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.
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 .
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.
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 .