Thread: Degree Problem

Let

f eR[x] where R is a commutative ring. If f has form

a(sub

n)x^n + ... + a1x + a0 with an a(subn) != 0 then the degree of f is defined to be n and the leading coefficient is defined to be an.

If

f, g eF[x] are non-zero polynomials where F is a field,
then,

deg(

fg) = deg f + deg g

Justify the above theorem. Explain why the proof does
not work if the coefficients are in Zm where m is composite. Hint: focus on the leading coefficients.

Any advice??

You know how to multiply polynomials right?

Let f(x), g(x) R[x]


for



Now if R is a field, fields have no zero divisors, so since you know thus the degree of this is the degree of f (n) plus the degree of g (m).

If the coefficients come from where m=pr (it is composite, so it has a prime factor p).

Consider
f(x) = px+1
g(x)=rx+1