# free associative algebra

Fix a commutative^{} unital ring $K$ and a set $X$. Then a $K$-algebra^{} $F$ is
said to be *free on $X$* if there exists an injection $\iota :X\to F$ such that for all functions $f:X\to A$ where $A$ is an $K$-algebra
determine a unique algebra homomorphism $\widehat{f}:F\to A$ such that
$\iota \widehat{f}=f$. This is an example of a universal mapping property for
free associative algebras and in categorical settings is often explained
with the following commutative diagram^{}:

$$\text{xymatrix}\mathrm{\&}X\text{ar}{[ld]}_{\iota}\text{ar}{[rd]}^{f}\mathrm{\&}F\text{ar}{[rr]}^{\widehat{f}}\mathrm{\&}\mathrm{\&}A.$$ |

To prove that free associative algebras exist in the category^{} of all
associative algebras we provide a couple standard constructions. It
is a standard categorical procedure to conclude any two free objects
on the same set are naturally equivalent and thus each construction
below is equivalent^{}.

## 1 Tensor algebra

Let $X$ be a set and $K$ a commutative unital ring. Then take $M$ to be any free $K$-module with basis $X$, and injection $\iota :X\to M$. Then we may form the tensor algebra of $M$,

$$T(M)=\underset{i\in \mathbb{N}}{\oplus}{T}^{i}(M),{T}^{i}(M)={M}^{\otimes i}=\underset{j=1}{\overset{i}{\otimes}}M.$$ |

[Note, $0\in \mathbb{N}$ and the empty tensor we define as $K$.]
Furthermore, define the injection ${\iota}^{\prime}:X\to T(M)$ as the map $\iota :X\to M$ followed by the embedding^{} of $M$ into $T(M)$.

###### Remark 1.

To make $M$ concrete use the set of all functions $f\mathrm{:}X\mathrm{\to}K$, or
equivalently, the direct product^{} ${\mathrm{\prod}}_{X}K$. Then the tensor algebra
of $M$ is the free algebra^{} on $X$.

###### Proposition 2.

$(T(M),{\iota}^{\prime})$ is a free associative algebra on $X$.

###### Proof.

Given any associative $K$-algebra $A$ and function $f:X\to A$, then
$A$ is a $K$-module and $M$ is free on $X$ so $f$ extends to a unique
$K$-linear homomorphism^{} $\widehat{f}:M\to A$.

Next we define $K$-multilinear maps ${f}^{(i)}:{M}^{i}\to A$ by

$${f}^{(i)}({m}_{1},\mathrm{\dots},{m}_{i})=f({m}_{1})\mathrm{\cdots}f({m}_{i}).$$ |

Then by the universal mapping property of tensor products^{} (used inductively)
we have a unique $K$-linear map ${\widehat{f}}^{(i)}:{T}^{i}(M)\to A$ for which

$${\widehat{f}}^{(i)}({m}_{1}\otimes \mathrm{\cdots}\otimes {m}_{i})=f({m}_{1})\mathrm{\cdots}f({m}_{i}).$$ |

Thus we have a unique algebra homomorphism ${\widehat{f}}^{(\mathrm{\infty})}:T(M)\to A$ such that $\iota \widehat{f}=f$. ∎

This construction provides an obvious grading on the free algebra where the homogeneous components are

$${T}^{n}(M)={M}^{\otimes n}=\underset{j=1}{\overset{n}{\otimes}}M.$$ |

## 2 Non-commutative polynomials

An alternative construction is to model the methods of constructing free
groups^{} and semi-groups, that is, to use words on the set $X$. We will
denote the result of this construction by $K\u27e8X\u27e9$ and we will
find many parallels to polynomial algebras with indeterminants in $X$.

Let $FM\u27e8X\u27e9$ be the set of all words on $X$. This makes
$FM\u27e8X\u27e9$ a free monoid with identity^{} the empty word and
associative product^{} the juxtaposition of words. Then define
$K\u27e8X\u27e9$ as the $K$-semi-group algebra on $FM\u27e8X\u27e9$.
This means $K\u27e8X\u27e9$ is the free $K$-modules oN $FM\u27e8X\u27e9$
and the product is defined as:

$$\left(\sum _{w\in FM\u27e8X\u27e9}{l}_{w}w\right)\left(\sum _{v\in FM\u27e8X\u27e9}{l}_{v}v\right)=\sum _{w,v\in FM\u27e8X\u27e9}{l}_{v}{l}_{w}wv.$$ |

For example, $\mathbb{Q}\u27e8x,y\u27e9$ contains elements of the form

$${x}^{2}+4yxy,-7xy+2yx,1+x+xy+xyx+{x}^{2}y+{x}^{2}{y}^{2}.$$ |

This model of a free associative algebra encourages a mapping to polynomial
rings^{}. Indeed, $K\u27e8X\u27e9\to K[X]$ is uniquely determined by
the free property applied to the natural inclusion of $X$ into $K[X]$.
What we realize this mapping in a practical fashion we note that this simply
allows all indeterminants to commute. It follows from this that $K[X]$ is
a free commutative associaitve algebra.

For example, under this map we translate^{} the above elements into:

$${x}^{2}+4x{y}^{2},-5xy,1+x+xy+2{x}^{2}y+{x}^{2}{y}^{2}.$$ |

We also note that the grading detected in the tensor algebra construction
persists in the non-commuting polynomial model. In particular, we say an
element in $K\u27e8X\u27e9$ is homogeneous^{} if it contained in
$FM\u27e8X\u27e9$. Then the degree of a homogeneous element^{} is the
length of the word. Then the $K$-linear span of elements of degree $i$
form the $i$-th graded component^{} of $K\u27e8X\u27e9$.

###### Remark 3.

We note that the free properties of both of these constructions depend
in turn on the free properties of modules, the universal property of
tensors and free semi-groups. An inspection of the common construction
of tensors and free modules^{} reveals both of these have universal properties
implied from the universal mapping property of free semi-groups. Thus
we may assert that free of associative algebras are a direct result of
the existence of free semi-groups.

For non-associative algebras such as Lie and Jordan algebras^{}, the
universal properties are more subtle.

Title | free associative algebra |
---|---|

Canonical name | FreeAssociativeAlgebra |

Date of creation | 2013-03-22 16:51:07 |

Last modified on | 2013-03-22 16:51:07 |

Owner | Algeboy (12884) |

Last modified by | Algeboy (12884) |

Numerical id | 10 |

Author | Algeboy (12884) |

Entry type | Definition |

Classification | msc 08B20 |

Related topic | Algebras |

Related topic | TensorAlgebra |

Defines | free associative algebra |