Content of Monic Polynomial

Theorem

Let $f$ be a polynomial with rational coefficients.

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

If $f$ is monic, then $\cont f = \dfrac 1 n$ for some integer $n$.

Proof

Since $f$ is monic, it can be written as:

$f = X^r + \cdots + a_1 X + a_0$

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

Let $d = \cont {n f}$.

Then by definition of content:

$d = \gcd \set {n, n a_{r - 1}, \ldots, n a_1, n a_0}$

Therefore, by definition of GCD, $d$ divides $n$.

So say $n = k d$ with $k \in \Z$.

Then:

$\cont f = \dfrac d {k d} = \dfrac 1 k$

as required.

$\blacksquare$