# Tensor algebra becomes an R-algebra. Theorem, Proof, Dummit and Foote

I have the definition of tensor algebra as follows: $T(M) = \bigoplus_{i =1}^{\infty} T^i(M)$, where $M$ is an $R$ module, where $R$ is commutative and contains the element $1$. Finally $T^k(M) = \bigotimes_{i =1}^{k} M$, Now i need to prove that $T(M)$ is an R-algebra. The book that im using is Dummit and Foote. I have a presentation about this topic and I need to proof this theorem in class. I will show how much I have with details to see if this proof is complete.

Definition:R-algebra

Let $R$ be a commutative ring. An $R$-algebra is a ring $A$ which is also an $R$-module such that the multiplication map $A \times A \rightarrow A$ is bilinear, that is, $$r*(ab) = (r*a)b = a(r*b), \ \mbox{for} \ a,b \in A, r \in R$$ where $*$ denotes the $R$-action on $A$.

To proof $(1)$, then it is enough to proof that multiplication is well defined and the map that defines multiplication is bilinear. with that being said, lets consider the map: $$\varphi: \underbrace{M \times \cdots \times M}_{i}\times \underbrace{M \times \cdots \times M}_{j} \longrightarrow T^{i+j}(M)$$ We define the multiplication as follows: $$\varphi(m_1, \ldots, m_i, m_1', \dots, m_j') = m_1\otimes \ldots \otimes m_i \otimes m_1' \ldots \otimes m_j'$$ Now such map is multilinear, since: $$\varphi(m_1, \ldots, rm_i+ r'\bar{m_i}, m_1', \dots, m_j') = m_1\otimes \ldots \otimes (rm_i+r'\bar{m_i}) \otimes m_1' \ldots \otimes m_j' = r\varphi(m_1, \ldots, m_i, m_1', \dots, m_j') + r'\varphi(m_1, \ldots, \bar{m_i}, m_1', \dots, m_j')$$ Clearly, this induces a bilinear map: $\phi : T^i(M) \times T^j(M) \longrightarrow T^{i+j}(M)$, since for $x \in T^i(M)$ and $y \in T^j(M)$ and $$\phi(x,y) = x \otimes y$$ we have that $$\phi(r_1 x_1 + r_2x_2, y) = (r_1 x_1 + r_2x_2)\otimes y = r_1\phi(x_1,y) + r_2\phi(x_2,y)$$ $$\phi(x, r_1 y_1 + r_2y_2) = x \otimes (r_1 y_1 + r_2y_2) = r_1\phi(x,y_1) + r_2\phi(x,y_2)$$ Hence, the map $\phi$ is bilinear, so extending the map to sums via the distributive laws, we obtain that $T(M)$ becomes an $R$-algebra that satisfies $$T^i(M) \otimes T^j(M) \subseteq T^{i+j}(M)$$ To prove that the action $\otimes$ is well defined, let $$\pi : T^i(M) \times T^j(M) \times T^i(M) \times T^j(M) \longrightarrow T^{i+j}(M)$$ by $\pi(a_1,b_1,a_2,b_2) = a_1a_2 \otimes b_1b_2$. Clearly, the map $\pi$ is multilinear, then by corollary 16 of the book, there exist a corresponding $R$-module homomorphism $$\bar{\pi}: T^i(M) \otimes T^j(M) \otimes T^i(M) \otimes T^j(M) \longrightarrow T^{i+j}(M)$$ such that $\bar{\pi}(a_1\otimes b_1\otimes a_2 \otimes b_2) = a_1a_2 \otimes b_1b_2$. On the other hand, we can view $T^i(M) \otimes T^j(M) \otimes T^i(M) \otimes T^j(M)$ as follows $$(T^i(M) \otimes T^j(M)) \times (T^i(M) \otimes T^j(M))$$ Using corollary 16 again, we have a map $$\Phi: (T^i(M) \otimes T^j(M)) \times (T^i(M) \otimes T^j(M)) \longrightarrow T^{i+j}(M)$$ such that $\Phi(a_1\otimes b_1, a_2 \otimes b_2) = a_1a_2 \otimes b_1b_2$ so multiplication is indeed well defined.

To prove $(2)$, lets assume that $\varphi: M \rightarrow A$ is an $R$-module homomorphism, then the map $(m_1,m_2, \ldots, m_k) \mapsto \varphi(m_1) \cdots \varphi(m_k)$ defines an $R$-multilinear map from $M \times \cdots \times M$ to $A$. Then by corollary 16, this induces a unique $R$-module homomorphism $\pi$ from $T^k(M)$ to $A$ which maps $$\pi(m_1 \otimes \cdots \otimes m_k) \mapsto \varphi(m_1) \cdots \varphi(m_k)$$ but then we can combine this linear maps that give a linear map: $$\Phi: T(M) \rightarrow A, \ \ \Phi(1_{T(M)}) = 1_A$$
that we define on simple tensors as $$\Phi(m_1 \otimes \cdots \otimes m_k) =\pi(m_1 \otimes \cdots \otimes m_k) = \varphi(m_1) \cdots \varphi(m_k)$$ Moreover, we have: $$\Phi(a,m,m_1\otimes m_2,...) = a\cdot1_A + \phi(m) + \phi(m_1)\phi(m_2) + ...$$

• Is there a question here? – Jonathan Rayner Apr 24 at 13:48