Polynomials over a field
In this context, we consider a commutative field (simply: field) . The polynomials will be constructed as a certain associative unital algebra over
(“
-aua”), together with a distinguished element called the formal variable.
itself, as a vector space over itself and the usual multiplication in
as the third law, is a
-aua,
Morphisms between two -auas are linear mappings that conserve the third law and map the unit of the first algebra onto the unit of the second one.
All that will be said of the polynomials over will be based on the universal property given below. The polynomials will thus be defined uniquely up to an isomorphism.
Definition
The polynomials over is any
-aua
together with a distinguished element
of
such that the following universal property holds:
For any -aua
and any
, there exists exactly one morphism
such that
.
Unicity up to an isomorphism
Let and
be two sets of polynomials over
. The universal property implies the existence of a morphism
such that
, and of a morphism
such that
.
We then have . Hence
is a morphism
that maps
to
. Now
is another such morphism. By virtue of the universal property of
, it follows that
, that is, is the
-aua category identity on
.
The same reasoning shows that is the
-aua category identity on
.
Hence is an isomorphism
that maps the formal variable of
to that of
.
Characterization
Let be an associative unital algebra over
, and
an element of
. Then
represents the polynomials over
if and only if the family
, with
and
, is a family-base of the
-vector space
.
Let us first suppose that the latter condition holds on , and prove the universal property.
Let be a
-aua and
an element of
.
Let us suppose that is a morphism
such that
. Then for all
, we have
. Since
as a
-aua morphism, it is in particular a linear mapping. Since a linear mapping is defined by the image of a basis, and
is a basis,
necessarily is the unique linear mapping
such that
. Now it is easy to check that this linear mapping does preserve the multiplication in
, and is hence a
-aua morphism.
Thus the universal property is proven for .
Let us now suppose the universal property for , and show that
is a family-base of the
-vector space
.
Let us first show that is linearly independent.
Let be a free
-vector space on set
. This entails that
is a family-base of
. To make
into an associative unital algebra, we must define a multiplication law that is bilinear, and also associative.
A bilinear law can be defined by the images of all the ordered pairs of base elements. Hence we may define the multiplication in by:
We then have
.
Hence the multiplication is associative with regard to the elements of the base; it is easy to check that this follow through to arbitrary elements, which makes an associative algebra.
is a unit; hence
is an associative unital algebra.
Applying the universal property of , we find that there exists exactly one
-aua morphism
such that
.
We then have
;
; and, for any
,
. Hence for all
,
.
Let us consider a linear combination of the family that vanishes; that is, a finite sequence of values of
,
, such that
. Taking the image of this linear combination by
, we obtain
. Since the
form a basis, all the
must be zero.
Thus the family is linearly independent.
Lastly, we must show that this family generates the -vector space
.
Let be the sub-aua of
generated by
.
There exists a unique aua morphism such that
.
Let then be
, but with all of
as codomain; in other words,
. It is clear that
too is an aua morphism, and it is such that
.
But there is exactly one aua morphism that maps
to
; and
is such.
Hence .
The image of is that of
, and is a subset of
. But the image of
is
itself. Hence
.
Thus the sub-aua of generated by
is
itself. This sub-aua is the set of linear combinations of all the powers of
; hence the sub-vector space generated by
is
.
Thus is a family-base of the
-vector space
.