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Revision difference : tensor algebra
Version 8 Version 7
Let $R$ be a commutative ring, and $M$ an $R$-module. Let $R$ be a commutative ring, and $M$ an $R$-module.
The \emph{tensor algebra} The tensor algebra
\[ \mc{T}(M) = \bigoplus_{n=0}^\infty \mc{T}_n(M)\] \[ \mc{T}(M) = \bigoplus_{n=0}^\infty \mc{T}_n(M)\]
is the graded $R$-algebra with $n$-th is the graded $R$-algebra with $n$-th
graded component simply the $n$th tensor power: graded component simply the $n$th tensor power:
\[ \mc{T}_n(M) = M^{\otimes n} =\overbrace{M\otimes \cdots \otimes \[ \mc{T}_n(M) = M^{\otimes n} =\overbrace{M\otimes \cdots \otimes
M}^{n\text{ times}},\quad n=1,2,\ldots,\] M}^{n\text{ times}},\quad n=1,2,\ldots,\]
and $\mc{T}_0(M)=R$. and $\mc{T}_0(M)=R$.
The multiplication $m:\mc{T}(M)\times \mc{T}(M)\to\mc{T}(M)$ is given The multiplication $m:\mc{T}(M)\times \mc{T}(M)\to\mc{T}(M)$ is given
by the usual tensor product: by the usual tensor product:
\[ m(a,b)=a\otimes b,\quad a\in M^{\otimes n},\; b\in M^{\otimes m}.\] \[ m(a,b)=a\otimes b,\quad a\in M^{\otimes n},\; b\in M^{\otimes m}.\]
\paragraph{Remark 1.} One can generalize the above definition to \paragraph{Remark 1.} One can generalize the above definition to
cover the case where the ground ring $R$ is non-commutative by cover the case where the ground ring $R$ is non-commutative by
requiring that the module $M$ is a bimodule with $R$ acting on both requiring that the module $M$ is a bimodule with $R$ acting on both
the left and the right. the left and the right.
\paragraph{Remark 2.} From the point of view of category theory, one \paragraph{Remark 2.} From the point of view of category theory, one
can describe the tensor algebra construction as a functor $\mc{T}$ can describe the tensor algebra construction as a functor $\mc{T}$
from the category of $R$-module to the category of $R$-algebras that from the category of $R$-module to the category of $R$-algebras that
is left-adjoint to the forgetful functor $\mc{F}$ from algebras to is left-adjoint to the forgetful functor $\mc{F}$ from algebras to
modules. Thus, for $M$ an $R$-module and $S$ an $R$-algebra, every modules. Thus, for $M$ an $R$-module and $S$ an $R$-algebra, every
module homomorphism $M\to \mc{F}(S)$ extends to a unique algebra module homomorphism $M\to \mc{F}(S)$ extends to a unique algebra
homomorphism $\mc{T}(M)\to S$. homomorphism $\mc{T}(M)\to S$.