If we do computational work with

algebras (for example
in

noncommutative geometry) we need a good way to
represent our algebras on the

computer. Generators and relations
allow us to do this.

Before going further, I should say what an algebra is.
Fix a base field *k*.
A *k*-algebra is just a ring that has *k*
as a subring If *R* and *S* are *k*-algebras
we say that a function *f:R-->S* is a *k*-algebra
homomorphism if it is a ring homomorphism and also
*f(a)=a* for all *a* in *k*.
For example the polynomial ring in two variables
*k[x,y]* is a *k*-algebra.

Now we define **generators**.
A subset *S* of an algebra *R* is a generating
set for *R* if every element of *R* can be written
as a finite sum of terms of the form

*
as*_{1}...s_{n} with *a* in *k* and
*s*_{1}, ..., s_{n} in *S*.

For example,

*{x,y}* is a generating set for

*k[x,y]*.
Now we know that the polynomial ring is

commutative so as well as these
two generators we also have a

**relation**
*
xy - yx = 0
*

Everything we know about the multiplication in the polynomial ring
can be reconstructed from these generators and the relation.
For example, we know that every element of the polynomial ring
is a finite sum of terms of the form

*
ax*^{i}y^{j}

We can see this by first using the generating set

*{x,y}*
to write an arbitrary element as a sum of terms each of which is
a product of various

*x*s and

*y*s in some order.
Then we can use the relation to shuffle all the

*x*s to the left.
For example

*
6 + 3xyxy + 7yx = 6 + 3xxyy + 7xy = 6 + 3x*^{2}y^{2} + 7xy

Furthermore to mutiply to polynomials together we only need to understand how
to mutiply two monomials

*
ax*^{i}y^{j}.bx^{m}y^{n}

But using the relation again we can shuffle the

*x*s to the left
to get the result (

*abx*^{i+m}y^{j+n}).

In other words we have discovered that mutiplying in the polynomial ring
is completely understood once we have the generators and relations.
We would like to formalise this and generalise it to. For this we need
the idea of the free algebra. Roughly speaking this is what you
get if you just have generators but no relations.

Suppose we are given a set *X*. A **word** in *X* is a formal expression
of the form

*
x*_{1}...x_{n} : each *x*_{i} in *X*

We write

*k<X>* for the

vector space with

basis consisting of all words in

*X*.
By convention we write

*1* for the empty word.
If we have two words

*w* and

*w'* then we can form
a product by

*w.w'=ww'*. i.e. just put one word after the other.
Since an arbitrary element of

*k<X>* is a linear
combination of such words we can extend this product to all of

*k<X>* in a natural way. It is easy to see that this
makes

*k<X>* into a

*k*-algebra. This algebra
is called

**the free algebra on** *X*. Let

*i:X-->k<X>*
be the natural inclusion that maps

*x* in

*X* to itself.

Lets think about a couple of examples. First of all, to simplify
notation, if *X={x*_{1},...,x_{n}} is finite
we write *k<x*_{1},...,x_{n}> = k<X>.
On with the examples. If we just have a single generator for the free algebra
then *k<x>=k[x]*. For, by definition, the elements of
*k<x>* are linear combinations of the words

*
1, x, xx=x*^{2}, xxx=x^{3}, ...

But when we have two variables the free algebra is a much more exotic
beast. This time

*k<x,y>* has a basis consisting of
all the words in

*x,y*. That is

*
1, x, y, x*^{2}, xy, yx, y^{2}, ...

So we can see that this is definitely different from the polynomial
ring in two variables. Because

*yx* is not the same as

*xy*,
this ring is non

commutative. Notice however that there is
a very natural surjective

*k*-algebra homomorphism

*f:k<x,y>-->k[x,y]*. This is part of a much more general
result.

**Theorem** Start with a set *X* and let
*i:X-->k<X>* be the natural inclusion.
If *h:X-->R* is
function and *R* is a *k*-algebra then there is
a unique *k*-algebra homomorphism *f:k<X>-->R*
with *fi=h*.

**Proof** Although this is not needed for the
proof, note that this is an example of a universal property.
So that *i* is uniquely determined up to unique isomorphism
by this property. The proof itself is fairly obvious.
A typical element *e* of
*k<X>* has the form

*
e = a*_{1}w_{1} + ... + a_{n}w_{n} :
*a*_{i} in *k* and *w*_{i} are words in *X*.

Lets's say that

*
w*_{i}=x_{i,1}...x_{i,mi} :
*x*_{p,q} is in *X*.

Define

*
f(e) = a*_{1}h(x_{1,1})...h(x_{1,m1})
+ ... + a_{n}h(x_{n,1})...h(x_{n,mn})

It's clear that

*f* is a

*k*-algebra homomorphism and has the
required property.

So now we can say what we mean by generators and relations.
Given a set *X* let *S* be a subset of *k<X>*.
Then **the algebra with generators** *X* **and relations** *S*
is the quotient ring *k<X>/I*, where *I* is
the ideal of the free algebra generated by *S*.
For example, it's not too hard to show that the polynomial ring in
two variables is isomorphic to the algebra given by generators
*x,y* and one relation *xy - yx*.
Another example, is
the Weyl algebra which is given by two generators *x,y*
and one relation *xy - yx - 1*. For some other examples
that arise in mathematical physics see noncommutative geometry.

We can similarly think about free objects, and generators and relations
in other categories. For example, see generators
and relations for groups.