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If $f(X), g(X) \in \mathbb{Q} [X]$ and $f(X) = g(X) h(X)$ , is $h(X)\in\mathbb{Q} [X]$ ?

Probably this is really really basic, but just in case I am missing something...

I think that in general $h$ may not be a polynomial, but if it is then is has to be a rational one. Is that right? Any ideas how to prove it?

Карпатський
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    Note that $\mathbb{Q}[X]$ is a Euclidean Ring, so has a "division algorithm" and unique factorization into irreducibles. As a result, $h(x)$ will be a polynomial, so $h(x)\in\mathbb{Q}[X]$. – Mark Schultz-Wu Nov 14 '17 at 08:02
  • $h(x)$ will be always a polynomial? Even if $\operatorname{gr}(f) < \operatorname{gr}(g)$ ? – Карпатський Nov 14 '17 at 08:08
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    If you don't know what $h$ is, how can you even define $g(X)h(X)$? The notation and lack of quantification over $X$ suggests that $h$ is a polynomial, in fact I probably should even write "$h(X)$ is a polynomial". And you need to know where $h(X)$ is in. If this isn't explicit the problem statement, then I'd assume $h(X)$ is in $\mathbb R[X]$. I'd try to prove by induction on the degree of $h(X)$, if it were me. – Git Gud Nov 14 '17 at 08:13
  • In principle, it is only a function. I think it may have terms like $3 \frac{1}{X^2}$ that could allow the equation be satisfied. By the way, there is no problem statement because it is just a question that I am making to myself in order to test if my understanding about polynomials is consistent. – Карпатський Nov 14 '17 at 09:33
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    If $h$ can be a function, then you shouldn't write $h(X)$ or $\mathbb Q[X]$, but rather talk about polynomial functions. Consider $x\mapsto x, x\mapsto x^2$ and $x\mapsto \frac 1 x$ as $f, g$ and $h$, respectively. If $h$, too, is a polynomial function, then its coefficients are rational, correct. – Git Gud Nov 15 '17 at 23:57

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$f(X) = g(X) h(X)$ gives a system of linear equations for the coefficients of $h$. If this system has a solution, than it has to remain in $\mathbb Q$ because it can be found by Gaussian elimination.

lhf
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