We can introduce addition and multiplication on the set of all polynomials over F as follows. Polynomials can be added by adding the coefficients of the terms of the same degree, and they can be multiplied by multiplying every term by every term so that the product of the terms and is , then we add the terms of same degree and rearrange the terms so that the power of x are in ascending order. For example, let
and . Then
and
About the degrees of the sum and product polynomials we can say the following: the degree of the sum polynomial cannot be greater than the degrees of the summands, that is
Polynomials
The equality holds exactly when f and g are of the same degree, and the sum of the coefficient of their terms of highest degree is zero. Furthermore,
that is the degree of the product polynomial is always equal to the sum of the degrees of the factors. Here we used that a product of elements of F can only be zero if one of the factors is zero. The truth of this statement for complex numbers will be proved later, under more general conditions.
The addition of the polynomial is commutative, associative, its additive identity is the zero polynomial, furthermore, every polynomial f has an opposite, namely the polynomial , whose coefficients are the opposites of the coefficients of f . Therefore, the set of all polynomials is closed under substraction, which means adding the opposite. The properties of commutativity and associativity hold for the multiplication of polynomials as well, the unity is the polynomial . None of polynomials of degree at least 1 has reciprocal, because by multiplying polynomials their degrees are added together, so a polynomial multiplied by a polynomial of degree at least 1 cannot result the polynomial of degree 0. Therefore, only the polynomials of degree zero have reciprocal, and they indeed have, because they are exactly the elements of F other than 0.
Because of the lack of reciprocal, we can only make the division as we have seen at the integers.
Theorem 3.1 (Division algorithm for polynomials). Let f and g be polynomials over F, such that . Then there exist unique polynomials q and r over F, such that
, and the degree of r is less than the degree of g.
Proof. We start with the proof of the uniqueness, that is we are going to show that there is only one pair of polynomials satisfying the conclusion of the theorem. Assume the contrary: there exist polynomials and such that
where and are less than , and . Subtracting the second equality form the first one, after rearranging we have the equality
Since , the degree of the polynomial on the left-hand side is at least than , whereas the degree of the polynomial on the right-hand side is less than the degree of g, which is a contradiction. Thus, , and so the polynomial on the left-hand side of (3.2) is zero. But then also has to be zero, whence .
Now we show how to construct the polynomials q and r. Let
and
Polynomials
be polynomials of degree n and m, respectively. If , then by and the theorem is satisfied: . In the case when , divide the term of highest degree of f by the term of highest degree of g, and then multiply the quotient
by g, finally subtract this product from f . Denote by this difference:
It is clear that after subtraction the term of highest degree of f falls out, so the degree of will be less than the degree of f . If , the we can stop: and
. Otherwise, we repeat the method with the polynomial instead of f : we construct
the polynomial , for which . If ,
the we stop, otherwise we repeat again, using instead of . Since the degrees of the polynomials get smaller and smaller, after finitely many steps (say at the kth step) we get to the case when holds. It is easy to see that
thus the division theorem of the polynomials is true.
The polynomials q and r are called the quotient and remainder of the division of f byg.
The demonstration may become easier to understand through a concrete example. Divide the polynomial by . First we divide the term of the highest degree of f by the term of the highest degree of g: , we write this quotient to the right-hand side of the equality:
We multiply g by the quotient just obtained (by ), and we write the product under f :
We subtract the product just obtained from f , and write the result under a line:
The degree of the polynomial under the line is not less than the degree of g, so considering it as , we repeat the previous steps. The quotient of the terms of the highest degree of andg is , this will be added to the polynomial on the right-hand side of the equality, then we write the product of g and under
, which finally will be subtracted form :
Polynomials
Following the notations of the proof, the degree of the undermost polynomial is less than the degree of g, so the
procedure is done. The result is: and . We may verify ourself by
checking the equality