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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.

QED

Reference: D. S. Dummit, R. M. Foote, Abstract Algebra, 3rd ed.

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