Polynomial has Integer Coefficients iff Content is Integer
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Theorem
Let $f$ be a polynomial with rational coefficients.
Let $\cont f$ be the content of $f$.
$f$ has integer coefficients if and only if $\cont f$ 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 $f \notin \Z \sqbrk X$.
$\blacksquare$