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.

Solution 1:

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 $R\in\mathbb{R}[X]$.

Solution 2:

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).

Solution 3:

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.