Let $R$ be a ring (commutative with unit), and $E$ a set. We denote $A$ the free $R$-algebra on $E$.
Note. $A$ is a non-commutative algebra.
I wanted to show that if we consider $A$ as a left $A$-module, then the set $E$ is $A$-linearly independant. That is, if $\sum_{e \in E} a_e e_e = 0$, then $a_e = 0$ for all $e \in E$ (Note. $a_e \in A$).
Since $A$ is free on $E$, we already know that the set of all words in E and the empty word form an $R$-basis of $A$. I think with that I have all I need for the proof, but I miss something on how to use this fact.
Any hint or help would be great.
Best regards.
K. Y.
Write $E = \{e_1, \ldots, e_n\}$ and assume $\sum_{i=1}^n a_ie_i = 0$ with $a_i \in A$. Every $a_i$ can be uniquely written as \begin{equation} a_i = \sum_{k=0}^\infty \sum_{i_1, \ldots, i_k \in \{1, \ldots, n\}} a^{(i)}_{i_1,\ldots, i_k} e_{i_1}\ldots e_{i_k}\tag{1}\end{equation} for some $a^{(i)}_{i_1,\ldots, i_k} \in R$ with at most finitely many of them nonzero. Plugging this in we get \begin{align} 0 &= \sum_{i=1}^n a_ie_i \\ &= \sum_{i=1}^n\left(\sum_{k=0}^\infty \sum_{i_1, \ldots i_k \in \{1, \ldots, n\}} a^{(i)}_{i_1,\ldots, i_k} e_{i_1}\ldots e_{i_k}\right)e_i\\ &= \sum_{i=1}^n\sum_{k=0}^\infty \sum_{i_1, \ldots i_k \in \{1, \ldots, n\}} a^{(i)}_{i_1,\ldots, i_k} e_{i_1}\ldots e_{i_k}e_i\\ &= \sum_{k=0}^\infty \sum_{i_1, \ldots, i_k,i \in \{1, \ldots, n\}} a^{(i)}_{i_1,\ldots, i_k} e_{i_1}\ldots e_{i_k}e_i. \end{align} Now the monomials $e_{i_1}\ldots e_{i_k}e_i$ are all different since they are parameterized by $(i_1, \ldots, i_k,i) \in \{1,\ldots, n\}^{k+1}$ so by definition it follows that all scalars are $0$: $$a^{(i)}_{i_1,\ldots, i_k}=0, \quad \forall i_1, \ldots, i_k,i \in \{1, \ldots, n\}.$$
Therefore $(1)$ implies $a_i = 0$ for all $1 \le i \le n$ so $E$ is $A$-linearly independent.