Thread: reducibility in Q implies reducibility in Z.

If , then factors into a product of two polynomials of lower degrees if and only if it has such a factorization with polynomials of the same degrees

.

How to prove it??

Originally Posted by abhishekkgp

If , then factors into a product of two polynomials of lower degrees if and only if it has such a factorization with polynomials of the same degrees

.

How to prove it??

Assume it factorises into two polynomials in Q, f=gh. Then f=(1/a)g'h', where a is the product of all the denominators of all the coefficients of g and h, with g' and h' in Z[x]. Then k=g'h' is in Z[x], so f=(1/a)k with k in Z[x]. Can you see where to go from here? The point is that you must be able to `spread out' the integer a between g' and h'.

Originally Posted by Swlabr

Assume it factorises into two polynomials in Q, f=gh. Then f=(1/a)g'h', where a is the product of all the denominators of all the coefficients of g and h, with g' and h' in Z[x]. Then k=g'h' is in Z[x], so f=(1/a)k with k in Z[x]. Can you see where to go from here? The point is that you must be able to `spread out' the integer a between g' and h'.

i will take a special case and do it.
let .

hence,.

this easily forces

doing the spreading out thingy you told.

this doesn't do it since all coefficients are not proved to be integers; the terms attached to 'x' are integers but the others i can't comment upon. i cannot figure out how to exploit

help!!!!