How can I set a Polynomial Algebra? I was reading about Algebra over a field, and I see the definition of $\mathbb{K}\langle X \rangle$ as follows:
Let $X \neq \emptyset$ a set, a word $w$ is an expression in the following way:
$w=x_{1}^{m_{1}}x_{2}^{m_{2}} \cdots x_{k}^{m_{k}}, \space \forall x_{i} \in X$ and $\forall m_{i} \in \mathbb{N}$, $\forall i \in \{1,2, \cdots, k\}$ where $k \in \mathbb{N^{*}}$.
Then the algebra $\mathbb{K}\langle X \rangle$ is based on each word $w$ defined above. The bilinear product is the concatenation of the corresponding words.
Does anyone know a pdf that has a more formal setting (including showing why $\mathbb{K}\langle X \rangle$ is a vector space, defining its operations, etc ...) or could answer me with a more formal definition?
Thanks!

If in addition also show how $\mathbb{K}\langle X \rangle$ generalizes the polynomials ring $\mathbb{K}\left[ X \right]$ would be great !!

Jim, I didn't know the definition of free vector space , so I searched on W. Greub, Linear Algebra, Springer-Verlag, Fourth edition, 1975. Let's see if I understand:
$\mathbb{K}\langle X \rangle := \{f: \langle X \rangle \longrightarrow \mathbb{K}/ \space f^{-1}(\mathbb{K^{*}}) \mbox{ is finite}\}$, where $\langle X \rangle$ is the set of words with letters in $X$. This is a vector space if we consider the usual sum of functions and the usual scalar multiplication.
So one basis for $\mathbb{K}\langle X \rangle$ is:
$B=\{f_{w}/ \space w \in \langle X \rangle\}$, where $f_{w}:\langle X \rangle \longrightarrow \mathbb{K}$ is a function, such that:
$f_{w}(x)=\begin{cases} 1_{\mathbb{K}} & \mbox{if } x=w \\ 0_{\mathbb{K}} & \mbox{if } x\neq w\\ \end{cases}$.
This is basis because:


*

*$B$ generate $\mathbb{K}\langle X \rangle$:
$\forall f \in \mathbb{K}\langle X \rangle$ we have $f^{-1}(\mathbb{K^{*}})=\{ w_{1}, w_{2}, \cdots, w_{n} \}$, for $n \in \mathbb{N}$ and $w_{i}\in\langle X \rangle$, $1 \leq i \leq n$. In this way:
$f=\sum_{i=1}^{n}f(w_{i})f_{w_{i}}\in [B] \Longrightarrow \mathbb{K}\langle X \rangle=[B]$.

*$B$ is L.I.:
Let $\{\lambda_{w}\}_{w \in B}$ be a family of scalars in $\mathbb{K}$ such that:
$g=\sum_{w \in B}\lambda_{w}f_{w}=0_{\mathbb{K}\langle X \rangle}$.
So $\forall w \in B, \space g(w)=0_{\mathbb{K}}\Longrightarrow \forall w \in B, \space \lambda_{w}=0_{\mathbb{K}}$.
I agree with you, to define a bilinear map (multiplication) one only needs to define what happens to basis vectors. In this case, how can I match the base $B$ with the words in $\langle X \rangle$? (I'm not able to see as a function $f_{w} \in B \subset
 \mathbb{K}\langle X \rangle$ can be a word in $\langle X \rangle$).
 A: The reason $\mathbb K\langle X\rangle$ is a vector space is because it's defined that way!  $\mathbb K\langle X\rangle$ is the free vector space with basis the words with letters in $X$.  Then to define a bilinear map $\mathbb K\langle X\rangle \otimes \mathbb K\langle X\rangle \to \mathbb K\langle X\rangle$ (multiplication) one only needs to define what happens to basis vectors.  Well, if $\mathbb K\langle X\rangle$ has basis corresponding to words in $X$, then $\mathbb K\langle X\rangle \otimes \mathbb K\langle X\rangle$ has basis corresponding to pairs of words in $X$.  So a map $\mathbb K\langle X\rangle \otimes \mathbb K\langle X\rangle \to \mathbb K\langle X\rangle$ can be defined by saying how to create a word in $X$ from a pair of words in $X$.  And for multiplication this is done using concatenation.
One reason $\mathbb K\langle X\rangle$ generalizes $\mathbb K[X]$ is simply because one is a quotient of the other.  You just mod $\mathbb K\langle X\rangle$ by the ideal generated by all commutators of letters.  Another reason is that $\mathbb K[X]$ satisfies a universal property with respect to commutative $\mathbb K$-algebras, whereas $\mathbb K\langle X\rangle$ satisfies essentially the same universal property with respect to all (not necessarily commutative) $\mathbb K$-algebras.
