The tensor algebra

April 18, 2012

In this post I will define the tensor algebra over a module and show that it satisfies a universal property which makes it the universal algebra with respect to algebras containing an homomorphic image of that module.

Let $M$ be an $R$ -module for $R$ a commutative and unitary ring. For each $k \geq 1,$ define

$T^k(M) = M^{\otimes _k} = \underbrace{M \otimes M \otimes \ldots \otimes M}_{\text{k times}}.$

where $\otimes$ denotes tensor product. Note that this definition is unambiguous considering the fact that taking tensor products is an associative construction. For $k = 0,$ define $T^0(M) = R.$ Define the tensor algebra to be

$T(M) = T^0(M) \oplus T^1(M) \oplus T^2(M) \oplus \cdots = \displaystyle\bigoplus_{k=0}^{\infty}T^k(M).$

Recall that the infinite direct sum of a family of modules $\{M_i\}_{i \in I}$ is the subset of the infinite product $\{(m_1, m_2, \ldots ) \mid \forall i \in I, m_i \in M_i \}$ such that all but finitely many of the $m_i$ ‘s are $0.$ For all $k,$ the elements of $T^k(M)$ are called k-tensors. Thus, every element of $T(M)$ can be viewed as a finite linear combination of k-tensors for various $k \geq 0.$ Identifying $M$ with $T^1(M),$ we can view $M$ as a submodule of $T(M).$

Now, defining addition componentwise in $T(M)$ gives it the structure of an $R$ -module. To get an $R$ -algebra structure, we define the multiplication in $T(M)$ as

$(m_1 \otimes \cdots \otimes m_k)(m_1' \otimes \cdots \otimes m_l') = m_1 \otimes \cdots \otimes m_k \otimes m_1' \otimes \cdots \otimes m_l'$

and extend it by linearity and distributivity to all of $T(M).$

Proposition: This multiplication is well-defined and makes $T(M)$ into an $R$ -algebra.

Proof: By the proof of the associativity of tensor products, there exists a unique isomorphism $\phi : T^k(M) \otimes T^l(M) \to T^{k+l}(M)$ such that

$\phi ((m_1 \otimes \cdots \otimes m_k) \otimes (m_1' \otimes \cdots \otimes m_l')) = m_1 \otimes \cdots \otimes m_k \otimes m_1' \otimes \cdots \otimes m_l'.$

Thus the multiplication map above is well defined. Indeed, suppose $m_1 \otimes \cdots \otimes m_k = n_1 \otimes \cdots \otimes n_k \in T^k(M).$ Then

$m_1 \otimes \cdots \otimes m_k \otimes m_1' \otimes \cdots \otimes m_l' - n_1 \otimes \cdots \otimes n_k \otimes m_1' \otimes \cdots \otimes m_l'$

$= \phi((m_1 \otimes \cdots \otimes m_k) \otimes (m_1' \otimes \cdots \otimes m_l'))$

$- \phi((n_1 \otimes \cdots \otimes n_k) \otimes (m_1' \otimes \cdots \otimes m_l'))$

$= \phi ((m_1 \otimes \cdots \otimes m_k - n_1 \otimes \cdots \otimes n_k) \otimes (m_1' \otimes \cdots \otimes m_l'))$

$= 0.$

QED

In fact, since $T^k(M)T^l(M) \subseteq T^{k+l}(M)$ , the $R$ -algebra $T(M)$ is a graded algebra. Here is the universal property satisfied by the tensor algebra:

Theorem (Universal property): Let $A$ be any $R$ -algebra, $M$ an $R$ -module. If $\varphi : M \to A$ is any $R$ -module homomorphism, then there is a unique $R$ -algebra homomorphism $\phi : T(M) \to A$ such that $\phi |_M = \varphi.$

Proof: For every $k,$ the map $\tilde{\varphi}_k : M^k \to A$ defined by $\tilde{\varphi}_k(m_1, \ldots, m_k) = \varphi(m_1)\cdots\varphi(m_k)$ is easily seen to be k-multilinear with respect to $R.$ Thus for every $k$ there exists a unique $R$ -module homomorphism $\tilde{\phi}_k : T^k(M) \to A$ such that $\tilde{\phi}_k(m_1 \otimes \cdots \otimes m_k) = \varphi(m_1)\cdots \varphi(m_k).$ With the definition of the multiplication in $T(M),$ it is easy to see that the induced map $\tilde{\phi} : T(M) \to A$ is an $R$ -algebra homomorphism satisfying the desired properties.