Polynomial has Integer Coefficients iff Content is Integer

Theorem
Let $f$ be a polynomial with rational coefficients.

Let $\operatorname{cont} \left({f}\right)$ be the content of $f$.

Then $f$ has integer coefficients iff $\operatorname{cont}\left({f}\right)$ is an integer.

Proof
If $f \in \Z \left[{X}\right]$ then $\operatorname{cont} \left({f}\right) \in \Z$ by definition of content.

Conversely, suppose that:


 * $f = a_d X^d + \cdots + a_1 X + a_0 \notin \Z \left[{X}\right]$

Let $m = \inf \left\{{ n \in \N : n f \in \Z \left[{X}\right] }\right\}$.

Then, by definition of content:


 * $\operatorname{cont} \left({f}\right) = \dfrac 1 m \gcd \left\{{m a_d, \ldots, m a_0}\right\}$

So $\operatorname{cont} \left({f}\right) \in \Z$ would mean that this GCD is a multiple of $m$.

This, however, means that for each $i$, $\dfrac {m a_i} m = a_i$ is an integer, which contradicts our assumption that $f \notin \Z \left[{X}\right]$.