Rational Polynomial is Content Times Primitive Polynomial

Theorem

Let $\Q \left[{X}\right]$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.

Let $f \left({X}\right) \in \Q \left[{X}\right]$.

Then:

$f \left({X}\right) = c_f f^* \left({X}\right)$

where:

• $c_f$ is the content of $f \left({X}\right)$
• $f^* \left({X}\right)$ is a primitive polynomial.

For a given polynomial $f \left({X}\right)$, both $c_f$ and $f^* \left({X}\right)$ are unique.

Proof

Proof of Existence

Consider the coefficients of $f$ expressed as fractions.

Let $k$ be any positive integer that is divisible by the denominators of all the coefficients of $f$.

Such a number is bound to exist: just multiply all those denominators together, for example.

Then $f \left({X}\right)$ is a polynomial equal to $\dfrac 1 k$ multiplied by a polynomial with integral coefficients.

Let $d$ be the GCD of all these integral coefficients.

Then $f \left({X}\right)$ is equal to $\dfrac h k$ multiplied by a primitive polynomial.

Proof of Uniqueness

Suppose that $a \cdot f \left({X}\right) = b \cdot g \left({X}\right)$ where $a, b \in \Q$ and $f, g$ are primitive.

Then:

$g \left({X}\right) = \dfrac a b f \left({X}\right)$

where $\dfrac a b$ is some rational number which can be expressed as $\dfrac m n$ where $m$ and $n$ are coprime.

Then:

$g \left({X}\right) = \dfrac m n f \left({X}\right)$

that is:

$m \cdot f \left({X}\right) = n \cdot g \left({X}\right)$

Suppose $m > 1$.

Then from Euclid's Lemma $m$ has a divisor $p$ which does not divide $n$ (as $m \perp n$).

So $m$ must divide every coefficient of $g$.

But this can not be so, as $g$ is primitive, so $m = 1$.

In a similar way, $n = 1$.

So $f = g$ and $a = b$, so demonstrating uniqueness.

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 6.31$: Theorem $61$