# nLab tensor product of algebras

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## Higher algebras

• symmetric monoidal (∞,1)-category of spectra

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# Contents

## Idea

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

$\mathsf{Alg}_R = {R \downarrow \mathsf{Ring}}$

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

$\mathsf{Alg}_R = {R \downarrow \mathsf{Rig}} .$

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

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

We write also $A \otimes_R B$ for the tensor product of algebras.

For commutative $R$-algebras, the tensor product is the coproduct in $Comm Alg_R$:

$A \otimes_R B \simeq A \coprod B \in Comm Alg_R = Comm R \downarrow \mathsf{Rig} ;$

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

$\array{ && R \\ & \swarrow && \searrow \\ A &&&& B \\ & \searrow && \swarrow \\ && A \otimes_R B } \,.$

## 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 $\boxtimes \colon Ab \times Ab \to Ab$:

$(A \otimes_k B) Mod \simeq (A Mod) \otimes (B Mod) \,.$

See at tensor product of abelian categories for more.

Last revised on March 23, 2016 at 03:22:29. See the history of this page for a list of all contributions to it.