2.14. DIVIDING POLYNOMIALS 35
Definition 2.14.4 Let all coefficients of all polynomials come from a given field F.For us, F will be the real numbers R. Let p(λ ) = anλ
n + · · ·+a1λ +a0 be a polynomial.Recall it is called monic if an = 1. If you have polynomials
{p1 (λ ) , · · · , pm (λ )} ,
the greatest common divisor q(λ ) is the monic polynomial which divides each, pk (λ ) =q(λ ) lk (λ ) for some lk (λ ) , written as q(λ )/pk (λ ) and if q̂(λ ) is any polynomial whichdivides each pk (λ ) , then q̂(λ )/q(λ ) . A set of polynomials
{p1 (λ ) , · · · , pm (λ )}
is relatively prime if the greatest common divisor is 1.
Lemma 2.14.5 There is at most one greatest common divisor.
Proof: If you had two, q̂(λ ) and q(λ ) , then q̂(λ )/q(λ ) and q(λ )/q̂(λ ) so q(λ ) =q̂(λ ) l̂ (λ ) = q(λ ) l (λ ) l̂ (λ ) and now it follows, since both q̂(λ ) and q(λ ) are monic thatl̂ (λ ) and l (λ ) are both equal to 1.
The next proposition is remarkable. It describes the greatest common divisor in a veryuseful way.
Proposition 2.14.6 The greatest common divisor of {p1 (λ ) , · · · , pm (λ )} exists and ischaracterized as the monic polynomial of smallest degree equal to an expression of theform
m
∑k=1
ak (λ ) pk (λ ) , the ak (λ ) being polynomials. (2.1)
Proof: First I need show that if q(λ ) is monic of the above form with smallest de-gree, then it is the greatest common divisor. If q(λ ) fails to divide pk (λ ) , then pk (λ ) =q(λ ) l (λ )+ r (λ ) where the degree of r (λ ) is smaller than the degree of q(λ ). Thus,
r (λ ) = pk (λ )− l (λ )m
∑k=1
ak (λ ) pk (λ )
which violates the condition that q(λ ) has smallest degree. Thus q(λ )/pk (λ ) for each k.If q̂(λ ) divides each pk (λ ) then it must divide q(λ ) because q(λ ) is given by 2.1. Henceq(λ ) is the greatest common divisor.
Next, why does such greatest common divisor exist? Simply pick the monic polynomialwhich has smallest degree which is of the form ∑
mk=1 ak (λ ) pk (λ ) . Then from what was
just shown, it is the greatest common divisor.
Proposition 2.14.7 Let p(λ ) be a polynomial. Then there are polynomials pi (λ ) suchthat
p(λ ) = am
∏i=1
pi (λ )mi (2.2)
where mi ∈ N and {p1 (λ ) , · · · , pm (λ )} are monic and relatively prime. Every subset of{p1 (λ ) , · · · , pm (λ )} having at least 2 elements is also relatively prime.