34 CHAPTER 2. THE REAL AND COMPLEX NUMBERS

for m+n≤M and suppose m+n = M+1. Then f (λ )g(λ ) =(a0 +a1λ + · · ·+an−1λ

n−1 +anλn)(

b0 +b1λ + · · ·+bm−1λm−1 +bmλ

m)

= (a(λ )+anλn)(b(λ )+bmλ

m)

= a(λ )b(λ )+bmλma(λ )+anλ

nb(λ )+anbmλn+m

Either an = 0 or bm = 0. Suppose bm = 0. Then it must be the case that you have

(a(λ )+anλn)b(λ ) = 0.

By induction, one of these polynomials in the product is 0. If b(λ ) ̸= 0, then this showsan = 0 and a(λ ) = 0 so f (λ ) = 0. If b(λ ) = 0, then g(λ ) = 0. The argument is the sameif an = 0.

Say the degree of r (λ ) is m≥ n where the degree of g(λ ) is n. Say r (λ ) = aλm+ l (λ )

with the degree of l (λ ) < m and g(λ ) = bλn + n(λ ) where the degree of n(λ ) is less

than n. Then r (λ )− ab λ

m−ng(λ ) has degree smaller than m. This is used in the followingfundamental lemma.

Lemma 2.14.3 Let f (λ ) and g(λ ) ̸= 0 be polynomials. Then there exist polynomials,q(λ ) and r (λ ) such that

f (λ ) = q(λ )g(λ )+ r (λ )

where the degree of r (λ ) is less than the degree of g(λ ) or r (λ ) = 0. These polynomialsq(λ ) and r (λ ) are unique.

Proof: Suppose that f (λ )− q(λ )g(λ ) is never equal to 0 for any q(λ ). If it is, thenthe conclusion follows. Now suppose

r (λ ) = f (λ )−q(λ )g(λ ) (∗)

where the degree of r (λ ) is as small as possible. Let it be m. Suppose m≥ n where n is thedegree of g(λ ). Say r (λ ) = bλ

m +a(λ ) where a(λ ) is 0 or has degree less than m whileg(λ ) = b̂λ

n + â(λ ) where â(λ ) is 0 or has degree less than n. Then

r (λ )− bb̂

λm−ng(λ ) =

r(λ )bλ

m +a(λ )−(

bλm +

bb̂

λm−nâ(λ )

)= a(λ )− ã(λ ) ,

a polynomial having degree less than m. Therefore, from the above,

a(λ )− ã(λ ) =

=r(λ )︷ ︸︸ ︷( f (λ )−q(λ )g(λ ))− b

b̂λ

m−ng(λ ) = f (λ )− q̂(λ )g(λ )

which is of the same form as ∗ having smaller degree. However, m was as small as possible.Hence m < n after all.

As to uniqueness, if you have r (λ ) , r̂ (λ ) ,q(λ ) , q̂(λ ) which work, then you wouldhave

(q̂(λ )−q(λ ))g(λ ) = r (λ )− r̂ (λ )

Now if the polynomial on the right is not zero, then neither is the one on the left. Hencethis would involve two polynomials which are equal although their degrees are different.This is impossible. Hence r (λ ) = r̂ (λ ) and so, the above lemma shows q̂(λ ) = q(λ ).