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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.

Bill Dubuque
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user144912
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4 Answers4

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

egreg
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  • I think this is the best way to see this. This works just the same in more general cases, for example when $\mathbb{C}$ and $\mathbb{R}$ is replaced by any field extension $E/F$. – spin Apr 22 '14 at 14:40
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    @spin Yes, and it's quite useful when dealing with algebraic extensions. – egreg Apr 22 '14 at 14:40
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    Alternatively, using division with remainder in $\mathbb R[X]$, we have $P=QR+S$ with $Q,S\in\mathbb R[X]$ and $\deg S<\deg R$. Then if $R\mid P$ in $\mathbb C[X]$, we conclude $R\mid S$, hence $S=0$ or $\deg S\ge \deg R$. – Hagen von Eitzen Apr 22 '14 at 14:48
  • @HagenvonEitzen It's not “alternative”: the uniqueness of the remainder immediately forces $S=0$, because $P=QR+S$ is also a correct division in $\mathbb{C}[X]$. – egreg Apr 22 '14 at 14:52
  • @egreg Yes, I only wanted to put this as an alternate formulation, not alternate argument. – Hagen von Eitzen Apr 22 '14 at 15:23
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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).

Najib Idrissi
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    Doesn't $QR = Q \bar R$ simply imply $R = \bar R$ (for $Q \neq 0$)? – user10676 Apr 22 '14 at 14:50
  • It isn't as simple as one might think, because $\mathbb{R}[X]$ is just an (euclidean) ring and not a field which means you can't divide. But I see now the uniqueness of the remainder works well. – user144912 Apr 22 '14 at 15:01
  • @user144912 I think he is right, because $\mathbb{C}[X]$ and $\mathbb{R}[X]$ are domain (? I'm not sure about the english terminology), but if $ab=ac$ with $a \neq 0$ then $b=c$, no ? – yago Apr 22 '14 at 15:07
  • @user10676: For some reason I forgot that $K[X]$ is an integral domain for any field... – Najib Idrissi Apr 22 '14 at 15:17
  • @Yann Hadaoui: Yes, he ist right. Forget what I said in the comment before. Multiplication with $Q$ is an injective endomorphism in $(\mathbb{C}[X],+)$, which allows to reduce. – user144912 Apr 22 '14 at 15:33
  • @user144912 Since $,\Bbb R[x],$ is a domain, nonzero elements are cancellable. However, for more general extensions, this method doesn't work so nicely as the method exploiting the uniqueness so preservation of the quotient and remainder (which is the essence of the matter, i.e if an equation has a unique root in a ring, then it cannot have a different root in some subring). – Bill Dubuque Apr 22 '14 at 15:35
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Hint $\ $ It follows from the uniqueness of the quotient (and remainder) in the division algorithm (which persist from $\,\Bbb R[x]\,$ up to $\,\Bbb C[x],\,$ using the polynomial degree as the Euclidean "size").

Namely, 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 $\,\Bbb R[x].\,$ In more detail

$$\begin{align} [\![1]\!]\ \ {\rm in}\ \Bbb C[x]\!:\,\ P\div Q\ \rightarrow\ &P =\color{#0a0} AQ\, +\, \color{#c00} 0,\ \ \ A \in \Bbb C[x]\\[.4em] [\![2]\!]\ \ {\rm in}\ \Bbb R[x]\!:\,\ P\div Q\ \rightarrow\ &P = \color{#0a0}BQ +\color{#c00} R,\ \ B,R \in \Bbb R[x],\ \deg{R} < \deg Q\end{align}$$

The equality and degree bound in $[\![2]\!]$ remain true in the extension ring $\Bbb C[x]\,$ so uniqueness of quotient & remainder in $\Bbb C[x]\,$ $\Rightarrow \color{#0a0}{A = B}\in \Bbb R[x],\ \color{#c00}{R= 0},\,$ so $\,Q\mid P\ \ \color{#0af}{{\rm in}\ \ \Bbb R[x]}$.

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.

Bill Dubuque
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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.

Michael Burr
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