Abstract- Polynomial's Ring F[x]

[WannaBe22]

Homework Statement

In this question we have to make use of the chinese remainder theorem and its applications:

1. Let F be a field and let p1(x), p2(x) two irreducible poynomials such as gcd(p1,p2)=1 over F.
Prove that: F[x]/[p1(x)p2(x)] Isomorphic to F1 x F2 where F1=F[x]/(p1(x)) and
F2=F[x]/(p2(x))...
2. Let F be Z3 (Z/3Z). Find a polynomial p(x)=p1(x)p2(x) such as F[x]/p(x) iso. to F1xF2 where F1,F2 are from order 9, 27 .

Homework Equations

The Attempt at a Solution

About 1- in order to prove this using the chinese remainder theo. we need to prove that
p1(x), p2(x) are two relatively prime (coprime) ideals i.e p1(x)+p2(x)=F[x]... But why this is true?
About 2- If I take p1(x)=x^2+1, p2(x)=x^3+2x+1...Both p1 and p2 are irreducible and F1,F2 are from order 9 and 27...gcd(p1,p2)=1 hence we know from 1 that F[x]/[p1(x)p2(x)] Isomorphic to F1 x F2 as needed...
But how can I PROVE 1?

Thanks a lot!

 

[rochfor1]

If two integers m and n are coprime, what can you say about their linear combinations? The same result generalizes to polynomials over a field.

 

[WannaBe22]

I need to say that the p1(x) and p2(x) spans F[x]..

If I have m,n such as gcd(m,n)=1 then we have a and b such as am+bn=1 and then we can get any integer x by the linear combination amx+bnx=x ...
So if we have 2 polynomials that their gcd is 1, we can produce any element in F[x] by a linear combination of them? Is it a theorem or anything because we don't know nothing about the dimension of F[x]...

Further guidance is needed

Thanks a lot!

 

[rochfor1]

The proof for F[x] is almost exactly the same as the proof for integers. The key fact is that both Z and F[x] are both principal ideal domains. In general, F[x] is a PID <==> F is a field. In fact, this result about linear combinations and the GCD holds in any PID. Briefly, you should look at the ideal generated by all linear combinations of p1(x) and p2(x) (with coefficients in F[x]), and think about what the generator of this ideal should be intuitively.

 

[WannaBe22]

Indeed...Tnx a lot