Free algebra

Remarks

Universal property: for every associative $k$-algebra $A$ with identity and every map $f \colon X \to A$, there is a unique homomorphism of $k$-algebras $\bar f \colon k\langle X \rangle \to A$ with $\bar f(x) = f(x)$ for $x \in X$. It sends a word $x_1 \cdots x_n$ to $f(x_1) \cdots f(x_n)$.

Example: for $X = \set{x}$, the words are the powers $x^n$, and $k\langle x \rangle$ is the polynomial ring $k[x]$.

Non-example: for $X = \set{x, y}$, the polynomial ring $k[x, y]$ is not free on $X$. The map $x \mapsto e_{12}$, $y \mapsto e_{21}$ into $2 \times 2$ matrices over $k$ has no extension to a homomorphism from $k[x, y]$, because $e_{12} e_{21} = e_{11} \ne e_{22} = e_{21} e_{12}$ while $xy = yx$ in $k[x, y]$.

The free algebra is the monoid ring $k[X^*]$. It is isomorphic to the tensor algebra $T(V)$ of the free $k$-module $V$ with basis $X$, graded by word length. For $\abs{X} \ge 2$ and $k$ a field, it is a domain that is not commutative.