1. 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.

2. You know how to multiply polynomials right?

Let f(x), g(x) $\in$ R[x]
$f(x)=a_nx^n + a_{n-1}x^{n-1}+... a_1x + a_0$
$g(x)=b_mx^m + b_{m-1}x^{m-1}+... b_1x + b_0$
for $a_m, a_n \not = 0$

$f(x)g(x)= a_nb_mx^{n+m} + ... + a_0b_0$

Now if R is a field, fields have no zero divisors, so since $a_m, a_n \not = 0$ you know $a_nb_m \not = 0$ thus the degree of this is the degree of f (n) plus the degree of g (m).

If the coefficients come from $\mathbb{Z}_m$ where m=pr (it is composite, so it has a prime factor p).

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

$f(x)g(x)=prx^2+(p+r)x+1=mx^2+(p+r)x+1=(p+r)x+1 \in \mathbb{Z}_m$