Polynomials

(a)
Let p be the arbitrary element of A
Then when r divides p
p=rq1+r1
where q1 is the quotient and r1 is the remainder with degree less than r
We argue that r1 should be 0. Otherwise, we have
r1=p-rq1 which is an element of A but this is a contradiction (because r is the element of A with least degree)
So r divides every polynomial in A
Now set m=1,n=0 , we can deduce that r|f
Similarly, set m=0,n=1, we can deduce that r|g
r is a factor of f and g
if h is also a factor of f and g, then h∈A,consider
r=hq2+r2
then r2 should be 0, also since deg(r)<=deg(h) we conclude that deg(r)=deg(h) and q2 is a constant polynomial. That is h divides r
So r is a GCD of f and g
(b)
From A , r divides every polynomial in A. So every polynomial in A can be written in the form hr where h is a polynomial∈P. This shows that A⊂B
To prove B ⊂ A, consider that r itself is an element of A, then it can be written in the form of af+bg, now for arbitrary element of B, We have hr=ahf+bhg∈A. This shows that B⊂A
Combine together, we prove that A=B
(c)
Because r=af+bg (a,b∈P) and degree r=0
We have 1=(a/r)f+(b/r)g=(m0)f+(n0)g
The result of A=P can be obtained by part b immediately. To see it more clearly, we see that since
Obviously, A⊂P
Consider arbitrary element of P, let say x
We know that x=x1=xr ∈A
So P⊂A. That is A=P