Thread: Chinese Remainder Theorem.

I need to prove problem 3.87 (The Chinese Remainder Theorem) on Rotmans book.

Prove that if k is a field and f(x), and f'(x) are in k[x] and are relatively prime, then given b(x) and b'(x) in k[x], there exists c(x) in k[x]. with

c-b in (f) and c-b' in (f'); more over if d(x) is another common solution then c-d is in (ff').

I have an good outline of the proof but I need help filling in major gaps.

Here is what I have: Every solution c-b in (f) has the form c=b+gf where g(x) is in k[x]. Hence we must find g such that gf - (b'-b) is in (f')

The first question I have is what is (f). then why a solution there has to have the form b+ gf? and lastly why is it enough to find such a g.

Thanks.

Originally Posted by jcir2826

I need to prove problem 3.87 (The Chinese Remainder Theorem) on Rotmans book.

Prove that if k is a field and f(x), and f'(x) are in k[x] and are relatively prime, then given b(x) and b'(x) in k[x], there exists c(x) in k[x]. with

c-b in (f) and c-b' in (f'); more over if d(x) is another common solution then c-d is in (ff').

I have an good outline of the proof but I need help filling in major gaps.

Here is what I have: Every solution c-b in (f) has the form c=b+gf where g(x) is in k[x]. Hence we must find g such that gf - (b'-b) is in (f')

The first question I have is what is (f). then why a solution there has to have the form b+ gf? and lastly why is it enough to find such a g.

Thanks.

I think something got a little muddled when you transcribed this problem, this isn't 3.87 in Rotman, what edition are you using?

The first edition is 3.87. The second edition is 2.95. Thanks.