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I was asked the following question:

$g\in \mathbb Q[x]$ is a polynomial (not the zero polynomial).

Find $f \in \mathbb Q[x]$ such that $f(x)^2=g(x)^2(x^2+1)$ or show that such an $f$ does not exist.

I really have no idea where to begin and would appreciate all help I can get to solve this.

Oria Gruber
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3 Answers3

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Hint: Can there exist integers $a$ and $b$ such that $a^2=b^2\cdot p$, where $p$ is a prime number?

Zev Chonoles
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  • No. That can't be. Can we say the same thing about polynomials? – Oria Gruber Feb 09 '14 at 20:09
  • Well, what was your reasoning for why it's impossible to do with integers? Presumably you used somewhere the idea of unique prime factorization in the ring $\mathbb{Z}$ - do you know whether the ring $\mathbb{Q}[x]$ has this property? – Zev Chonoles Feb 09 '14 at 20:14
  • My reasoning was this, if $a^2=b^2p$ then $a=b \sqrt p$ but $\sqrt p$ isn't an integer, and $b$ is an integer, so $b \sqrt p$ can't be an integer. – Oria Gruber Feb 09 '14 at 20:22
  • @OriaGruber: It isn't enough to know only that $\sqrt{p}$ isn't an integer; for example, $\frac{1}{2}$ is not an integer, but $$3=6\cdot\frac{1}{2}.$$ What you need to know is that $\sqrt{p}$ is irrational (see Bill's answer, for example). – Zev Chonoles Feb 09 '14 at 20:27
  • So in essence, I need to show that $(x^2+1)$ is prime and that should suffice to show that such an $f$ does not exist. Will eisenstein's criterion work here? http://en.wikipedia.org/wiki/Eisenstein's_criterion – Oria Gruber Feb 09 '14 at 20:29
  • @Oria: No, Eisenstein's criterion will not apply here because $1$ is not divisible by any prime number. But it is easy to see that $x^2+1$ is irreducible in $\mathbb{Q}[x]$ because if it factored, it would of course have to factor into linear terms (since $x^2+1$ is only of degree $2$). But a polynomial $h\in\mathbb{Q}[x]$ having a linear factor $(x-a)$ - that is, having $h=(x-a)\cdot k$ for some $k\in\mathbb{Q}[x]$ - is the same as having the number $a$ as a root. But the roots of $x^2+1$ are $\pm i$, which are not rational numbers. – Zev Chonoles Feb 09 '14 at 20:36
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Hint $\ $ Most all proofs showing $\sqrt{p}$ irrational also work to show that $\sqrt{x^2+1}$ is an irrational polynomial, i.e. not a quotient of polynomials, e.g. compare parity of exponents of the prime $\,p = x^2+1\,$ in $\, f^2 = pg^2\,$ using the uniqueness of prime factorizations. Or apply the Rational Root Test to deduce any rational (function) root $\,Y\,$ of $\ Y^2 - (x^2+1)\, =\, 0\,$ must have denominator dividing $\,1,\,$ so $\,x^2+1 = Y^2\,$ is the square of a polynomial, contradiction. Or employ Bezout's gcd identity as in the proof below. Most proofs using unique prime factorization generalize to any ring with this property (a UFD or possibly also to gcd domains), and most proofs using Bezout's gcd identity generalize to any PID.

Theorem $\quad \rm r = \sqrt{f}\;\;$ is a polynomial, $ $ if a rational function, $ $ for polynomial $\:\rm f\in\mathbb{Q}[x]$

Proof $\ \ $ Let $\rm\,\ \color{#0a0}{r = \large{\frac{a}b}},\ a,b\in{\Bbb Q}[x],\,\ \gcd(a,b) = 1\ \Rightarrow\ \color{#C00}{ad\!-\!bc \,=\, \bf 1}\;$ for $\:\rm c,d \in \mathbb{Q}[x]\,\ $ by Bezout, since $\,\Bbb Q[x]\,$ has Division with smaller remainder, so has a Euclidean gcd algorithm.

$\rm\color{#C00}{That\,}$ and $\rm\: r^2\! = \color{#90f}{\bf f}\:\Rightarrow\ \color{#0a0}{0\, =\, (a\!-\!br)}\, (c\!+\!dr) \ =\ ac\!-\!bd\color{#90f}{\bf f} \:+\: \color{#c00}{\bf 1}\cdot r \ \Rightarrow\ r \in \mathbb{Q}[x]\ \ \ $ QED

The proof easily generalizes to roots of monic quadratic polynomials (and to higher degrees).

Community
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Bill Dubuque
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Hint If $P(x) \in \mathbb Q[x]$ and $x^2+1|P^2(x)$ then $P(\pm i)=0$. Use this to deduce that $X^2+1 |P(X)$.

N. S.
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