free associative algebra


Fix a commutativePlanetmathPlanetmathPlanetmath unital ring K and a set X. Then a K-algebraMathworldPlanetmathPlanetmathPlanetmath F is said to be free on X if there exists an injection ι:XF such that for all functions f:XA where A is an K-algebra determine a unique algebra homomorphism f^:FA such that ιf^=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 diagramMathworldPlanetmath:

\xymatrix&X\ar[ld]ι\ar[rd]f&F\ar[rr]f^&&A.

To prove that free associative algebras exist in the categoryMathworldPlanetmath 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 equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath.

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 ι:XM. Then we may form the tensor algebra of M,

T(M)=iTi(M),Ti(M)=Mi=j=1iM.

[Note, 0 and the empty tensor we define as K.] Furthermore, define the injection ι:XT(M) as the map ι:XM followed by the embeddingPlanetmathPlanetmath of M into T(M).

Remark 1.

To make M concrete use the set of all functions f:XK, or equivalently, the direct productMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath XK. Then the tensor algebra of M is the free algebraMathworldPlanetmath on X.

Proposition 2.

(T(M),ι) is a free associative algebra on X.

Proof.

Given any associative K-algebra A and function f:XA, then A is a K-module and M is free on X so f extends to a unique K-linear homomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath f^:MA.

Next we define K-multilinear maps f(i):MiA by

f(i)(m1,,mi)=f(m1)f(mi).

Then by the universal mapping property of tensor productsPlanetmathPlanetmathPlanetmath (used inductively) we have a unique K-linear map f^(i):Ti(M)A for which

f^(i)(m1mi)=f(m1)f(mi).

Thus we have a unique algebra homomorphism f^():T(M)A such that ιf^=f. ∎

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

Tn(M)=Mn=j=1nM.

2 Non-commutative polynomials

An alternative construction is to model the methods of constructing free groupsMathworldPlanetmath and semi-groups, that is, to use words on the set X. We will denote the result of this construction by KX and we will find many parallels to polynomial algebras with indeterminants in X.

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

(wFMXlww)(vFMXlvv)=w,vFMXlvlwwv.

For example, x,y contains elements of the form

x2+4yxy,-7xy+2yx,1+x+xy+xyx+x2y+x2y2.

This model of a free associative algebra encourages a mapping to polynomial ringsMathworldPlanetmath. Indeed, KXK[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 translateMathworldPlanetmath the above elements into:

x2+4xy2,-5xy,1+x+xy+2x2y+x2y2.

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 KX is homogeneousPlanetmathPlanetmathPlanetmathPlanetmath if it contained in FMX. Then the degree of a homogeneous elementPlanetmathPlanetmath is the length of the word. Then the K-linear span of elements of degree i form the i-th graded componentMathworldPlanetmathPlanetmath of KX.

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 modulesMathworldPlanetmathPlanetmath 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 algebrasMathworldPlanetmathPlanetmath, 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