Content of Monic Polynomial
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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.
$\blacksquare$