gcd of two monic polynomials

Question

I am trying to answer the following question but I'm having some trouble:

If $f$ and $g$ are monic polynomials in $Z[x]$, does their (monic) greatest common divisor in $Q[x]$ necessarily have coefficients in $Z$?

I've tried using Gauss's lemma to get an answer but haven't really gotten anywhere. Any help would be appreciated, Thanks!

Answer 1

First, we show that a monic factor of a monic polynomial is always in $\mathbb Z$

Suppose, $f(x)=g(x)\cdot h(x)$

Choose $k,l\in \mathbb Z$, such that $g'(x):=k\cdot g(x)$ and $h'(x):=l\cdot h(x)$ are primitive. Then, we have $kl\cdot f(x)=g'(x)\cdot h'(x)$

Because of Gauss's Lemma, $kl\cdot f(x)$ is primitive. But $kl$ is a common factor of the coefficients of $kl\cdot f(x)$. Hence, $kl$ must be a unit and therefore $g(x)$ and $h(x)$ must be in $\mathbb Z[X]$

This implies that a monic greatest common divisor of monic polynomials in $\mathbb Z[X]$ must be in $\mathbb Z[X]$.

Answer 2

Let $d$ be a monic polynomial dividing $f$ in $\mathbb{Q}[x]$.

Then $f(x) = d(x)h(x)$ for some monic polynomial $h \in \mathbb{Q}[x]$.

Let $a \in \mathbb{N}$ be the lcm of the denominators of the coefficients of $d$, and similarly let $b \in \mathbb{N}$ be the lcm of the denominators of the coefficients of $h$. Then $abf(x) = ad(x)bh(x)$ and $ad, bh \in \mathbb{Z}[x]$.

The content of $abf$ is $ab$, the content of $ad, bh$ is 1 since $f, ad, bh$ are all polynomials with gcd of their coefficients equal to 1.

By Gauss' lemma, the content of $abf$ is the product of the contents of $ad$ and $bh$, thus $ab = 1$ and this implies $a, b = \pm 1$.

Therefore $d$ is in $\mathbb{Z}[x]$ and we are done, since the gcd of $f, g$ is obviously a divisor of $f$ and is monic by the hypothesis.