Why do we take $R$ when constructing the tensor algebra?l

Question

Let $R$ be a commutative ring and $M$ be an $R$-module.

Define $T^0(M)=R$ and $T^n(M)=M\otimes...\otimes M$(n-times) for $n\in\mathbb{Z}^+$.

Then we take $T(M)\triangleq \oplus T^n(M)$ and give an operation to make it an $R$-algebra.

My question is why do we take $T^0(M)$? What's the role of $T^0(M)$ in the tensor algebra $T(M)$?

And I don't get how to define an operation on $T(M)$.

It's written in my text that we take an operation as $(m_1\otimes...\otimes m_i)(n_1\otimes...\otimes n_j)=(m_1\otimes...\otimes m_i\otimes n_1\otimes...\otimes n_j)$ but what does this mean when $i$ or $j$ is $0$?

Answer 1

You want $T(M)$ to be an $R$-algebra. Thus, the role of $T^0(M)$ is to give you an algebra homomorphism $R\stackrel{\sim}{\to} T^0(M)\to T(M)$.

When $i=0$ and $j>0$ you define $r\cdot (n_1\otimes \dots\otimes n_j)=(rn_1)\otimes \dots\otimes n_j$, similarly for $j=0$. When $i=0=j$, then the multiplication in $T(M)$ is just the same as in $R$.