Polynomial has Integer Coefficients iff Content is Integer

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

Let $\cont f$ be the content of $f$.

Then $f$ has integer coefficients $\cont f$ is an integer.

Proof
If $f \in \Z \sqbrk X$ then $\cont f \in \Z$ by definition of content.

Conversely, suppose that:


 * $f = a_d X^d + \cdots + a_1 X + a_0 \notin \Z \sqbrk X$

Let $m = \min \set { n \in \N : n f \in \Z \sqbrk X}$.

Then, by definition of content:


 * $\cont f = \dfrac 1 m \gcd \set {m a_d, \ldots, m a_0}$

So $\cont f \in \Z$ 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 $f \notin \Z \sqbrk X$.