Divisor in $\mathbb{C}[X]$ $\implies$ divisor in $\mathbb{R}[X]$? let $P \in \mathbb{R}[X]$ be a real polynomial divisible by a polynomial $Q \in \mathbb{R}[X]$
in $\mathbb{C}[X]$. How can I easily show that $P$ is also divisible by $Q$ in $\mathbb{R}[X]$?
A simple argument without using higher algebraic theorems is desirable. If I could use instruments of higher algebra, the exercise I have to do in whole would be done in two lines. But I'm not allowed to use. I think there would be an easy argument which I can't see yet because of my mental fogginess that I have sometimes.
Thank you beforehand.
 A: The division algorithm uses only the field operations on the coefficients of the polynomials. If $P$ and $Q$ have real coefficients, all the computations take place in $\mathbb{R}$; so, if $P=QR$, then $R\in\mathbb{R}[X]$.
A: Say you have $P = QR$ where $R \in \mathbb{C}[X]$. Then $\overline{P} = \overline{QR} \Rightarrow P = Q \bar R$ (this is complex conjugation). If $Q$ is the zero polynomial, then so is $P$ and you are done. Otherwise, there is an infinite number of points $x \in \mathbb{R}$ where $Q(x)$ is nonzero, and for every such $x$, $\bar{R}(x) = \frac{\bar P(x)}{\bar Q(x)} = \frac{P(x)}{Q(x)} = R(x)$. The two polynomials $R, \bar{R}$ agree on an infinite number of points and are therefore equal. This means that $R$ has real coefficients, and so $Q$ divides $P$ in $\mathbb{R}[X]$.

More generally the technique I used works for any Galois extension. Suppose $K \subset F$ is a Galois extension, and that $Q \neq 0$ divides $P$ in $F[X]$, ie $P = QR$ with $R \in F[X]$, $P, Q \in K[X]$. Then for every $g \in \operatorname{Gal}(F/K)$, $P = Q R = g(P) = g(Q) g(R) = Q g(R)$ (where $g(P)$ is the polynomial where you apply $g$ to every coefficient). Since $Q$ is nonzero and $F[X]$ is an integral domain, it follows that $g(R) = R$ for all $g$, and therefore all the coefficients of $R$ are in $K$ (by general Galois theory).
A: Hint $\ $ It follows from the uniqueness of the quotient (and remainder) in the division algorithm (which is the same in $\,\Bbb R[x]\,$ and $\,\Bbb C[x],\,$ using the polynomial degree as the Euclidean "size"). 
Therefore, since dividing $\,P\,$ by $\,Q\,$ in $\,\Bbb C[x]\,$ leaves remainder $\,0,\,$ by uniqueness, the remainder must also be $\,0\,$ in $\,\Bbb R[x].\,$ Thus $\ Q\mid P\, $ in $\,\Bbb C[x]\ $ $\Rightarrow$ $\ Q\ |\ P\ $ in $\,\Bbb R[x].$  
This is but one of many examples of the power of uniqueness theorems for proving equalities.
Remark $\ $ More generally, $ $ it follows from  persistence of Euclidean gcds in extension domains since, by Bezout, the gcd may be specified (up to unit factor) via the solvability of a system of (linear) equations over $D,\,$ and such solutions  persist in extension domains of $D,\,$ i.e. roots in $D\,$ persist as roots in $E\supset D.\,$ Note $\, Q\nmid P\,$ in $\,\Bbb R[x]\,$ iff their gcd $\,(Q,P) = AQ+BP\:$ has smaller degree than $\,Q.\,$  If so, the Bezout equation persists as a witness that $\,Q\nmid P\,$ in $\,\Bbb C[x]$.
Such uniqueness is a characteristic property of polynomial domains over fields. Namely, if $D$ is a Euclidean domain with division algorithm having unique quotient and remainder, then either $D$ is a field or $D = F[x]$ for a field $F.\,$ For proofs see e.g.
M. A. Jodeit, Uniqueness in the division algorithm, Amer. Math. Monthly 74 (1967), 835-836.
T. S. Rhai, A characterization of polynomial domains over a field, Amer. Math. Monthly 69 (1962), 984-986. 
A: This can also be done by contradiction as follows: 
Suppose that $P=QR$ where $P$ and $Q$ are in $\mathbb{R}[x]$ and $R$ is in $\mathbb{C}[x]\setminus\mathbb{R}[x]$.  Let us write 
$$
R=\sum_{i=0}^kr_ix^i.
$$
Since $R$ is not a polynomial with real coefficients, there is some $r_i$ which is not real.  Let $j$ be the largest index where $r_i$ is not real.  Then, 
$$
R=\left(\sum_{i=0}^{j-1}r_ix^i\right)+r_jx^j+\left(\sum_{i=j+1}^kr_ix^i\right).
$$ 
I claim that, for $x$ sufficiently large, $R(x)$ is not a real number.  


*

*For any real number $x$, the third summand is real since all of the coefficients are real.  

*For all nonzero real numbers $x$, $r_jx^j$ is not a real number since $r_j$ is not real.

*For all real numbers $x$ sufficiently large, the first summand is less than the imaginary part of $r_jx^j$.  For a sketch, let $r$ be the maximum of the absolute values of the $r_i$'s.  Then, using a geometric sum, the absolute value of the first sum is bounded from above by 
$$
r\left(\frac{x^j-1}{x-1}\right).
$$
A short calculation will show that if $x$ is sufficiently large, 
$$
r\left(\frac{x^j-1}{x-1}\right)<\Im(r_j)x^j.
$$
Combining all of this results in the conclusion that the imaginary part of $r_jx^j$ cannot fully cancel, so $R(x)$ is not real.
This leads directly to our contradiction.  After fixing $x$ sufficiently large from above, note that $P(x)$ and $Q(x)$ are real numbers (and we may choose $x$ sufficiently large so that these are nonzero).  We then have $P(x)=Q(x)R(x)$, but it is impossible for this equality to hold while $P(x)$ and $Q(x)$ are real, but $R(x)$ is not.
