nLab
tensor product of algebras

Idea

Let $R$ be a commutative ring. The category of associative algebras over $R$ is the category

$${\mathrm{Alg}}_R={\mathrm{Ring}}^{R/}$$Alg_R = Ring^{R/}

of rings under $R$. If $R$ is a commutative rig, we can do the same with

$${\mathrm{Alg}}_R={\mathrm{Rig}}^{R/}.$$Alg_R = Rig^{R/} .

The tensor product of $R$-algebras has as underlying $R$-module just the tensor product of modules of the underlying modules, $A{\otimes }_{R}B$. On homogeneous elements $(a,b)\in A\times B\stackrel{\otimes }{\to }A{\otimes }_{R}B$ the algebra structure is given by

$$(a_1,b_1)\cdot (a_2,b_2)=(a_1\cdot a_2,b_1\cdot b_2).$$(a_1, b_1) \cdot (a_2, b_2) = (a_1 \cdot a_2, b_1 \cdot b_2) \,.

We write also $A{\otimes }_{R}B$ for the tensor product of algebras.

For commutative $R$-algebras, the tensor product is the coproduct in $\mathrm{Comm}{\mathrm{Alg}}_R$:

$$A{\otimes }_{R}B\simeq A\coprod B\in \mathrm{Comm}{\mathrm{Alg}}_R=\mathrm{Comm}{\mathrm{Rig}}^{R/};$$A \otimes_R B \simeq A \coprod B \in Comm Alg_R = Comm Rig^{R/} ;

hence the pushout in Comm Ring? (or Comm Rig?)

$$\begin{array}{ccc}& & R\\ & \swarrow & & \searrow \\ A& & & & B\\ & \searrow & & \swarrow \\ & & A{\otimes }_{R}B\end{array}.$$\array{ && R \\ & \swarrow && \searrow \\ A &&&& B \\ & \searrow && \swarrow \\ && A \otimes_R B } \,.

Properties

Relation to tensor product of categories of modules

For $A$ an associative algebra over a field $k$, write $A$Mod for its category of modules of finite dimension. Then the tensor product of algebras corresponds to the Deligne tensor product of abelian categories $⊠:\mathrm{Ab}\times \mathrm{Ab}\to \mathrm{Ab}$:

$$(A{\otimes }_{k}B)\mathrm{Mod}\simeq (A\mathrm{Mod})\otimes (B\mathrm{Mod}).$$(A \otimes_k B) Mod \simeq (A Mod) \otimes (B Mod) \,.

See at tensor product of abelian categories for more.

Related concepts

• tensor product

• Hochschild cohomology