# Relatively Prime Polynomials in Extension Fields

Homework Statement
Recall that two polynomials f(x) and g(x) from F[x] are said to be relatively prime if there is no polynomial of positive degree in F[x] that divides both f(x) and g(x). Show that if f(x) and g(x) are relatively prime in F[x], they are relatively prime in K[x], where K is an extension of F.

The attempt at a solution
I'm guessing this will be a proof by contradiction where the contradiction will be that f(x) and g(x) are not relatively prime in F[x]:

Let p(x) be an irreducible factor of the divisor of f(x) and g(x) in K[x]. How can I show that p(x) divides f(x) and g(x) in F[x]?

## Answers and Replies

You can't do what you ask for in your last sentence: p(x) isn't even necessarily in F[x]. But the question doesn't ask you to show that.

Is there another way of getting a contradiction? I can't think of anything else.

Why have you stopped what you're doing? The only thing wrong was your assumption that any factor p(x) in K[x] must be in F[x].

Consider an example if it helps you. E.g. x^2+1 and x^4-1 over Q[x]. These are not relatively prime, but you don't prove that by considering *one* irreducible linear factor in C[x], since x-i and x+i are not in Q[x].

I got the impression that my whole approach was wrong.

So are you saying I have to consider all the irreducible factors in K[x]? In what way? Perhaps the product of some of the irreducible linear factors of the divisor is a member of F[x], e.g. (x - i)(x + i) = x2 + 1 in your example?

Let's suppose that I have a polynomial f(x) with rational coefficients. And suppose that, passing to the complex numbers, I know x-i is a root of f. What other linear factor must be a root of f(x)?

What do you know about things like Aut(K:F)? The field automorphisms of K that fix F. Such as complex conjugation in Aut(C:Q).

Let's suppose that I have a polynomial f(x) with rational coefficients. And suppose that, passing to the complex numbers, I know x-i is a root of f. What other linear factor must be a root of f(x)?
I'm guessing x + i. I don't really know. To find another linear factor I would try to find a zero of f(x)/(x - i).

What do you know about things like Aut(K:F)? The field automorphisms of K that fix F. Such as complex conjugation in Aut(C:Q).
I've never seen that notation before.

My problem is from a chapter on extension and splitting fields.

Yes, that makes sense - you can assume that K is a splitting field for f(x).

You do know the first part: complex roots of a real polynomial occur in complex conjugate pairs.

You do know the first part: complex roots of a real polynomial occur in complex conjugate pairs.
That's an interesting fact. So in the case of Q and C, the contradiction works. But in general? Does p(x) have a conjugate?

I was told that since f(x) and g(x) are relatively prime in F[x], then there are s(x) and t(x) in F[x] with f(x)s(x) + g(x)t(x) = 1. This equation is also valid in K[x] and this supposedly implies that f(x) and g(x) are relatively prime in K[x]. I don't understand this implication. Why is this?

It's just euclid's algorithm. Just think of Z rather than a polynomial algebra.

You can do it with a contradiction for any extension. But you don't need to.

Does f(x)s(x) + g(x)t(x) = 1 imply that f(x) and g(x) are relatively prime? Isn't it the other way around?

It is exactly the same as it is for the integers. You should be able to answer your own question.