Linear Algebra, Theory And Applications, 2012a

8.3. LOTS OF FIELDS 207
and this is not zero by the assumption that f (λ) − g (λ) l (λ) is never equal to zero for any
l (λ) yet has smaller degree than r (λ) which is a contradiction to the choice of r (λ).
Now with this lemma, here is another one which is very fundamental. First here is a
definition. A polynomial is monic means it is of the form
λ n + c n−1 λ n−1 + ···+ c 1 λ + c 0 .
That is, the leading coefficient is 1. In what follows, the coefficients of polynomials are in
F, a field of scalars which is completely arbitrary. Think R if you need an example.
Definition 8.3.4 A polynomial f is said to divide a polynomial g if g (λ) =f (λ) r (λ) for
some polynomial r (λ). Let {φ i (λ)} be a finite set of polynomials. The greatest common
divisor will be the monic polynomial q such that q (λ) divides each φ i (λ) and if p (λ) divides
each φ i (λ) , then p (λ) divides q (λ) . The finite set of polynomials {φ i } is said to be relatively
prime if their greatest common divisor is 1. A polynomial f (λ) is irreducible if there is no
polynomial with coefficients in F which divides it except nonzero scalar multiples of f (λ)
and constants.
Proposition 8.3.5 The greatest common divisor is unique.
Proof: Suppose both q (λ) andq ′ (λ) work. Thenq (λ) divides q ′ (λ) and the other way
around and so
q ′ (λ) =q (λ) l (λ) ,q(λ) =l ′ (λ) q ′ (λ)
Therefore, the two must have the same degree. Hence l ′ (λ) ,l(λ) are both constants. However,
this constant must be 1 because both q (λ) andq ′ (λ) aremonic.
Theorem 8.3.6 Let ψ (λ) be the greatest common divisor of {φ i (λ)} , not all of which are
zero polynomials. Then there exist polynomials r i (λ) such that
ψ (λ) =
p∑
r i (λ) φ i (λ) .
i=1
Furthermore, ψ (λ) is the monic polynomial of smallest degree which can be written in the
above form.
Proof: Let S denote the set of monic polynomials which are of the form
p∑
r i (λ) φ i (λ)
i=1
where r i (λ) is a polynomial. Then S ≠ ∅ because some φ i (λ) ≠0. Thenletther i be chosen
such that the degree of the expression ∑ p
i=1 r i (λ) φ i (λ) is as small as possible. Letting ψ (λ)
equal this sum, it remains to verify it is the greatest common divisor. First, does it divide
each φ i (λ)? Suppose it fails to divide φ 1 (λ) . Then by Lemma 8.3.3
φ 1 (λ) =ψ (λ) l (λ)+r (λ)
where degree of r (λ) is less than that of ψ (λ). Then dividing r (λ) by the leading coefficient
if necessary and denoting the result by ψ 1 (λ) , it follows the degree of ψ 1 (λ) is less than
thedegreeofψ (λ) andψ 1 (λ) equals
ψ 1 (λ) =(φ 1 (λ) − ψ (λ) l (λ)) a