Krdytkyu

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.

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 $Rinmathbb{R}[X]$.

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).

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 $ Qmid 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 $Esupset D.,$ Note $, Qnmid P,$ 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 $,Qnmid P,$ 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.

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 $mathbb{C}[x]setminusmathbb{R}[x]$. 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,
$$
R=left(sum_{i=0}^{j-1}r_ix^iright)+r_jx^j+left(sum_{i=j+1}^kr_ix^iright).
$$
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
  $$
  rleft(frac{x^j-1}{x-1}right).
  $$
  A short calculation will show that if $x$ is sufficiently large,
  $$
  rleft(frac{x^j-1}{x-1}right)<Im(r_j)x^j.
  $$

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.