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Hometensor product of algebras
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tensor product of algebras
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:
$\xymatrix{A\ar[dr]^{f}\ar[d]&\\ D\ar[r]^{h}&C\\ B\ar[ur]_{g}\ar[u]&}$ 
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