# Thread: reducibility in Q implies reducibility in Z.

1. ## reducibility in Q implies reducibility in Z.

If $f(x) \in \mathbb{Z}[x]$, then $f(x)$ factors into a product of two polynomials of lower degrees $r \, and \, s \in \mathbb{Q}[x]$ if and only if it has such a factorization with polynomials of the same degrees

$r \, and \, s \in \mathbb{Z}[x]$.

How to prove it??

2. Originally Posted by abhishekkgp
If $f(x) \in \mathbb{Z}[x]$, then $f(x)$ factors into a product of two polynomials of lower degrees $r \, and \, s \in \mathbb{Q}[x]$ if and only if it has such a factorization with polynomials of the same degrees

$r \, and \, s \in \mathbb{Z}[x]$.

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'.

3. 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 $f(x)=(\frac{a_1}{b_1}x-\frac{a_2}{b_2})(\frac{a_3}{b_3}x-\frac{a_4}{b_4})\, where\, a_i,\,b_i \in \mathbb{Z}\, with \, gcd(a_i,b_i)=1$.

hence,$f(x)=\frac{1}{b_1b_2b_3b_4}(a_1b_2a_3b_4x^2-(a_1b_2b_3a_4+b_1a_2a_3b_4)x+b_1a_2b_3a_4)$.

this easily forces $b_2|a_4, \, b_4|a_2, \, b_1|a_3, \, b_3|a_1$

doing the spreading out thingy you told. $f(x)=(\frac{b_2a_1}{b_2b_3}x-\frac{b_1a_2}{b_2b_3})(\frac{b_4a_3}{b_1b_4}x-\frac{a_4b_3}{b_1b_4})$

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 $(b_1b_2b_3b_4)|(a_1b_2b_3a_4+b_1a_2a_3b_4)$

help!!!!