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:

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 $K⟨X⟩$ and we will find many parallels to polynomial algebras with indeterminants in $X$.

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

$$\left(\sum _{w\in FM⟨X⟩}{l}_{w}w\right)\left(\sum _{v\in FM⟨X⟩}{l}_{v}v\right)=\sum _{w,v\in FM⟨X⟩}{l}_v{l}_{w}wv.$$

For example, $\mathbb{Q}⟨x,y⟩$ 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, $K⟨X⟩\to K[X]$ 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:

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

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 $K⟨X⟩$ is homogeneous if it contained in $FM⟨X⟩$. 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 $K⟨X⟩$.

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