Rational Polynomial is Content Times Primitive Polynomial/Existence
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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.
Proof
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.
$\blacksquare$
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 $61$