A\otimes B$ and $B\to A\otimes B$ , given by $$a\mapsto a\otimes 1,\qquad b\mapsto 1\otimes b$$ turn $A,B$ into subalgebras of $A\otimes B$ (up to algebra isomorphism). In fact, $A$ and $B$ are commuting subalgebras, in the sense that $$(A\otimes 1)\cdot (1\otimes B)=(1\otimes B)\cdot (A\otimes 1).$$

Like tensor products of modules, there is also a universal property associated with the tensor products of algebras: let $A$ and $B$ be $R$ -algebras and $f:A\to C$ and $g:B\to C$ be algebra homomorphisms such that $f(A)g(B)=g(B)f(A)$ and $f(R\cdot 1)=g(R\cdot 1)$ . Then there is a unique $R$ -algebra $D$ (=$A\otimes B$ ) and algebra homomorphism $h: D\to C$ , such that the following diagram of algebra homomorphisms commutes: