How prove this $g(x)|f(x)$ on $Q[x]$ Let $C,Q$ is  complex numbers Field and Rational number
 Field,respectively,if $f(x),g(x)\in Q[x]$,
if $g(x)|f(x)$ on $C[x]$,show that
$$g(x)|f(x)$$ on $Q[x]$
My try: since 
$g(x)|f(x)$,then we have
$$f(x)=g(x)h(x)$$
where $h(x)\in C[x]$.
Then I can't prove also have
$$g(x)|f(x)$$ on $Q[x]$.
Thank you
 A: Hint $\ $ It is an immediate consequence of the uniqueness of the quotient (and remainder) in the Division Algorithm (which is the same in $\rm\:\Bbb Q[x]\:$ and $\rm\:\Bbb C[x]).$ Since dividing $\rm\:f\:$ by $\rm\:g\:$ in $\rm\:\Bbb C[x]\:$ leaves remainder $\,0$, by uniqueness, the remainder is $\,0\,$ in $\rm\:\Bbb Q[x],\:$ i.e. $\rm\:g\ |\ f\ $ in $\rm\:\Bbb C[x]\,$ $\, \Rightarrow\, $ $\rm\, g\ |\ f\ $ in $\rm\:\Bbb Q[x]$.  
This is but one of many examples of the power of uniqueness theorems for proving equalities.
A: Induct on the degree of $h$ : if $deg(h) = 0$, then $h = a_0$, a constant, which must be rational.
If the result is true for $deg(h) \leq n-1$, assume $deg(h) = n$, then
$$
h(x) = a_0 + a_1x + \ldots + a_nx^n
$$
Then, compare the leading coefficients on both sides to conclude that $a_n \in \mathbb{Q}$. Now consider
$$
f_1(x) = f(x) - a_nx^ng(x) \in \mathbb{Q}[x],\text{ and }  h_1(x) = h(x) - a_nx^n
$$
Then $deg(h_1) < n$, and
$$
f_1(x) = g(x)h_1(x)
$$
and $f_1, g \in \mathbb{Q}[x]$. By induction, it follows that $a_i \in \mathbb{Q}$ for all $i$.
A: A generalization of the stated problem from $Q$ a subfield of $C$ to arbitrary fields $F \subset E$ is resolved by means of the following 
Proposition:  Let $E$ and $F$ be fields with $F$ a subfield of $E$.  Let $f(x), m(x) \in F[x]$ with $f(x) = q(x)m(x)$ in $E[x]$.  Then in fact $q(x) \in F[x]$.
Proof:  Let $f(x) = \sum_0^{\deg f} f_i x^i$, $m(x) = \sum_0^{\deg m} m_i x^i$, and $q(x) = \sum_0^{\deg q} q_i x^i$ with the $f_i, m_i \in F$ and the $q_i \in E$.  Then since $f(x) = q(x)m(x)$, we have
$f_i = \sum_{j + k = i}q_j m_k \tag{7}$
with $j, k \ge 0$; this if you will is the definition of $q(x)m(x)$.  Since $\deg f = \deg q + \deg m$, it follows that  $f_{\deg f}$, the coefficient of the leading term of $f(x)$, satisfies
$f_{\deg f} = q_{\deg q} m_{\deg m}; \tag{8}$
this shows that the leading coefficient of $q(x)$, $q_{\deg q} = m_{\deg m}^{-1} f_{\deg f} \in F$.  Next, we have that
$f_{\deg f - 1} = q_{\deg q - 1} m_{\deg m} + q_{\deg q} m_{\deg m - 1}, \tag{9}$
yielding 
$q_{\deg q - 1} = m_{\deg m}^{-1}(f_{\deg f - 1} -  q_{\deg q} m_{\deg m - 1}) \in F \tag{10}$
as well.  Now (7) shows that
$f_{\deg f - l} = \sum_{j + k = \deg f - l}q_j m_k = m_{\deg m} q_{\deg f -l - \deg m} + \sum_{j + k = \deg f - l, k < \deg m} q_j m_k$
$= m_{\deg m} q_{\deg q -l} + \sum_{j + k = \deg f - l, k < \deg m} q_j m_k, \tag{11}$
for $0 \le l \le \deg q$, since $\deg q = \deg f - \deg m$.  Scrutiny of (11) shows that (8) and (9) are respectively the $l = 0$ and $l = 1$ cases of this equation.  Now suppose that $q_{\deg q}, q_{\deg q - 1}, . . ., q_{\deg q - r} \in F$ for some $r$ with $0 \le r < \deg q - 1$.  Then by (11), 
$f_{\deg f - r - 1} = m_{\deg m} q_{\deg q - r - 1} + \sum_{j + k = \deg f - r - 1, k < \deg m} q_j m_k, \tag{12}$
and since $k < \deg m$ in the summation on the right, it follows that $j > \deg f - \deg m -r - 1 = \deg q -r - 1$ for every coefficient $q_j$ appearing in this expression.  If we now assume that each such $q_j \in F$, we may conclude that 
$q_{\deg q - r - 1} \in F$ as well, since (12) yields
$q_{\deg q - r - 1} = m_{\deg m}^{-1}(f_{\deg f - r - 1} - \sum_{j + k = \deg f - r - 1, k < \deg m} q_j m_k), \tag{13}$
and every coefficient appearing on the right of (13) is in $F$.  Thus we have completed an inductive demonstration that $q(x) \in F[x]$.  END:  Proof of Proposition.
If we apply this proposition to the problem at hand, taking $E = C$, $F = Q$, $m(x) = g(x)$ and $q(x) = h(x)$, it follows immediately that $h(x) \in Q[x]$.  QED!!!
Hope this helps!  Happy  New Year,
and as always,
Fiat Lux!!!
