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Welcome to edit my post to revise any mistakes, especially English, thanks.

Proposition

Assuming $f(x),g(x)$ are integer polynomials, and $g(x)$ is primitive.

If $f(x)=g(x)h(x)$, where $h(x)$ is rational polynomial, then $h(x)$ must be polynomial with integer coefficients.

Proof:

$f(x)=g(x)h(x)=g(x)a h_1(x)=a g(x)h_1(x)=a f_1(x)$,

where $a$ is rational number and $h_1(x)$ and $f_1(x)$ are primitive polynomials.

If $a$ is not an integer number, then $f(x)$ should not be an integer polynomial which lead to contradiction with the assumption.

so $h(x)$ must be integer polynomial.

amWhy
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HyperGroups
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  • Are you asking if your proof is correct? – Antonio Vargas May 31 '13 at 15:15
  • @AntonioVargas Yes, I am not sure. English's problems are also welcome to edit. – HyperGroups May 31 '13 at 15:17
  • It looks fine yet (1) I'd love to see a little (little!) more explanation why $,a\notin\Bbb Z\implies f(x),$ is not an integer polynomial, and (2) you may also want to describe a little more how is $,a,$ obtained: you assume at least one coefficient of $,h(x),$ isn't integer, so in order to get $,h_1(x),$ you first must "take out" acommon denominator of all those coefficients of $,h(x),$ to get an integer pol., and then you take out a common divisor of all the coefficients to make it primitive...etc. – DonAntonio May 31 '13 at 15:36
  • @DonAntonio yeah, 2)I obtain $a$ just consider it an common denominator to get primitive $h_1(x)$, but without the thought of assuming at least one coefficient of h(x) isn't integer, your suggestion is useful. 1)$a\notin \mathbb{Z}$ so get an rational poly $\text{fNew}(x)\neq f(x)$ – HyperGroups May 31 '13 at 15:59
  • @HyperGroups I concur with Don, you certianly need to justify the claim (1). I cannot parse the reason given above. – Key Ideas May 31 '13 at 17:09
  • @KeyIdeas well,Don say a litte more.... I wonder if this is enough. a not belongs to Z, a belongs to Q, so a=r/s, s not equal to 1, and (r,s)=1, f_1 does not have common denominator s, so $a f_1$ with rational coefficients. – HyperGroups Jun 01 '13 at 01:51

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