Suppose $p(t)=r(t)s(t)$ where $r(t)$ is a polynomial with coefficients in $\Bbb C$. Show $r(t)$ belong to the field of polynomial over $\Bbb R$. Here is the entire problem: "Let $p(t)$ denote a polynomial with real coefficients. Suppose that $p(t)=r(t)s(t)$ where $r(t)$ is a polynomial with coefficients in the complex numbers (i.e $r(t)$ belongs to the set of polynomials over $\Bbb{C}$ and $s(t)$ belongs to the set of polynomials over $\Bbb{R}$). Show that $r(t)$ belongs to the set of polynomials over $\Bbb{R}$."
I've made a few attempts but they all hit dead ends. If I assume $r(t)$ has degree $0$, then It follows trivially that $p(t) \in$ $\Bbb{R}$. But I'm not sure how to proceed at a degree that is not $0$. Also, taking into account whether our functions are monic or not comes to mind. Any help would be greatly appreciated for this problem. I feel like I'm just seriously over complicating this.
 A: Hint $\ $ It follows from the uniqueness of the quotient and remainder in the division algorithm (which is the same in $\rm\,\Bbb R[x]\,$ and $\rm\,\Bbb C[x],\:$ using the polynomial degree as a measure of "size"). 
Thus divide $\,p\,$ by $\,s\,$ in both $\,\Bbb C[x]\,$ and its subring $\,\Bbb R[x]\,$ and equate the quotients and remainders. We are given $\, p = rs + 0\,$ in $\,\Bbb C[x].\,$ In $\,\Bbb R[x]\,$ let the division be  $\,p = t s + u\,$ for $\,t,u\in \Bbb R[x].\,$ By the uniqueness of the quotient and remainder in $\,\Bbb C[x]\,$ the remainders are equal $\,u=0\,$ and the quotients are equal $\,r=t\in\Bbb R[x],\,$ which is the sought inference. 
The uniqueness theorem has a simple proof:  $ $ if $\,f = q\, g + r = q' g + r'\,$ then $\,(q-q')g = r'-r\,$ has degree $< g$ since both $\,r',r\,$ do. Hence $\,q-q' = 0,\,$ therefore $\,r'-r = (q-q')g =0.$
This is but one of many examples of the power of uniqueness theorems for proving equalities.
Remark $\ $ 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
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: Let $F$ be a field and $E\supset F$ an extension field.
Let  $p,q\in F[X]$, $r\in E[X]$ such that $p=rq$ holds in $E[X]$. The if $p$ is not the zero polynomial, $r\in F[X]$.
Indeed, $p\ne0$ implies $q\ne 0$. Thus we can use polynomial division in $F[X]$ to write $p=aq+b$ with $a,b\in F[X]$, $\deg b<\deg q$. Then $b=(r-a)q$ holds in $E[X]$. If $r\ne a$, the degree on the right hand side is at least $\deg q$, hence $>\deg b$, contradiction. Therefore $r=a\in F[X]$ (and $b=0$).
A: Lauds to Hagen von Eitzen and Bill Dubuque for simple, elegant solutions based upon the division algorithm.  Below is a perhaps less elegant, more computationally intensive, demonstration directly based upon the coefficients of the polynomials.  Since I cut this snippet of text from another answer which I never posted, it is stated in terms of arbitrary fiels $E$, $F$ with $F \subseteq E$.  And the notation for the polynomials differs from the OP's.  But if this proposition is a applied to the present question with $F = \Bbb R$, $E = \Bbb C$, $t = x$, $f(x) = p(x)$, $m(x) = s(x)$ and $r(x) = q(x)$, then the sought-for result pops right out!
Proposition:  Let $E$ and $F$ be fields with $F$ a subfield of $E$.  Let $p(x), m(x) \in F[x]$ with $p(x) = q(x)m(x)$ 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.  Since we assumed 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.
QED.
Note:  I think it is worth observing that the above result generalizes to the case in which $E$ and $F$ are taken to be commutative rings with unit $1_F$ such that $1_F \in F \subseteq E$ provided the leading coefficient of $m(x)$, $m_{\deg m}$, is a unit in $F$.  For all the formulas in the proof carry over since the hypothesis that $m_{\deg m} \mid 1_F$ in $F$ allows us to affirm that $q_{\deg q} = m_{\deg m}^{-1} f_{\deg f} \in F$ is meaningful, as well as (10) and (13).  End of Note.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
