Abstract Algebra/Polynomial Rings

The degree of a polynomial $$a_0+a_1X+...+a_nX^n$$ is defined to be $$n$$. If $$R$$ is a field, and $$f$$ and $$g$$ are polynomials of $$R[X]$$, then we can divide $$f$$ by $$g$$ to get $$f=gq+r$$. However, we can also do this for any arbitrary ring if the leading coefficient of $$g$$ is 1.

Take real numbers R for the ring and adjoin two indeterminants X and Y. The free algebra R over R is the collection of sums and products involving X, Y, and real numbers. The polynomial ring R[X,Y] is the algebra reduced by XY=YX, commutativity of the two indeterminants. In terms of quotient rings, with ideal generated by XY&minus;YX, R[X,Y] = R/(XY&minus;YX). The polynomial ring is central to commutative algebra, such as in the discussion of bibinarions and tessarines to follow.

In non-commutative algebra, the anti-commutative property XY=&minus;YX is illustrated in the quaternions. So-called "imaginary units" correspond to irreducible binomials XX+1 and YY+1. The elements of the quotient algebra R/(XY+YX, XX+1, YY+1) multiply as quaternions.

Exercises: What are the common names of these quotients ?
 * Polynomial quotient R[X]/(XX&minus;1)
 * Free algebra quotient R/(XY+YX, XX+1, YY&minus;1).

Application: Bibinarions & Tessarines
Commutative four-dimensional hypercomplex number systems were put forward in the nineteenth century. James Cockle presented the tessarines and Corrado Segre the bicomplex numbers, called Bibinarions in the study of Associative Composition Algebra. The isomorphism of these systems can be demonstrated through quotients of polynomial rings:

Consider the polynomial ring R[X,Y], where XY = YX. The ideal $$A = (X^2 + 1,\ Y^2 - 1)$$ then provides a quotient ring representing tessarines. In this quotient ring approach, elements of the tessarines correspond to cosets with respect to the ideal A. Similarly, the ideal $$ B = (X^2 + 1,\ Y^2 + 1 )$$ produces a quotient representing bicomplex numbers.

A generalization of this approach uses the free algebra R⟨X,Y⟩ in two non-commuting indeterminates X and Y. Consider these three second degree polynomials $$X^2 + 1,\ Y^2 - 1,\ XY - YX$$. Let A be the ideal generated by them. Then the quotient ring R⟨X,Y⟩/A is isomorphic to the ring of tessarines.

To see that $$(XY)^2 + 1 \in A$$ note that


 * $$X Y^2 X = X(Y^2 - 1) X + (X^2 + 1) - 1,$$ so that
 * $$X Y^2 X + 1 = X(Y^2 - 1)X + (X^2 + 1) \in A.$$ But then
 * $$ XY(XY - YX) + X Y^2 X + 1 \in A.$$ as required.

Now consider the alternative ideal B generated by $$X^2 + 1,\ Y^2 + 1,\ XY - YX$$. In this case one can prove $$(XY)^2 - 1 \in B$$. The ring isomorphism R⟨X,Y⟩/A ≅ R⟨X,Y⟩/B involves a change of basis exchanging $$Y \leftrightarrow XY$$.

Alternatively, suppose the field C of ordinary complex numbers is presumed given, and C[X] is the ring of polynomials in X with complex coefficients. Then the quotient C[X]/(X2 + 1) is another presentation of bicomplex numbers.