Polynomial division over an extension 
Suppose $F \subseteq K$ are fields. I'm trying to show that if $P,Q \in F[X]$ and $P \mid Q$ over $K$, then $P \mid Q$ over $F$ as well.

Denote the field of fractions of $F$ with $F(X)$. Then I have come to the point that it suffices to prove
$$F(X) \cap K[X] = F[X]$$
The inclusion $\supseteq$ is obvious. The other one should be straightforward, but I'm clearly forgetting something.
So, let $P \in F(X) \cap K[X]$, then $P = \frac{F_1}{F_2}$ with $F_1,F_2 \in F[X]$. I feel like this brings me back to where I started.
Other methods are appreciated as well.
 A: Let $F$ be a field and $K$ an extension field.
Let $f,g\in F[X]$. Then the highest common factors of $f,g$ in $F[X]$ and $K[X]$ coincide. This is because if we carry out the classical Euclidean Algorithm on $f,g\in F[X]$ the h.c.f. and all the intermediate polynomials lie in $F[X]$. But  this calculation (thought of as in $K[X]$) is just the Euclidean Algorithm in $K[X]$, and yields the same h.c.f.
Now suppose $p,q\in F[x]$ and $p|q$ in $K[X]$. Then $p$ is the h.c.f. in $K[X]$ of $p$ and $q$. Hence $p$ is the  h.c.f. in $F[X]$ of $p$ and $q$. So $p|q$ in $F[X]$. 
A: A demonstration which illustrates exactly how the coefficients of the quotient in $K[x]$ lie in the base field $F$:
Given fields
$F \subset K, \tag 1$
with
$P(x), Q(x) \in F[x] \tag 2$
and
$P(x) \mid Q(x) \; \text{in} \; K[x], \tag 3$
then
$\exists D(x) \in K[x], \; Q(x) = P(x) D(x) \; \text{in} \; K[x]; \tag 4$
from this it follows that the term of $Q(x)$ of least degree $q$, that is
$Q_q x^q, \; Q_q \ne 0, \tag 5$
where
$Q(x) = \displaystyle \sum_q^{\deg Q} Q_jx^j, Q_j \in F,\tag 6$
satisfies
$Q_qx^q = P_px^pD_dx^d = P_pD_dx^{p + d}, \tag 7$
where again
$P(x) = \displaystyle \sum_p^P P_jx^j, P_j \in F, P_p \ne 0, \tag 8$
and
$D(x) = \displaystyle \sum_d^{\deg D} D_jx^j, D_j \in K, D_d \ne 0; \tag 9$
(7) is directly obtained by taking the product of (8) and (9) according to the usual rules of polynomial multiplication, and it affirms that
$q = p + d \ge p, \tag{10}$
since
$d \ge 0. \tag{11}$
At this point we pause to observe that, in light of (10), (7) yields
$Q_q = P_qD_d, \tag{12}$
whence
$D_d = P_p^{-1} Q_q \in F; \tag{13}$
we will extend this result to all $D_j$, $d \le j \le D$.  Setting
$Q_1(x) = \displaystyle \sum_q^{\deg Q} Q_jx^{j - q},  \tag{14}$
we have
$Q(x) = x^qQ_1(x), \tag{15}$
and similarly setting
$P_1(x) = \displaystyle \sum_p^P P_jx^{j - p}, \tag{16}$
so that
$P(x) = x^pP_1(x), \tag{17}$
we write (4) as
$x^qQ_1(x) = x^pP_1(x)D(x); \tag{18}$
whence again invoking (10),
$x^dQ_1(x) = P_1(x)D(x); \tag{19}$
we perform this maneuver one last time with
$D_1(x) = \displaystyle \sum_d^{\deg D} D_jx^{j - d}, \tag{20}$
$D(x) = x^dD_1(x); \tag{21}$
then (19) yields
$x^dQ_1(x) = x^dP_1(x)D_1(x), \tag{22}$
or
$Q_1(x) = P_1(x)D_1(x); \tag{23}$
comparing the degree $0$ terms of either side of this equation once again gives (12); when we equate first-degree terms we obtain, via 
(14), (16), and (20),
$Q_{q + 1} = P_{p + 1}D_d + P_pD_{d + 1}, \tag{24}$
from which
$D_{d + 1} = P_p^{-1} (Q_{q + 1} - P_{p + 1}D_d) \in F, \tag{25}$so we now have
$D_d, D_{d + 1} \in F. \tag{25}$
Further scrutiny of (23), again in light of (14), (16) and (20), reveals that the coefficients of the powers of $x$ in general satisfy
$Q_{q + k} = \displaystyle \sum_0^k P_{p +j}D_{d + (k - j)} = P_pD_{d + k} + \sum_1^k P_{p +j}D_{d + (k - j)}, \tag{26}$
and thus
$D_{d + k} = P_p^{-1} \left ( Q_{q + k} - \displaystyle \sum_1^k P_{p +j}D_{d + (k - j)} \right); \tag{27}$
from this formula it is easily seen that $D_{d + k}$ depends on the $D_{d +j}$, $0 \le j \le k - 1$; therefore, since the all of the coefficients
$P_j, Q_j \in F, \tag{28}$
it follows from a simple inductive argument (the details of which are left to the reader) that
$D_j \in F, \; d \le j \le \deg D, \tag{29}$
that is,
$D(x) \in F[x], \tag{30}$
and hence that
$P(x) \mid Q(x) \; \text{in} \; F[x]. \tag{31}$
A: The equation $Q = PU + V$ of Euclidean division can be written as a linear system where the unknowns are the coefficients of $U$ and $V$ and the knowns are in $F$. By Cramer's rule, the solution, if any, must be in $F$.
