# 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 $\iota:X\rightarrow F$ such that for all functions $f:X\rightarrow A$ where $A$ is an $K$-algebra determine a unique algebra homomorphism $\hat{f}:F\rightarrow A$ such that $\iota\hat{f}=f$. 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:

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

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 $\iota:X\rightarrow M$. Then we may form the tensor algebra of $M$,

 $T(M)=\bigoplus_{i\in\mathbb{N}}T^{i}(M),\qquad T^{i}(M)=M^{\otimes i}=% \bigotimes_{j=1}^{i}M.$

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

###### Remark 1.

To make $M$ concrete use the set of all functions $f:X\rightarrow K$, or equivalently, the direct product $\prod_{X}K$. 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 $f:X\rightarrow A$, then $A$ is a $K$-module and $M$ is free on $X$ so $f$ extends to a unique $K$-linear homomorphism $\hat{f}:M\rightarrow A$.

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

 $f^{(i)}(m_{1},\dots,m_{i})=f(m_{1})\cdots f(m_{i}).$

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

 $\hat{f}^{(i)}(m_{1}\otimes\cdots\otimes m_{i})=f(m_{1})\cdots f(m_{i}).$

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

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

 $T^{n}(M)=M^{\otimes n}=\bigotimes_{j=1}^{n}M.$

## 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 $K\langle X\rangle$ and we will find many parallels to polynomial algebras with indeterminants in $X$.

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

 $\left(\sum_{w\in FM\langle X\rangle}l_{w}w\right)\left(\sum_{v\in FM\langle X% \rangle}l_{v}v\right)=\sum_{w,v\in FM\langle X\rangle}l_{v}l_{w}wv.$

For example, $\mathbb{Q}\langle x,y\rangle$ contains elements of the form

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

This model of a free associative algebra encourages a mapping to polynomial rings. Indeed, $K\langle X\rangle\rightarrow K[X]$ 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:

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

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 $K\langle X\rangle$ is homogeneous if it contained in $FM\langle X\rangle$. 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 $K\langle X\rangle$.

###### 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 FreeAssociativeAlgebra 2013-03-22 16:51:07 2013-03-22 16:51:07 Algeboy (12884) Algeboy (12884) 10 Algeboy (12884) Definition msc 08B20 Algebras TensorAlgebra free associative algebra