Basis of a free algebra: linear independence of tensor powers

Question

It is said that the tensor $T(M)$ of an $R$-module $M$ is a free $R$-algebra on $X$ when $X$ is a basis of $M$. Rotman in his book Advanced Modern Algebra, Part I claims that if $X = (x_i)_{i \in I}$ is a basis of $M$, every element of $T(M)$ has a unique expression as $$\sum_{p \geq 0} \sum_{i_1,...,i_n \in I} r_{i_1...i_p}(x_{i_1}\otimes ... \otimes x_{i_p}).$$ I personally don't understand why the expression would be unique. It is clear that every $m \in T(M)$ uniquely decomposes as a sum $\sum_{p \geq 0} m_p$ where $m_p \in \bigotimes^p M$ since $T(M)$ is the direct sum $\bigoplus_{p \geq 0} \bigotimes^p M$. However, it's not clear to me that $\{x_{i_1}\otimes ... \otimes x_{i_p} \mid i_1,...,i_p \in I \}$ is a linear independent subset of $\bigotimes^p M$ for $p \geq 1$. Indeed, it is known that the canonical tensor map $\otimes^p\colon \prod^p M\to \bigotimes^p M$ which sends $(m_1,...,m_p)$ to $m_1\otimes ... \otimes m_p$ is not injective.

Answer 1

It follows from two main facts.

$(1)$ If $M$ and $N$ are free $R$-modules for a commutative ring $R$ with bases $(e_i)_{i \in I}$ and $(f_j)_{j \in J}$, then $M\otimes_R N$ has a basis $(e_i\otimes f_j)_{i \in I, j \in J}$.

For a proof, see these notes by Keith Conrad (specifically, page $16$).

From this, due to the associativity of tensor product and the fact that $R$-module isomorphisms send bases to bases, it follows, by simple induction, that if $M_1,...,M_n$ are free $R$-modules for a commutative ring $R$ with bases $(e_{1i})_{i \in I_1}, ..., (e_{ni})_{i \in I_n}$, then $M_1\otimes ... \otimes M_n$ has a basis $(e_{1i_1}\otimes ... \otimes e_{ni_n})_{i_1 \in I_1,...,i_n \in I_n}$.

$(2)$ If $(M_i)_{i \in I}$ is a family of free $R$-modules with bases $(e_{ij})_{j \in J_i}$, then $\bigcup_{i \in I}\{e_{ij} \mid j \in J_i\}$ is a basis of $\bigoplus_{i \in I} M_i$.