# Rational Polynomial is Content Times Primitive Polynomial

## Theorem

Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.

Let $\map f X \in \Q \sqbrk X$.

Then:

- $\map f X = \cont f \, \map {f^*} X$

where:

- $\cont f$ is the content of $\map f X$
- $\map {f^*} X$ is a primitive polynomial.

For a given polynomial $\map f X$, both $\cont f$ and $\map {f^*} X$ 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 $\map f X$ 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 $\map f X$ is equal to $\dfrac h k$ multiplied by a primitive polynomial.

$\Box$

### Proof of Uniqueness

Suppose that $a \cdot \map f X = b \cdot \map g X$ where $a, b \in \Q$ and $f, g$ are primitive.

Then:

- $\map g X = \dfrac a b \map f X$

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

Then:

- $\map g X = \dfrac m n \map f X$

that is:

- $m \cdot \map f X = n \cdot \map g X$

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

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Theorem $61$